NASA TT F-44
TECHNICAL TRANSLATION
F-44
THE ATTAINABILITY OF HEAVENLY BODIES
By Walter Hohmann
Translation of "Die Erreichbarkeit der Himmelskorper. "
R. Oldenbourg (Munich - Berlin), 1925.
NATIONAL AERONAUTICS AND SPACE ADMINISTRATION
WASHINGTON November 1960
CONTENTS
Page
I. LEAVING THE EARTH , . 1
II. RETURN TO EARTH . l6
III. FREE-SPACE FLIGHT k8
IV. CIRCUMNAVIGATION OF CELESTIAL OBJECTS j6
V. LANDING ON OTHER CELESTIAL OBJECTS 89
PREFACE
Tlie prese.it work will contribute to the recognition that space
travel is to be taken seriously and that the final successful solution
of the problem cannot be doubted^ if existing technical possibilities • !
are purposefully perfected as shown by conservative mathematical
treatment.
In the original work 10 years ago^ the author believed that
2^000 m/sec was the highest gas velocity attainable for the foreseeable
future by our technical means. Calculations originally were only carried
out for this value. But in the meantime three works concerning the
rocket problem were published, which make it apparent that far higher F
exhaust-gas velocity can be reached with a suitable array: ^
k
Robert H. Goddard: "A Method of Reaching Extreme Altitudes"
(mainly on the basis of practical experiments);
Herman Oberth: "The Rocket Into Interspace" (especially valuable
for detailed suggestions on the basis of theory);
Max Valier: "The Advance Into Space" (a simple presentation of
the problem), •
Therefore and for direct comparison with Oberth' s results, calcula-
tions were extended to higher gas velocity (2,500, 5,000, U, 000, and
5,000 m/sec), leaving the original 2,000 m/sec now as the lower limit.
This makes conditions much more favorable; however, the following
observations apply:
The use of low gas velocities required avoidance of any deadweight.
This led to arranging the fuel to be carried with the rocket in the shape
of a cylinder of solid fuel, whose combustion would automatically result
in the escape of the gas at the required velocity. This arrangement
represents the ideal solution, because no deadweight is Involved, but
is conceivable only at low gas velocities. According to Oberth the
higher velocity can be achieved only by means of gases escaping through *
narrowed jets during combustion; carrying the jets and the container
for the liquid fuel means a more or less large deadweight, which is
easier to carry at higher gas velocity.
The weights used in the last two sections do not include these
unavoidable deadweights, because estimation without practical experiments
with jets and containers is hardly possible. Quoted weights to be lifted
Gq therefore represent the lowest value while using an ideal fuel.
Taking higher velocities into consideration as well as later sup-
plements - particularly investigations concerning possibility of landing
without deceleration ellipse at the end of section II, and concerning ^
11
Intersecting ellipses at the end of section V^ also considerations con-
cerning heating during landing - are due to suggestions by Mr. Valier
and Professor Oberth.
Since the writer is an engineer^ not a mathematician^ clumsy
approximations in place of mathematical formulas occasionally appear in
the calculations; this should not affect the results.
Essen, October 1925 W, Hohmann
iii
MTIOmL AERONAUTICS ADTD SPACE AEMINISTRATION
TECHNICAL TRANSLATION F-kk
THE ATTAINABILITY OF HEAVENLY BODIES^
By Walter Hohmann
LSAVDIG THE EARTH
In free space we could impart to oui- vehicle an arbitrary
velocity Av by expelling a part Am of the vehicle's roass m with
the velocity c in opposite direction relative to the vehicle.
Since the center of gravity (CG) of the total inass m rennins at
rest^ we have after time t according to Figure 1:
Am
-0-
^
^v'i
m -Am
C t-
or
Figure 1
Ai2i(c • t " Av • t) = (m - Arn) • Av • t;
m -Am -, c - Av
Am Av ^
or
Am Av ^
(1)
thus
Am
Av = c • — ■ ;
m
i.e.^ once Am moves with velocity c^ the remaining mass (m -Am)
moves with velocity , Am opposite to Am, as long as nothing
Av = C • "■"
m
else liappens.
"Die Erreichbarkeit der Himmelskorper. " R. Oldenbourg (Munich
Berlin), 1925.
If mass is expelled or "radiated" at the rate dm , then the
dt
remaining iihss has acceleration:
^= ^ . dm r^^
dt in dt ^
while the nass decreases steadily.
If we arrange operations such that the radiated mass dm is
dt
always proportional to the mass still present, so that
dm ^ ^
rr-:m = c;< = constant F
then the acceleration is uniform and independent of the mass; ,
4
dt = ^ • ^> (1^)
providing c is uncimnged.
The mass decreases according to
dm /^ V
^-^ -o<m (Ic)
(negative, since m decreases with time), thus
- cm/ (dt
and after integration
Ijn m = - o^t + C •
K raQ is the mass at t = 0, then
In mo = + C;
C = In m^;
thus
In m = - o<,t + In mQ,
or
dm
Im
In
Iflr
= - c<t.
and
e or —
m
.«<t.
(2)
I.e., after time t there renains
mr
m =
.-<t
F
1+
If now a gravitational acceleration g opposes the acceleration
Qoc of a vehicle such as described above, then the resulting accele-
ration is
dv
dt
= c^ -
If f .i. the vehicle moves radially outwards at a distance r
from the center of the Earth and if gQ is the earth^s acceleration,
corresponding to Tq (Figure 2), then in accordance with the law of
gravity (derived at the end of section III):
v^-- -^
,*.;
li)
n
r 1
fgj
1 \.< — J
Figure 2
g = So • ■?■
thus the total acceleration is :
dv
dt
Ca(
SoX
(3)
and since
dr
dt =^
there follows
dv
o=<
~ r2
dr
V
So^o^ X v2 So^c^
; jvdv = \ (ctK 2 — ^^^^' 2~ " ^^ "^ ~ — "** ^'
If acceleration is to begin from rest (v = O) at r = r^^ then
= co<ro + — + C,
^o
thus
C = -OKTo - So^o = -^o(c^ "^ So)^
therefore in general
_2 ST* IT
--. = cc<i^ + _2 ro(c^ + go) = (^ - ro)(co( - g^ —) (ii-)
If at ri^ after reaching nHximum velocity v-^y acceleration c^
ceases^ then the vehicle vill subsequently behave like a body throvra
upfward with velocity vi, i.e., its instantaneous velocity at r' > r]_
V
1 =
dr
at
diminishes at the rate
dvl „ _. ^
r
dt ^o 2 '
these two equations give
vMv» = -So^o "^ >
r*
So^o^
thus
where
2 2
^ ~ "2 ri '
or
If the vehicle is to reach its highest velocity v^^ at r^ such
that it does not return because of gravity after acceleration cv.
ceases, then" the final velocity v* = must be reached only at
r* = 00, so tliat according to equation (5)
2 2
n ^ £0^0
2 " ^1
(6)
on the other hand by equation (k)
therefore
or
-^ = ccxri + ~- ro(o-< + g^);
co<ri = ro(c<^ + Sq),
c°< + £ .^ ^ SOv ,„v
and
The duration t^, after vhich r;^ ^"<^ Eia:;irjur.i velocity vj, are
attained, follows from
dr
dt =^
in the general form
^^
dr
2goro2 „ - .
2co(r + — 2ro(ca< + go)
Since the solution of this integral preser^ts difficulties^
the variation of the gravitatiorial acceleration g during the time
t-^ will be neglected and g will be replaced by a mass value g^.^ be-
tween So ar)d gn and^ to be conservative, g^ will be taken not as
gp ^ gl , but
^^0 -^ gl
gm - 3 >
or taking into account equation (3)
^°'
g, 3 -(2 - ^)
(for small values of o<c, this mean is too favorable, more correct
vould. be the general form
- ^ • 60 -^ 61
Sm ^-^-J >
where f .i. % might be taken £ _ ^o , so that f or o<c = g^ the
2ro - ri
entire acceleration /S would actually = 0.) i.e., the duration is to
be calculated for an acceleration
/3 =. ccK - f^(2 -^ ^) (9)
T. 2
in place of ^o . We have then approximately, making use
c°< - So "2
of (7) and (8),
r"
V
1 ^1
co< - —(2 + --^, .=, - 2 .- ■ 2'
(10)
|°(2 + 1^) OCX - ^{2 . ^^)
(1 . 1^)
If the value t3_, thus formed, is used in (2), there follows
— «e or — - e^ -^ (llj
mo mi
giving the relation between the laass m^ at the beginning of the ac-
celeration period t^^ and the bbss m^^ rennining at the end of that
duration. The difference m^ - mi is so to say dead weight and has
to be expelled with uniform velocity c during this time to impart
to mi the velocity v-i, which it must have at r^.
T/IHLE I
P
k
k
Acceleration ca (in/sec^)
ri = r,(l.|s) {^)
2go^o
(m/sec)
P = ca - ^ 2 + -2^ (m/sec2)
t-, = -f (sec)
Ratio ^ = e -L
for velocity c
c = 1,000 m/sec .
c = 1,500 m/sec .
c = 2,000 m/sec .
c = 2,500 m/sec .
c = 5,000 m/sec .
c = U,000 m/sec .
c = 5,000 m/sec .
c = 10,000 m/sec
15
10, 600
8, 66o
7.27
1;192
58,700,000
li+9, 000
7,570
1,270
588
87-5
55-7
6.0
20
9,510
9,150
12.00
762
k, l60, 000
25,000
2,010
1+58
159
kk.B
20.9
k.6
25
8,860
9,1+70
16.76
565
1,51+5,000
12,000
1,160
282
110
51+.1
16.7
j+.l
50
8,i+90
9,680
21.61
41+6
675,000
7,750
825
216
88
28.7
1I+.6
5.8
1+0
7,950
10,000
52.55
519
5I+6, 000
i+,950
587
i6i+
70
24.2
12.8
5.6
50
7,640
10, 200
41,18
21*8
240, 000
5,840
495
145
62
22.2
11.9
5-5
100
7,000
10, 650
90.76
117
120, 500
2,1+00
5^7
108
49
18.7
10.4
5.2
200
6,680
10,890
190.46
57
89,150
2,000
299
95.5
44.7
17.2
9.8
5.1
Thus raj constitutes the payload liberated from the Earth's
gravitation. If it is chosen according to practical considerations
together with c and acceleration c<^ ~ and thus also the value ^,
then r]_, vi^ ti and mo follow from equations (7), (8), (lO) and
(11).
T&ble I permits a s\arvey of the influence of values c and cq(
on the relation of tuq to m^^. The assumptions were
8
Tq = 6,380 km and g^ = 9.8 m/sec^ = O.OO98 Ion/sec^
(the results represent only approxiinations ).
The table shows, that the effect of co<: is relatively sinaller
than that of c, therefore the highest possible jet velocity c is
of prime importance, vhile the acceleration c<^ may be chosen as
high as is possible to withstand # This acceleration is felt by
the occupants as an increase in gravity and is thus limited by
physiological considerations. The following consideration will
serve to find a suitable va lue ^ A man jumping from a height of
h = 2 m has a velocity v =v/21igQ, when touching the ground. From
that moment he decelerates this velocity to zero within a distance
^^ ^' ~ Q»5 J ^ ai^d consequently will feel the acceleration j3, given p
by V = /2h^, as additional gravity. These two equations give ^
k
^ =So|r-So|f = ^go==-^^0m/sec2.
If one takes into consideration, that during space travel the
acceleration will act over a span of several minutes, while in our
example its duration is limited to a fraction of a second, tlien an
acceleration co( of about 20 to 30 m/sec^ would seem just feasible
(More detailed investigations concerning the physiological effect
of acceleration are carried out in "The Rocket Into Space" by
Oberth).
More difficult to meet is the requirement of the highest pos-
sible velocity c. The highest velocity now available to man is
that of an artillery shell, say 1,000 to 1,500 m/sec. But because
of the high value of mo/mi, this cannot be considered here, as nay
be seen from Table I; rather we liave to demand at least a value c
of about 2,000 m/sec.
Thus the ratio m^ with c = 2,000 m/sec and co< = 30 m/sec^
r^ = 825
^1
is the least required by these considerations.
Tliis lower limit (co( = 3O; c = 2,000) is used for the follow-
ing calculations, with higher values of c being occasionally used
for comparison.
The HBss to be expelled/sec at the beginning of departure is
in accordance with (Ic)
dm^
dt ^ '
where
"^ "^ c 2^000 m/sec "^ sec '
and
so that
dm^
eIq = 825 mi;
T~ = 0.015 • 825 inj_ = 12,i|- rai.
dt
At the beginning^ considerable amounts of mass liave to be radi-
ated, coiTipared to the payload» If one were to e^pel this mass by
firing artillery, one would liave to carry correspondingly heavy
pieces, thus in turn increasing the pay load mj^ and again in turn
the total m^. To avoid this, the operational mass m^ - m-^ may be
arranged so to say like a rocket, tliat is burning slowly, while
the combustion products are expelled with the required velocity c
into a space, assuDied empty. Since the combustion mass rate is now
proportional to the rocket cross section and since -- by (ic). — it
also must be proportional to the mass remaining above the instantane-
ous cross section, we must imagine the cross section to be propor-
tional to the mass resting above it, so that the fuel would be air -
ranged in the form of a tower of constant weight/area ratio (see
Fig-ure 3),
Figure 3
The mass to be expelled/sec at F is, according to (Ic) and
Figure 3:
~=o4m - J
dt go '
where gQ and Y* denote density of the fuel and gravitational ac-
celeration at the earth sxirface respectively. Therefore
10
or, since
dh _ am . Sq
dt F y '
F " Fl Fq ■
a^^^l So
(12)
and
ami
*i
So ( ^^ ami _ ^
__ . \ dt = =; • -T- • t-
" FT V J ^^-fT y ^1'
o
or if G2_ = Ki-i • Sq denotes the veight at the Earth surface of the
payload ui, then:
at-L Gi
^-':r*^^ (12a)
Further, by (12):
If f .1. the weight to be lifted G]_ = 2t and of the fuel density
>• = 1.5t/ni3, then for the assumed case (co( =» 30 m/sec^; c = 2_,000
m/sec; o< = Q«Q15 : ti » i+if8 sec; !^ = 825 ) there follows the rela-
sec ^1
tion:
0,015 ^ ^^8 2>0 8.96
1.5 ' Fi " Fi ^
Fo = 825 • Fi;
and if one assumes an upper cross section for the tower of Fj_ =
0.332 m^ corresponding to a O.65 iQ dia.:
Fq = 825 • 0*332 = 273 m^, corresponding to I8.7 m(p,
h - T^^H = 27 m (see Figure k ) .
11
(mi)
(fTjQ'mf)
- fS7mf'
Figure k
The force on the material^ cansidering co( = 30 in/sec^ instead
of the usual Sq " 9»8 ra/sec^, is:
= £21 .
2t
go Fi %d 0.332 in^
= 18*5 t/m^ =1.85 kg/cm^.
To produce a material^ having the necessary strength as well
as the necessary energy of combustion to produce velocity c, is a
question for the technology of explosives.
So far, air resistance has been ignored. Even if the shape of
the vehicle (Figure 4) is favorable for overcoming atmospheric fric-
tion and even if higher velocities are achieved only at heights,
where very little or no atmosphere is present, the influence of the
lower and denser regions must at least approximately be taken into
consideration.
According to von Loess 1, the resistance W of air of specific
density > with respect t"o a body, moving with velocity v at right
angles to its cross section F, is :
Vv2
s
F • i\/ (eq. {Ik), section II),
where g is the Earth acceleration and tf has a value depending on
the body*s geometry (for vertically effected areas 4^= l). The
resulting retardation is
12
A^
m s m
^.
In the present case according to (12)
F 4. ^ -^1 0.332 1 in3
- - constant = 55- = g^OOO/lO = Wo kg/sec^ '
further^ according to Flgvire 5> we may approximate by a cone
>f; -sin2.p =. -(J44f)2 = 0.12,
so tlmt
Aft, -'^ 9-^ - >i£ . 1
^r ' a ' 500~ " g 5,000 •
g
(13)
Figiire 5
In the range considered, we have with sufficient accuracy
g
10 m/i
sec'
and according to (k):
v^ = 2(r " Vq){c^ - go ~).
Tlie values for > may be found in Table III (page 19).^ section
II, and accordingly in Table II the values for Vvf^ in kg/m^ are
e
calculated for various values of the distance r.
J
*
13
Table
II
•
r
km
(r - ro)
Ion
(c^ - go ~)
km/sec^
v2^
km^^sec^
y (from
O^ble III)
Wm3
)V2
kg/m2
6,380
0.02020
0.00
1.30
6,381
1
0.02020
0.04
1.15
4,600
6,382
2
0.02020
0.08
1.00
8,000
6,383
3
0.02020
0.122
0.90
11,000
6,384
k
0.02020
0.162
0.80
13,000
6,385
5
0.02020
0.202
0.70
14,200
6,386
6
0.02020
0.243
0.62
15,100
6,388
8
0.02021
0.323
0.48
15,500
6,390
10
0.02021
o.4o4
0.375
15,200
it
6,395
15
0.02022
0.606
0.215
13,000
6, It 00
20
0.02023
0.810
0.105
8,500
6,ino
30
0.02024
1.214
0.0283
3,440
6,i^20
40
0.02026
1.620
0.OC74
1,200
6,U30
50
0.02027
2.028
0.00187
370
6,1^^0
60
0.02028
2.434
0.00045
110
6,460
80
0.02032
3.250
0.000023
7.5
6,480
100
0.02035
4.0/0
0.000001
0.4
Ab<
Dve 50 km from the Karth's
surface the air resistance ac-
cording
to (13) is
no longer significant at -
the velocities
3 attained
there .
To take an
unfavorable case, we will
assume an average value
of
'■
~ = 12,000 kg/m^
Ik
for the region between and 50 km, so that the iiiean retardation
from (13) i$
aS = 12^000 ^ 2A m/sec2
^^ 5,000
leaving an effective acceleration of only-
ox - Ad = 30 - 2. it = 27.6 m/sec^
belot/ the first 5 km in place of c^ » 30 la/sec^.
In r = 6,it30 km or r - r^ = 50 km height, we have therefore
according to (k):
I" = 50(0.0276 " 0.0098 1^) = 0.355 km^/sec^
in place of
50(0.03 - 0.0098 • l^||§) = l-Ol^i- kxr^/sec^
or
V = 72 • O.S95 = 1.3^0 km/sec
in place of
J 2 • l^Olif = l.li-25 km/sec
and the time up to that point
^' " — ¥'<- fe0' ""^^
in place of
>^>^^^ . . M =» 70.3 sec
the time difference therefore /it =» ^.7 sec.
Since fiirther the final velocity turns out too low by
^v< ^ I.I4-25 - 1.3^-0 ^ 0.085 km/sec
15
the acceleration has to act longer for a time
resulting in a propulsion time
t^« =. iOi-S + h.^ + 3.5 =» 1^56 see-
in place of t^ = hhd sec; therefore
^ti^ - 0.015 • ^56 = 6M
P and the ratio
k
h -2. =. e^"*^!* » 933 in place of 825.
mi
The result becomes somewhat more favorable, if the acceleration
below the first 50 km is simply increased by ^(3 = 2.4 m/sec^, the
propulsion period then remaining at khQ sec, diiring the first 70. 3
sec of which ex c =» 32.4 m/sec^ with 32 >4 n mto > while
° 2,000 ^ ^•^•^*^
dxiring the remaining 377 •? sec o^c « 30 m/sec^, corresponding to an
o^ » 0.015, so that
£2. ^ ^o<t ^ ^0.0162 • 70.3 ^ 0,015 • 377*7 « 898.
°1
The table below shows the air resistance calculated similarly
but for different <^c and c:
o<c = 30 m/sec^ ofc « 100 m/sec2 o<c « 200 m^sec
(t-L» « 456 in (ti* =» 123 in (ti* = 64 in
place of 448 place of II7 place of 57
m/sec sec) sec) sec)
c = 2,000 933 in place of 468 in place of 602 in place of
825 3^2 299
c = 2,500 235 in place of I38 in place of I66 in place of
216 108 95.5
c « 3,000 95 in place of 60 in place of 7I in place of
88 49 44.7
16
o<c = 30 m/sec'^ <KC = 100 m/sec^ ^c = 200 m^sec
(t;^' = ^56 in (t-L« = 123 in (t^^* = &1- in
place of kkQ place of II7 place of 57
m/sec sec) sec) sec)
c = i+^OOO 30 in place of 22 in place of 25 in place of
28.7 18.7 17*2
c ^ 5,000 15 in place of 12 in place of 13 in place of
. ik.6 io>^ 2j3. /
mo c 1
-— a e
mi
The effect of tbe air thus rapidly increases with increasing
o<c^ so that finally high values here can become more unfavorable
than lower values of o<c, because of the early attainirent of high
velocities •
The above concept, to impart to a body an acceleration op-
posing gravity by expelling parts of its mass, is not new in itself •
It is contained unwittingly in Jule Verne *s "Journey Around the Fioon,"
where mention is made of rockets, which are taken along to decelerate
the vehicle, and it is continuously used in Kurd Lasswitz's "On Two
Planets." Here however under the very favorable assumption, that
the velocity of expulsion is that of light, resulting in only a
minor decrease of the vehicle's mass.
The more recent works of Goddard, Oberth and Valier have al-
ready been mentioned. Also the pioneer of aviation, Hermann Gans-
wiudt, has proclaimed the idea of a rocket vehicle in public talks
around I89O, at the same time also, the Russian Cielkowsky. Finalily,
Newton mentioned the possibility of travel in empty space on the oc-
casion of a lecture concerning impulse reaction.
II
RETURN TO EARTH
To decelerate a vehicle, as described in the last section
(Figure k), falling toward the center of attraction within the
range r^ and r^ (Figure 2) from velocity v^ to a state of i-est, we
require the same t^, given by (10 ), during which now however the
particles are expelled in the direction of motion at the rate dm •
dt
17
Thus for a return trip tlie propulsion period would become
twice as great^ resulting in a niass ratio ^"-^ o _ o^ti • 2, i.e.,
Ho
the second power of the values quoted in Table I for the ratio ^ ,
f.i. in place of c^ = 30 m/sec and c = 2^000 m/sec:
m.
t
2^= 825^ = 630,625.
^1
This type of retardation -- at least for presently feasible
jet velocity c -- would nake natters extremely unfavorable. A
different type of landing, namely retardation by the iilarth atmos-
phere, niust be attempted.
According to von Loessl, the resistance to a body, entering
the atmosphere, is
P
W « w • Fv^ = > ♦ — • F4/, (14)
o
where v == instantaneous velocity
g = gravity
V = density of the air
w = pressure/unit area, vertically to the direction of motion
F = area of the body at right angles to the motion
4^ = a constant, depending on the surface geometry, f .i.
plane area ^ = 1, convex hemisphere ^ = 0.5.
Setting atmospheric pressure at sea level p^ and zero at
height h and assuming (Figure 6) a law
P=Po(^)" (15)
then the pressure rises with height dy as
dy i^n y
but
18
dp = ydy or ^ = >,
so that
y =
^.-=-^- (16)
i ^
W y
/
b
X
F
^
^X
i^
h
sea level
Figure 6
Since at sea level y = h and p = Pq^ we have
nPo
^0 h '
'
or
.-^°-h (17)
and according to (l6):
>-^- h • ^ • y°"^ = ^oC^)"-! (16a)
Smpirically
Vq = 1.293 kg/in3
Po = 0.76 m • 13,600 kg/m^ =« 10,330 kg/iii2 (weight of Ifcrcury column)
^o ,^ 1.293 Wm3 .^ 1
; - ris 'ifa)
Po 10,330 lcg/m2 8,000 11
Results of balloon experiments show the atmospheric pressure
to be 210 ma l^fercury at h - y = 10 km, or that
19
£_ _ 210 _ _1_
Po " f^ " ""3.6 '
which follows from (15) independently of h, providing this height
is assumed to lie between 100 and 1,000 km. Meteor observations
and theoretical reflections suggest an atmosphere of at least
h = i<-00 Ion (c.f. Trabert "Lehrbuch der kosmischen Physik, " page
304). In the following this value is adopted; from (I7) and (17a)
x^ - ^ = 50; n - 1 = lt9;
and the value of > corresponding to h - y is given in Table III
below.
F
1^ Table III
h - y y ' ^-"^h
km km lqg/m3
55 3^3 0.000 915
60 3i^o 0.000 kkQ
65 335 0.000 217
70 330 0.000 102 5
75 325 0.000 o!+9 7
80 320 0.000 023
85 315 0.000 010 6
90 310 0.000 och 9
95 305 0.000 002 2
100 300 0.000 000 98
105 295 0.000 000 1+23
110 290 0.000 000 185
150 250 0.000 000 000 13
200 200 0,000 000 000 000 002 3
l<-00 0.000 000 000 000 000 000
h - y
y
v = 1.293(^)^9
km
km
It 00
lcg/m3
1.3
1
399
1.15
2
398
1.00
3
397
0.90
k
396
0.80
5
395
0.70
10
390
0.375
15
385
0.215
20
380
0.105
25
375
0.055
30
370
0.028 3
35
365
o.oii^ Oi-
1^0
360
0.0C7 k
i^5
355
0.003 76
50
350
0.001 87
20
i^OO km abcfve sea level at r = 6,780 1^ from the Earth cen-
ter, a body, falling tovard the Earth under gravity alone, has
by (6) a velocity
V
^fi>o ^ -J2- 0.0098 • l^^l^ = 10.9 Wsec.
It is clear that such a velocity cannot be stopped within
i^-OO 1cm without damage to the vehicle and its passengers . The
retardation distance can hov;-ever be arbitrarily increased with
a tangential approach.
A body, subject only to gravity of the Earth, moves at large
distances radially or approxiniately parabolically, the Earth center
being the focus of the motion, the velocity at distance r being p
(using the notation of Figure 2), i.e., a tangential velocity,
when passing the Earth, of
^nax = y^i^ = 72 • 0*0098 • 6,380 = 11.2 km/sec,
while, when passing higher up, at the limit of the atmosphere, the
tangential velocity is
V - /2 • 0.0098 • ^^l^Q = 10»9 km/sec,
thus the mean is
V* = 11.1 km/sec.
To determine within which limits of the atmosphere a usable
retardation may be obtained; Table IV shows the air resistance
>v2 acting at right angles to the motion in kg/m2 as a re-
e
suit of a velocity of 11.1 km/sec.
Table IV
h -y
y
r
6 «go;2-
y'>o
(Z)h9
v= y.^'"
km
km
km
in/sec2
ks/m3
kg/m2
1+00
6,780
8.69
0.000 000 000
000 000 000
0.000 000 000
200
200
6,580
9.21
0.000 000 000
000 002 3
0.000 000 03
150
250
6,530
9.36
0.000 000 000
13
0.001 7
110
290
6,1+90
9A8
0.000 000 185
2.1+
105
295
6,1+85
9.50
0.000 000 1+23
5.5
100
300
6,480
9.51
0.000 000 98
12.7
95
305
6,1+75
9.53
0.000 002 2
28.5
90
310
6,1+70
9.5^
0.000 001+ 9
63.1+
85
315
6,1+65
9*56
0.000 010 6
137
80
320
6,1+60
9.57
0.000 023
297
75
325
6,1+55
9.59
0.000 01+9 7
6>+0
70
330
6,1+50
9.60
0.000 102 5
1,320
0^
335
6,1+1+5
9.62
0.000 217
2,780
60
3^+0
6,1+1+0
9.63
0.000 1+1+8
5,720
55
3U5
6,1+35
9.65
0.000 915
11,800
50
350
6,1+30
9.66
0.001 870
23,900
ro
22
Tbe atmosphere above 100 km cannot be considered suitable for
retardation for this velocity. On the other hand, one cannot ex-
pose the vehicle, vhich now has only its sioall final nass m -- con-
trary to the situation when penetrating the atmosphere from below,
as investigated at the end of the last chapter -- and which is in-
tentionally designed to utilize and not minimize friction to exces-
sive resistances >V^ , but rather with regard to laaneuver ability,
"" " g
things Imve to be so arranged, that as with an airplane, the velocity
in the lov^er atmospheric strata, where g = 9-8 m/sec^ and y = 1.3
kg/m3^ is 50 m/sec approx.imately, so that
According to Table IV, a height between 75 and 100 km above tlie
ground gives a good mean value.
The atmosphere should thus be entered with a tangential peri-
gee of 75 km above sea level or
6,380 + 75 * 6,1^55 km
from the earth center •
The length of the retardation paths between 75 and 100 km,
according to Figure 7:
Limit of
the at-
mosphere
Earth
surface
Figure 7
Gene3rally for a parabola
cos^ o<^*.
23
i.e..
^^^ ^^ =iP" = jl^l = ^-998975;
and with siifficient accuracy
s^ = r' sin 2o<' = 6,^^-80 • 0.12^^-28 = 805 ion;
i.e., between 75 ^^nd 100 km, the effective distance, during which
retardation is experienced, is
2s^ = 1,610 km,
if in the first order we neglect changes in the trajectory due to
retardation (these will be dealt with at the end of the section).
Within s^ the retardation P of the mass mi, through the air
resistance w, has the variable value
or (by (Ih) and (l6a) with g =» -^go)
dt go^i h
further
and approxiinately
ds_
dt
ds ^a Sa
dy Ar r» - r '
a
therefore
dy dt ds dy g^m^ Ar^ ^h' '
2k
or
(|)^5 . ay;
VY sa y50
when entering the retardation stretch, i.e., for y = y' we have
Vo^t sa v'5°
50 g^nii ^a b^
in the center of the stretch, where y = y^^,
Vp^'f Sa ya50
a 50 g^mi ^^ ^TTF ''^
therefore during the passage of the first half:
If one uses the values :
y^ = 1.3 kg/m^; ^^a " ^' " ^a " ^^^ - 75 = 25 km;
s^ = 805 km; |2-. « |21 = 32.2;
h = i|-00 km = 1+00,000 m; y^^ = 325 km; y' = 300 Ian;
and further as before uses Sq^t = weight of the vehicle G-]_ on the
earth = 2,000 kg and F'f = 6.1 m^ corresponding say to a parachute
of 2.8 m dia., moving at right angles to its area, so that the max,
value of the retardation at 75 km above sea level is
/^m^=l^-Ff=i§- 6.1 =. 19.5 Bi/sec2,
then one finds the velocity Vg^ at the perigee from
^" ^ ■ fS'rsf^ • 32.2 • ^00,000 /J^)^° - (22a,50j , 0.031,
or
vl , gO.031 =, 1.
Va
032,
25
thus
^a "" 1.032 •
Similarly for the second half 3^ of the stretch we find the
exit velocity
^1 = 1:032 = r:^ = iTSsi^- = 10-^ Wsec.
In consequence of the retardation, there occurs an alteration
in the trajectory, in particular the parabola is replaced by an
ellipse, so that the vehicle returns to the same retardation area
after completing an orbit, but this time with an entrance velocity =
v^ = 10.^ loii/sec. Since within the short breaking distance the el-
liptic orbit is only minutely different from the parabolic orbit,
one may again use 2Sg^ = 2 • 805 = l,6lO km. After a second passage,
the new velocity
^1 v^ . 11 q. n A 1 /
^2 = 1:032? = 1:032^ = 17032^ == 9-8 Wsec.
This further lowering of velocity causes a smaller elliptic
orbit, at the end of which a fiurther breaking period follows with
entrance velocity vg = 9»8 km/sec. If the same distance 2sg^ =
1,610 km is assumed (actually it increases a little every time,
thus increasixig the retardation), then
^3 " 1:032^ =" 9.2 km/sec,
and so on;
11.1
vi| " 1T032H ° ^'^ Ian/sec,
^5 " 1.QJ2IU " ^'^ km/sec,
until finally^ after a crossing of half the distcince Sg^, a velocity
^5 11.1 ^ pn 1 /
^a = T:B32 ^^ 1.032ii = ^'^ ^/^^^
is reached. This however is the velocity
26
yggl-a =1
sa-^a
r.2
rn2
'go
2:*a
0.0098 • 6^555^ == 7.85 Wsec,
corresponding to a circular orbit at the distance r^ = 6^^55 km
from the Earth center or 75 km above the Earthy providing friction
is neglected. The vehicle would thus remain within the atmosphere,
permitting a continuation of the landing in the form of a glide.
To calculate the time necessary for the passage through the
various orbits during retardation, we first liave to find their di-
mensions (see Figure 9)»
A body at distance r from the Earth center E and of mass m
experiences attraction
Jr = - p »
r
At sea level, r " x^f P = mgo^ i.e., the weight of the body
mgr
ro'
so that
/^ = g^ro^ = 0.0098 • 6,380^ =» i^00,000 kmS/sec^.
If the body (Figure 8) has a velocity Vg^ _[_ r^ at distance Vg^
(perigee or apogee), then it describes an ellipse with semi -axis
Jk.
^
and b =
va^
2/u. 2
■p a
(See derivation at the end of section III.)
Figure 8
27
If the corresponding exit velocities are assumed to occur with
sufficient accuracy all at the same point r^ = 6,^55 ^^, then ap-
proxiixitely 2^ 800. OOP _ ,p] :
for Vq_ = 10, i| Ion/sec:
^i^l^^^^^-^^^ooo^,
, l OA ■ 6,4p^ , ,g 800 ion;
for V2 = 9.8 km/sec:
for Vn = 9.2 km/sec:
If 00. 000 ,,, _p,p, ,
. = ? '8 • 6,^^^ , 11,950 km;
2 yi2i^ - 9.8^
ij-00.000 ,^ „cr, w,
b, - ^•^, ' ^'^^l - 9,500 ion;
^ ^22k - 9.2^
for v^j^ = 8.6 km/sec:
1^00.000 = Q nnn wn
8.6 • 6M
bi^
7,850 km;
for vc =8.1 km/sec;
lt-00.000 X- _,„p, ^
^5 ° I2i^ i 8.1'^ 6,900 km,
5 yi2i^ - 8.1''^
28
The period of the orbits follows from the theorem (39)^
of section III:
end
•
rr = constant
at
^a • ^a
' 2 '
dP = 2"' * ^"^^
^a^a
thus
t = 2ab.t ^
^a • ^a
(18a)
s^ retarding distance
^ BrtmutrKJ*
Sarth- .^S^'Sfc^
/ ///if • 1 v\\ \\.
'
/ l^^fV
\
!
1 Eui^
\-U -paraDOJLci
\U \
Figiire 9
The time required for the 5 orbits is therefore:
, 2 • 25,000 • 16,800 • 1L _
39,300 sec = ^^10.9 hours
^1" 10.l^ • 6,i+55
2 • 11^300 . 11.950 'fC _
16,900 sec = ^k*^ hovrs
*"
^2- 9.8 . 6,H-55
i
29
% - ' ' %7-'i ] i%° ' " - T,100 oec - -2.0 hou...
*5 ' '''^?1°.'^%T'^ - 5,700 sec = ^1.6 hours
a total of tg^ = 79^300 sec = ^^2*1 hoiirs
Now commences a glide, vhich may be thought of as follows: It
begins at height h - y^^ =» 75 ^^^ with a tangential velocity v^ =*
7.85 Ian/sec, with the centrifugal force _ "^a^ , is exactly equal
to ga, since, according to page 25, v^^ =" ga * ^a* Because of the
resistance, a constant deceleration /3 abts and therefore v and with
it the centrifugal force v^ , decrease steadily, while the
gravitational fcxrce g remains about constant. In addition to the
tangential retardation, there has thus to be provided a steadily
increasing radial acceleration € to balance g versus a centrifugal
force z :
e » g - z = g(l - |),
or, since _ v^ and, within the range of and 75 ^1 with suf-
^ 2
f icient accuracy also ^a :
s = -^
v2
e = g(l - ^) (19)
va
This radial acceleration e may be provided by a wing F^,
which is gradually brought to bear by moving its plane frora hori-
zontal toward an inclination as shown in Figure 10:
r^ ^
T
TV
L - - ^r^i^^v.
Figure 10
30
^ " ^ * -^o * sin^ o( • cos o<;
m
(20)
the simultaneoois tangential componeijt *L - Q * tan oi may be still
neglected with respect to the large deceleration y^.
To maintain an equal maneuverability, the resistance v must
not increase over its initial value, i.e., by (l^) and (l6a):
g
So ^ ii
i.e., things have to be arranged so that always
v^ ^ (~')'^-9
(^^5
(21)
applies. A definite height y may only be taken up, when the ve-
locity V has been decreased correspondingly.
In Figure 11, the values of v^ belonging to each height y,
are shown together with the values of . v^ , which by (I9) rep-
resent the required increase in G on a scale l:g.
jOOOO
, iUfl
/'STS^ ^aWflO
yr W '^ fl'^ .
L
sea level
Figure 11 •
Further the distance s, covered when a velocity v has been
reached at constant deceleration jB ^ ^a> ^^ •
51
2^ 2fi^ ^^ va2^ 2^a ^ V ^ -^ ' ^^^^
SO that 3 also is given by ^ v^ in Figure 11^ on a scale i^^SL, *
From this it nay be seen that, if /3 reneined constant, the glide
would turn from its initial favorable course into a crash. Thus
may remain constant only to the point, where the trajectory begins
to deviate more strongly from the horizontal.
Now, the inclination of the trajectory, according to (22), is
which gives
Va^ ya 2y9^ ^^ ^
If the retardation corresponding to height h - y^ = 75 km
(ya =* 325 ^^^ and v^ « 7.85 km/sec) resulting from F = 6.1 m^ and
given by
(feo) = 9.3 m/sec2 = O.OO93 km/sec2
is maintained, then (23) gives a limiting inclination of d^ _ 1 ,
ds " 10
which occurs at the height
(2:k)5o = iiS -h^L-. . i- = 50,
Va 325 2 • 0.0093 10 -^ '
-2
Cff
or at
1^
yb = ya * 50^° = 325 • 1.0811^ = 352 km.
h - yb = ^00 - 352 = 48 km
52
above the ground, with a velocity v^, according to (21):
or
v^ = Va /0. 02163 ='7.85 * 0.11^7 - 1.15 Wsec
and after a distance according to (22):
^b = 2^ (1 - ^2") " 2 .0.0093 ^^ " 0-02163) = 3,250 km
and after a time
t, . Z5g^ . 7.8^0^-^1,1^0 , ^3^ ^^^^
The radial deceleratioo is by (I9)
^b " s(i ?) " ^(1 - 0,02163) - 0.97837 • g.
or nearly equal to the total gravitational acceleration g and may
be thoi;ight resulting from a wing Fq^ obeying (20):
w p
g a — . F . sln"^ <y • cos o< =* /--g.
where ^ according to our assumptions, w still has the value
So
30 that
,.h. .^2(^)^9 . i^ . ,,85^ (g^)^9 - ^310 W.^
^ . 2 ^1^ 2,000 ^ ^ 2
Fq • sm^ o< • cos c< = -jp = /x^ ""310" " ^'5 ^*
^should be chosen as small as possible, say max o< - 20^, to
get a value t = £ • tan c)( not excessively large with respect to
^^, so that
max T = 0.36!4- • 9.8 = 3*56 m/sec^
compared to
35
Po^^ 9 m/sec^
and
" 0.3^2^ • O.9J+O
59 ii]2 (^5 m • 12 m).
I.e., the angle o< of the wing has to increase from 0^ to 20^
to the horizontal for a constant wing area Fq =» 59 i^^ Q-i^^ constant
breaking area F « 6.1 m^, while the distance s-^ = 3,250 km is
covered and the height dj.ops from h - y « 75 down to ^4-8 km, in
order to have the radial deceleration g increase from zero to g,
while the velocity v^^ =» 7,650 km/sec decreases down to v^^ =
1,150 km/sec as a result of the constant resistance w = 310 km/m^
(see Figures 12A-B).
From the height h - y^ = i|-8 km, ^ must be decreased to avoid
an excessive descent, say, by eliminating the parachute type break-
ing area F, leaving onJy the component obtained last ^ = 3*56
m/sec2 « 0,00356 kra/sec2, produced by the wing for the further re-
tardation. This value may also not be maintained to the end, since
it would result in too steep a trajectory after a short time.
ip decreasing from 3*56 to 0.102 m/sec^)
(e ^remaining ^ g) -(6 increasing from to g)
{S- gltichbleibend ^g)
T*S56 ^^ zunehmend von Ofysg)
^/ ^
Figure 12.
If e remains constant (equal to g), then the retardation must be
made to decrease, say, by changing the inclination of the wing Fq
from B gradioally to D and finally to the horizontal position at F
(see Figure 12).
At all points of the trajectory:
3h
v2
-^ds = d(2-)
or, since
v2 = v,2 . (^)*9 :
-fi.. • f • d(^'')^9 - - f • ^ • C')50 . ay.
SO that generally
ds _ ^af. . Ij^ . .ya>50 ,24)
where y5 is constant*
If the flight is to terminate at ^-5°, then for 7 = Vq "=" ^00 tan.
we must have:
|2: » i^ (see Figi;ire 13),
/7)>
/ ^-J
/y<y
rr
1
Figure 13.
the final value of p therefore is :
/^min 2 y^ Vo ^^ 2 325 ^^^^ 72
0.000102 km/sec2 = 0.102 in/sec2.
At the end of the trajectory (Figure 12F), the tangential com-
ponent T of the wing is zero and the retardation /^^^ is no\^ solely
caused by the atmospheric resistance against the nose cone, the
form of which with reference to Figure 13 is thus
55
w d'^^ A_ 2
I.e.,
,2
1 _ d. / w • 7t_
^ * 7«il • Pmin '
or putting in the values
w = 310 lan/m^ (according to our still valid assTiniption);
mi = ^^QQO ^ = ^200 fe V^^^^ ;
9.8 m/sec^ m
d = 1«5 (practical minimum diioensions of the vehicle):
1 = !:£ pUZSI = 3.88 m.
it ^200 •0.102
Tlie final speed is given by
V = v^ • (^) = 7,850 • 0.062 = lj-8.5 m/sec,
so that actually the resistance is
^ = ^ ^ = h4 • ^8.5^ = 310 kg/in^
So -^•^
permitting a sinrple landing.
For simplicity, a stepwise reduction in deceleration replaces
the gradual decrease from fi =» 3»56 to /3 = 0.102 m/sec^. We assume
this to take place in k steps: B-C, C-D, D-E and E-F (see Figure
12), withyS^., ^g^, ^Q and /Sf being 3.5, 1.0, 0.2 and 0.102 m/sec^
respectively. These are to lead to trajectory inclinations d^
ds
7- ^ — ^ o ^^ "7= ^ there follows from these assumptions at the
terminations of the respective sections :
36
Section B-C:
by equation i2k):
dy - 2^^ y^ V,
c
or
fe50 . Zaf. . 1+2 . d^r 7.8^^ . !tl. . 1 222;
Va^ 2y3^ y^ ds 2 • 0.0035 325 ^
therefore
y^ = ya • 222^° = 325 * l.H^ = 362 fan; h - y^ = 38 km;
further by (21):
^= (^)^9 = ^ = 0.00502;
Va'^ y^ 222
Vc = Va yo. 00502 =7.85 • 0.0706 = 0.555 km/sec;
and by (22):
Vb^ - vc^ ^ 1.1^^ - O-'P^f = 11,6 km.
^c 2/2. 2 • 0.0035
and
^ PC ^--^
Section C-D:
Va^ 2^ ya <is 2 • 0.001 325 3 '^^ '
1_
yd = ya • 1.550^° = 325 • 1,158 - 377 km; h - y^ = 23 km;
!^= (^)^9 = L^8_ 0.00(775;
va^ yd 1,550
Vr, =7.85 70.00(775 = 0.215 km/sec
37
td = ^V^ = ^^^^ = 3U0 sec;
Section D-3:
r{s.)5o = !^ . !i2 . ^ , .I-8£ !t£_ . 1 = 11 600-
Va 2^ ya <^s 2 • 0.0002 325 2 -^-^^o^'
ye = ya • 11,60050 = 325 • 1.206 = 392 km; h - yg = 8 km;
^e^ /ya^kQ 1.206
f ''' - ^ " °-°°°^*'
Vg = 7.85 /O. 000104 = 0.080 km/sec;
_ ^d^ - ^e^ 0.215^ - 0.08o2
^e 2^ 2 • 0.0002 " ^^ ™'
Section E-F:
y = it-00 km; h - y = 0; v^ = ^^^9 m/secj
s^ = ^e^ - Vf^ =, O.OSpg - 0.0i^9^ . 20 km;
^ 2/3^ 2 • 0.0001
^e - Vf 80 - 1^9 :,Tn
tf---^ 0J-^ = 310 3ec.
The total gliding distance is theref core
s^.^ = 3^250 + lif6 + 131 -^ 99 -^ 20 =^ 3^6^6 km
and lasts
^b-f ^ T20 + 170 + 3^0 + 675 + 310 = 2,215 sec = --37 min.
The entire landing period from the first entrance into the
atmosphere to the final touch-down is
79,300 + 2,200 = 81,500 sec » --^22.6 hours.
38
Id the treatment of the breaking ellipses it was assmned that
to first order there is a sudden change of the orbit from one el-
lipse (or parabola) to the next. Actually this change is gradixal
because of the gradual resistive action and the change occurs in
a transition spiral. The transition forces the vehicle into lower
and therefore denser layers, which in turn cause a greater friction
than was assumed. In consequence, the true exit ellipse has an in-
clined as well as shortened axis. To get an idea of the possible
error, we shall in the following determine the transition spiral
between the entrance parabola and the first ellipse by fitting el-
liptic sections together.
To this end one may divide the angle i|-o(» = ll|^l6*, which is
the extent of the parabola within the effective layers into six
sections, each of Zi^= 2*^22 2/3', each of which cavers a distance
^^ As = 1 . 5 610 3 ^270 km on the assumed transition spiral. As re-
6
quired, further angles may be added to those on the left of Fig-
ure 7, At the termination of each section the retarding action of
the following section -^s is assumed to occur in a sudden decrease
of velocity _ P • As , where v is the velocity for the preceding
Av - ^
section and ^ may be f oimd from table IV through the relation (i «
— . F • C^)^ • Values, not directly given, have been linear ily
extrapolated for better comparison. For the beginning of each sec-
tion the values r^, vj_, o<-j^ are given by Av and an investigation of
the preceding section and from these there follows by means of the
equations
a. - A . ,2 » vi^ri2cos2o<i 2.
^1 "^ ri 1
(c.f. (^5) and (i^■6) in connection with the area theorem) and
(c.f. eqiiation of an ellipse)
the angle ^ between the initial direction and the corresponding
major axis of the elliptic section considered; further, since A(p =
2^2 2/3' is known, the angle ^2 "= fl + ^f between the final direc-
tion and the major axis a and finally the values
59
To =
b2
a + /a^^bS • cos ^2
(c.f. eqiiation of an ellipse);
2 ^rg Vi 1 ^
(see equation (k ) )
aiKL
cos ^Q =» COS cs<-, • — —
p 2 2
, (see area theorem^ (39))/
belonging to the end points of an elliptic sector. This is re-
peated ixntil again a distance r > 6,^80 km is reached^ being the
entrance ray of the exit ellipse.
Tlie trajectory elements, calculated in this manner, are com-
piled in the table on page kO.
Sector
r^ (icm)
Vj^ (km/sec)
»1
a= ii—
^- v2
e-^/^2T^
^--
(p^ from cos q)^ ==
^
(pg = cpj^ ± Ap
r,= ^^-
^ a + e cos cpg
a2 from cos Eg = cos a-y X ^ ^ . . .
2
P = 5^ >< ^ X (^)
As
Av = L^
2
Vg - ^v
;5
6,kQo
11.10
jO^i,.
0.00058
270
0.01
-11.09
6,i+80
11.09
505I,.
2022|'
6,k66
11.10
2022|'
0. 00^^12
270
0.10
-11. 00
II
6,k66
11.00
2°22|*
1^,050
1,869.1 X 10^
11^1,580
2022|'
20i^li'
6,1^57
11.01
1°17'
O.Olif
270
0.55
-10.66
m
6,1+57
10.66
1*^17'
58,987
461. 56 X 10^
52, 55^^
2^221'
0°27i'
10. 66^
>0°0'
oo°l6'
0.018
270
0.46
-10. 20
IV
6,454
10. 20
>0*^0'
<0Ol6'
20,080
217.525 X 10"
15,627
2022|'
2022|
6,456
10. 198
X>°55'
<0O59'
0.016
270
0.40
-9.80
6,456
9.80
X)055'
<0°59'
14, 547
145.570 X 10°
7,894
2^50'
2O22I'
4°52|
6,462
9.794
1^45'
0.0067
270
0.19
-9.60
VI
6,462
9.60
1°45'
12,641
121. 500 X lO"-
6,188
2O22I'
7°57|
6,472
9. 5905
2*^50'
0.00064
270
0.018
--9. 57
YH
6,472
9.57
20 50'
12,486
119.500 X 10°
6053. 5
7°4l'
2°22|'
10°2|'
P-P-*^
For comparison, the values corresponding to a sudden transi-
tion from the parabola to the first ellipse are shown below in
contrast to those due to gradual transition:
41
Limiting
Point
O-I
I-II
II-III
III-IV
Parabola
and first
ellipse
r
V
6,kQ0
11.10
6,kG6
U.n
6,458
11.12
6,455
10.40
°<
3031,.
2 022 2/3'
1°11 1/3 •
0^0 '
Spiral
r
6^80
6,466
6,457
6,454
V
11.10
11.00
10.66
10.20
•^
3°3^'
2°22 2/3'
1017«
>0°0'
<0°16'
Limiting
Point
IV-V
V-VI
VI-VII
VII-VIII
Parabola
and first
ellipse
r
V
6,457.5
10.1<.0
6,464.5
10.39
6,476.3
10.38
:;
°<
1^0 •
2°1'
3°2'
—
Spiral
r
6,456
6,462
6,472
6,485
V
9.80
9.60
9.57
°<
>o°55'
<0°59'
1°45'
2°30'
—
The
resulting ellipse
accordingly ]
3as an a =» 12
,486 km in
place of 25,000 km and b = v/ll9>500,000 = 10,931 km in place of
16,800 Ion and is thus much smaller than the first ellipse calcu-
lated previously; the major a:xes being displaced by 7*^41* -
yogi = 331 fpom that of the former. Perigee would be at
+ e " 12,486 + 6,033 ^^"^52-7 km
in place of 6,455 km.
This makes it probable that actually a passage through two
ellipses, instead of the earlier estimate of five, will be
k2
sufficient to attain the correct orbitting velocity^ particu-
larly, if the area F is slightly increased.
Finally, it is to be investigated, whether one could not
force a circular orbit at the first entrance into the atmosphere,
without employing ellipses. This requires altitude control.
Since this however is required anyway for the subsequent glide,
this is no handicap.
The first, xinfavorable approximations give a vertex of
r^^ = 6,455 km for the parabola with an already diminished ve-
locity of 11.1 -^r^ ^^ ^ / '^ ^^^ vehlcle Is to be
Va = i7o32 " -^^'^^ km/sec.
forced into a circular orbit with this velocity and at this
distance, a centripetal acceleration
:£
^a
10,750^ / ?
^^55,000 = 17.9m/sec2
is required there in place of the gravitational acceleration of
Sa = 9.8 . (|^)2 = 9.6 m/secS.
The corresponding radial added acceleration
e = Zg^ - g^ = 8.3 m/sec^
may be produced through the effect of the air resistance on the
wing Fq, which, in accordance with Figure 13a, has to be placed
at an angle o< with respect to horizontal, so that by (20):
e « - • Fo • sin^ a • cos o<.
m
Figure 13a.
With decreasing trajectory velocity v, the required radial
acceleration t slowly decreases, which may be achieved by a cor-
responding decrease in the angle ^.
1^5
For Vg^ = 10,75 lofl/sec and Tq^ = 6,^55 i^ we have, maintain-
ing F at the same area as previously, namely 59 ^i^ and a vehicle
o
mass
^^ ^2,000 kfi ^sQQkfi ' aec^ ^
10 m/i
sec^
m
w - 61fO • (jiTio)^ " 6^^ Wm^
and
„2
I. Fo- ^^g^Vl^^ • 59 102 = 177 in/sec^;
200
^ ^ POO ^S_L.sec^
zn
therefore for circular orbit we must have
p ^ 8.3
sin'^^oc • cos o< = ^ « 1Y7 "^ 0«0^T;
m o
o^ « ^12 2/3 •
The angle o< is gradually to be decreased to zero^, when the
free orbitting velocity 7,85 km/sec is reached.
The retardation at 75 km height, where Vjj^^ « 11.1 km/sec
and for the parachute area F - 6.1 m^ was previously given as
^max ^ 0-0193 km/sec^. During the forced circular orbit at 75 km
height, the retardation for an instantaneous velocity v:
max
is therefore, further
therefore
ds
dt = ^'
dv ,
r— = - vk;
da ^
dv
ks = In V + C
kh
at the parabola vertex for s = 0:
therefore
or
= In v^ + C; C =» - In Vg^;
V
ks = In V - In v^ = In TT' :
1 -, V
s = — • 1X1 .
Accordingly, we have, at the end of the forced orbit, i.e.,
at V = 7.85 km/sec, which corresponds to the free circular orbit,
covered a total distance
p
max s = o^lll^ ' 2x1 ^^ = 6,1+00 • (6.98OO8 - 6.66568) - 2,000 km.
•
The tiTTip required to cover this distance follows from
^
i=-v^-^^
dv
kdt - - ^ ;
!
kt = + i. + C;
for t = 0, i.e., at the vertex of the parabola;
0=~+C;C = -~;
^a ^a
therefore
and
•
1 ,1 1 \ 1 /^iiHx^ ^max^v
^ " k ' ^v " V ^ 3 ( ■ V - V ^'
^ ^ ^a riTiRX ^ ^a
t =
^5
( ^ A'^ •- 1 n vc: ) ^ "^n mno — ^ = ^18 sec = 3.63 min.
0.0193 ^ 7.85 ^ 10.75 0.0193
The entire landing tiioe, including the subsecLuent glide, is
only
218 + 2,200 = --2,1^.00 sec = UO min.
A landing withcait breaking ellipses is therefore very well
possible. However, the forced orbit, during which the passen-
gers, because of centrifugal force, are pressed against the up-
per wall, represents an inverse flight, during which safety of
maneuverability is perhaps lnipaired. The pilot however will have
to see to it that he does not get into too low a strata, since
this, according to Figure 11, could lead to a crash. If he, how-
ever, renains too high, then he will in the worst case bring his
vehicle out of the atmosphere temporarily and enter a smaller or
larger elliptical orbit, after which he can again, duly relaxed,
attempt a landing.
In apparent contradiction to the proposed landing method,
there stands the fact that n^teorites burn up when entering the
earth atmosphere, indicating strong frictional heating. Here it
may be replied that these meteorites have a much higher entrance
velocity than our vehicle. We have after all specified that it
is subject only to the attraction of the earth indirectly, there-
fore that it is following the movement of the earth around the
sun, which amounts to 30 km/sec, representing the unavoidable
attraction of the sun, while meteorites, because of the gravi-
tational field of the sun, generally reach the Earth^s orbit at
42 km/sec. To this must be added the velocity of the Earth,
30 km/sec, when the velocity vectors are opposed, so that in
this unfavorable case a velocity of ^2+30= 72 km/sec results,
compared to the 11.1 km/sec of our vehicle. Since the resistive
forces are proportional to the square of the velocity, we have
in this unfavorable case a resistance /22_v2 _ l^ times as large
as that against the vehicle. Of course, it must not be over-
looked that during the retardation of v* = 11,100 m/sec to v =»
an energy mv*^ becomes available. This for our assumed mass of
2
approx.
^ , .2 ,000 kg ^ 200 M.:L£££
10 m/sec2 ^i
gives
k6
ffiv'^ 200 o
^ — = ~ • 11,100^ = 12,300,000,000 mkg.
This energy must either be transformed into turbulence or
heat or both. Up to now, we indirectly assumed transf ornation
into turbulence. The other extreme — total transformation into
heat — using a mechanical equivalent 1 , would lead to a value
for the heat
12,300.000,000 r. o
q = — ^J^f — == 28,800,000 heat units.
During the retardation, so far assumed as high as possible,
the parachute would become hot and b\irn up. It would therefore
be necessary to bring along for the several passes through the
retarding range and for the glide up to point B in Figure 12 a
whole series of suitable parachutes to be used one after the
other. (Since at point B the vehicle is already down to
1,150 m/sec, a further heating need not be feared.)
If however any kind of combustion is to be avoided, re-
tardation would have to be siofficiently reduced, so that heated
surfaces would have sufficient time to radiate or conduct the
generated heat to the surroundings.
Generally, the energy becoming available between a given
velocity v* to the instantaneous velocity v is
mv*^ mv^
2 " 2
E =
its increase/sec thus :
dE dv
dt ^ ^^ • dt ^
the corresponding heat/sec therefore
d^ _ mv dy
dt " ^ ' dt '
or, if the permissible heat intake/sec d^ is known, then the re-
dt
tardation for an instantaneous velocity v can at most be
dv ^ dg^ ^ k21
dt dt mv *
^1
The heat intake/sec is equal to the rate of transfer through
radiation and conduction aj3d may, if necessary, by means of cool-
ing rims on the surface of the vehicle, be assumed to be 5OO heat
units/sec, so that, using again j^ _ 200 ^ * sec^ :
m
dv 500 » k27 1>000 / . / \
f .1., the nHximum retaxdation can be
for V = 10,000 m/sec: ^ = li222_ = 0.1 m/sec^,
' dt 10,000
"-= 5,000 ■■ g = i^ = o.a " ,
" -- i'~° " dT = i^ - 1.0 " ,
" -- 1°° • if-%5r ■''■° ■■ ■
To produce such a small deceleration a parachute is hardly
necessary, the air resistance against the body and vings alone
should suffice.
The entire landing distance s is now composed as follows :
dv ^ 1,000
dt " V
■ ^ ^ dv 1,000
therefore r— « ■ ^ a
ds I ds ^
— — ss Y
dt
v^dv
ds =
1,000 '
11,100
^ " ijfeo • ( "^^"^ ' 3^1^1^000 " ^10.T00,000 m - 1^10,700 to
approximately 10 Earth circvunferences]
Of these the travel between v = 11,100 and 7^850 m/sec
(forced circular orbit) takes up:
i
ka
3 • 1 000 * 249^450^000 m = approx. 6 Earth circumferences;
between v = 7,850 and if, 000 m/sec:
7,8503 - U,0003
^ • 1 000 "* 139,920,000 m = approx. 3.5 Earth circumferences;
between v = i+^OOO and m/sec :
k ooo3
V ■ i o ' OQ "* 21,330,000 m = approx* O.5 Earth circumferences.
All this under the incorrect assumption, that the entire
breaking energy is transformed into heat.
The reality lies between the two extremes considered. In
any case, the following has to be taken into consideration during
landing :
1. Retardation must not be too strong, i.e., the parachute
must not be chosen too large.
2. The parachute must have a form suitable for the forma-
tion of turbulence (the requirements 1 and 2 are best met by
following Valier*s suggestion of replacing the parachute by a
number of concentrically arranged cones, whose vertices are di-
rected forward).
3. Beca\ise of the possibility of combustion, a number of
spare parachutes (or cones) has to be taken along.
if. The vehicle must not only be provided with wings, but
also with cooling fins, made of metal.
Furthermore, these conditions being subject to such unusually
high velocities and such unusixally low atmospheric densities, re-
quire further experimental clarification.
Ill
FREE- SPACE TRAVEL
In the sections so far we have considered departure from
the Earth up to the velocity, where no return to Earth occurs,
and the return to Earth from the moment of entering the atmos-
phere, separately. The question now arises whetloer, after the
Earth laas been left behind, the vehicle can be directed in such
a manner, that a return in the desired manner, i.e., tangen-
tially, is possible.
^9
After acceleration ceases, the vehicle moves in a radial di-
rection away from l^rth, providing we ignore for the sake of sim-
plicity the lateral initial velocity due to the Earth rotation
(approx. ^63 m/sec at the equator). It rises or falls with a
steady decreasing velocity into space and beyond doubt, the pas-
sengers will with the sudden cessation of gravity first of all
sense in all probability the fear of a steady fall, which after
some experience will go over into the more pleasant feeling of
floating. Vfliether the velocity zero is finally reached at in-
finity depends on the highest velocity v^, reached at the dis-
tance r-,, when acceleration ceases, whicn after all has not been
exactly determined because of the uncertain air resistance. In
any case, let the velocity, to be determined at an arbitrary dis-
tance rp from the center of the earth by means of distance meas-
urements taken in definite time intervals, be V2*.
Generally, the retardation due to gravity at the distance r
is
r 2
dt ° ■ ^'
and the velocity
therefore
dr
dt = ^^
dv So^o
dr
r^
>
or
vdv =
- Sc^o^
dr
from which
f-*
goTo^
r
+ C;
thus
at the distance
^2
•
•
2
So^o^
r2
+ C;
and
therefore
(25)
50
2 " ^2 " ~^ '
from this tlie distance i*o', at which velocity becomes zero,
follows
2goro2
^3* = 2i^^ — : • (26)
r2 ■^2"'
If the height is to be not r^* but ro then at the point r2
the velocity would have to be in accordance with (25a):
'2
y^^^?"|^-/w"^lpf (^T)
in place of V2 * •
The velocity V2* must therefore be corrected by AVg «
V2 - ^2*. This can be done- by firing a directional shot of
mass Am vith a velocity c from the present vehicle mass m, so
that according to (l):
^m . ^^2
m c
with a + sign, according to whether Ay is positive or negative.
If f .i, at the distance r2 = ^4-0, 000 km the measured velocity
is V2' = h,h6 km/sec (which would give ro* « 00), and if r3 is to
be only 800,000 km (say about twice the distance to the moon),
then according to (27) with
2goro^ =» 2 • 0.0098 • 6,380^ » 800,000 km3/sec2
we must have:
therefore
AV2 = V2 - V2* = ^.35 - ^.^6 = - 0.11 km/sec.
51
and for a velocity of the projectile of c = 1.0 km/sec;
M ^ 0.11
m " 1.0
0.11;
i.e., a sheU of approx. I/9 of the present vehicle mass m would
have to be fired in the direction of travel with a velocity of
1,000 m/sec. This shot is the more effective, the sooner it is
fired.
F
k
k
After reaching the desired height r3, the vehicle, if left
to itself, would return to earth radially. In order to achieve
the tangential entrance into the atmosphere, required in ac-
cordance with section II, it must receive at the point ro, when
the rate of velocity is zero, a tangential velocity vo (see Fig-
ure ik). The return trajectory will in that case however not be
a parabola as provided in section II, but a very elongated el-
lipse of major axis
Figure 1^4-.
To + r«
■A
on the other hand however, by the law of gravity (see (U5 ) at the
end of this section):
So^o'
^ = 2goro2
—^ —
V32
therefore
Ho^^o^
2goro2
V2^
ro + re
_ii
from which
52
2 _ !§^ ^gp^o^ 2 ^a
or
2 ^^
-3-/^^°-° r3(r3.rj > (=8)
2 . ,„ , 2 . __I2 . . „ 2 . i£
and similarly
or
a 3 r^
F.i. for r^ = 800,000 km; r^^ = 6,il55 km and go^o^ "^ 400,000:
V3 =y800,00Q . ^ ^ooo'-^k^^^ = °-°9 Wsec = 90 in/sec.
The tangential velocity may again be imparted by means of a
directing shot with
m 1.0 ^'^^^
i.e.^ a shell of say l/ll of the present mass has to be fired at
1,000 m/sec at right angles with respect to the present direction
of travel.
Tlae velocity v^^ close to the Earth (r^) is then
8 00,000 ^, - . /
Vg^ = 0.09 ♦ ■ / 1! re = --'ll^l km/sec,
thus approximately the same as calculated for a parabolic path
before*
Since the measurement of velocity, to be done en route, as
well as measurements of distance are possibly subject to error,
a check during the travel and possibly a correction of the tra-
jectory is desirable in the following way (see Figure 15 ):
8J
53
Earih
F
k
k
Figvire I5.
At the distance r let us assmne that subsequent measurements
have established v* and the direction of the path (=><) and that
these lead to some vmdesirable perigee r^*. If instead a perigee
at the distance r^ is desired, then the following relations exist
between r^, r, o< and the required velocities v^ and v (see the end
of the last section);
1« By Gravitation
P = - go^o ^ >
2. Conservation of Energy
(Pdr
or
for r = r^:
R r 2
^00
+ C = 0;
thus
5^
or
r r^ 2 2 '
^ ^a r
3. Acccording to the Area lav:
V • r • sin *h = v^ • r^
or
2 v^r^ sJD^ o<
we must therefore have
2
or
v2 (E:^- ^^^°< - 1) - 2goro2 (i- - i) (29)
_2 gSoTp^ r - Ta
'^ "rS sin2<^ -r 2 * ^a * r '
and
2gorQ^ P"^^
T"^ sin^ o^ - ra'
in place of v*.
^ '/rS aii.2^ . :^^2 ra r , (30)
If f .i. at the distance ri^ = ii-OO^OOO km the velocity vi^* =
1.1+15 km/sec and the direction <k\^ = 7O5O* are determined (both
of which would correspond to a parabola with a perigee of ra' "
7,500 km), then
r42 s1b2^;, ^ 1.00,0092 . 0.1^72 ^ ^
ra 6,1.55
and to achieve a perigee of r^ = 6,lt-55 km, we must in accordance
with (30)
55
I 2gc>ro^ ri^ - ra / 800,000 t^OO.OOO - 6.^33
^^ ^Jrk^ siD2<^U - ra^ ""^ ri, 'Jk6^,000 - 6,1+55 * ^00,000
1.31 Ion/sec,
therefore
Avi^ = Vi^ - V]^' = 1.310 - I.U15 - 0.105 km/sec,
i.e., the correction of the journey may again be done by a cor-
recting shot with
^ = f!i = 0^ = 0.105
m c 1.0
or approiximately vith l/9»5 of the vehicle laass — to be fired
directly forward.
By means of (29) finally one can also take into account the
influence of the Earth rotation^ neglected so far. It imparts to
the rising vehicle an initial velocity v^ at the equator ^0,000 km
86^UOO sec "
0.463 lai]/sec and at our latitude of approximately 50^ about 0.463 *
cos 50^ = ^0.3 km/sec. In consequence the vehicle does not rise
in a straight trajectory, and the direction of motion at the dis-
tance r-L, after v^ is reached, is not exactly radially, but in-
clined to the radius by an angle c^i^ given by
^u
sm o<-, = —
(see Figure I6).
With the previously found values of r^ = 8,^90 and v^^ =
9.68 km/sec, the further trajectory would now be a shallow parabola
with a very close perigee of about 8 km. At r2 = 40,000 km the
velocity is
y^i a /- a 4.46 km/sec
T2
and according to the area theorem
V2r2 sin <K2 " ^l^i ^^ ^1^
and therefore
sin 0^0 = Bin «(i . lili = :^i^ » Q,'3 • 8,l^gQ » 0.OII+3.
sm 0(2 Bin ^i ^^^^ ^^^^ ^^^^ ^ ^Q^QQQ 3
56
ascent transit
parab ola-~__jellipse
» ^return ellipse
Figure 16.
If the orbit is again corrected by a shot of c = 1 km/sec and
— = 11 ^^°^ "^2* " ^'^^ ^° "^2 "^ ^'35 Wsec, there results a
m
transition ellipse with maximun perigee and apogee given by (29):
V2^r2 sin^ «< 2
- V2'
2 =
Sgo^c
SSo^c
^3' ('
2SoV
r2
SoJ^o'^
^2
2^ 2
vg'^) - r3 • 2goro'^ = - vg^rg'^ sin'^o^gj
max J,- =
min ^ 2goro2 __ 2
^2
- ^2
^1 t /I - ( 2
V2r2 siDC(2^p^goro2
)2(E
r2
^2
^)J;
max
1^00.000
min ^3 ~ 800.000 _ 1^ 352
1+0,000
^li
/
"■, . U.^5 ' 1+0.000 ♦ 0.0143 NO / 800. 000 , P^ y
1 - ^-^ 1^06,000 ^^ 1+0,000 ^'35 -'-/
max ^3 = 370,500 /I t 0.9999^;
i.e., the perigee is on^y 1+ km or nearly zero, and the apogee is
about 71+1,000 km or near]^ equal to the previous height of 800,000
km. On the other hand, at this distance r3 = 71+1,000 km, the
57
velocity is not now zero, but, given by the area theorem
^3 ; ' Till 000 " 0.003if km/sec =3-^ m/sec^
tangentially.
To enter the desired return ellipse, we now need in place of
the previous value Vo = 0,09 Ion/sec according to (28)
1 V3
1
-J^^^o^ 7^
^■^a'
= /SOO, 000
6,i^5,5
7^1
,000 . 7k7,k^^
4 c
0.0964 km/sec
= 96. ll- m/sec.
l.€
"*>
Av =
3GX - 3.h
- 93
m/sec.
SO
that
•
Am
a ^ = n DQ-i
=s ^.^3
1
m c "-"^^ ^10:^
in place of the previous l/ll; the Earth rotation is therefore of
no great significance.
An arbitrary course of the Journey between launching and land-
ing, according to what has been said above, presents no great dif-
ficulties .
If the velocity corrections, as assuioed up to now, aj:e done by
means of individual shots, and if m^ and m^^ denote masses before
and after the shot, then by (1)
or
Am
E
'0 -
"1
M_
m
mi
™o
1
C
1
c
(31)
To avoid sudden shocks and to lower the weight of the cannon^
it is desirable to replace the shots by several consecutive ones.
In the limit this resembles the procedure of section I^ i.e., mass
radiation, then
58
or generally
dm dv
■ ■ ■ ^ •^^
m c
V
In 22 = ^ + C.
If, at the beginning of the correction, the mass is m^ and
the velocity Vq^ while at the end these are mj^ and v^, then
IjQ mi - ~ + C
onsequently
^„ ""o . v^ - n
=» ^
□d
BU C
c
„1 (32)
Since only a hbss decrease, never an increase, occurs, the
sign of Av refers to direction of the shot or radiation.
For smaller values of |dv the results of (31 ) and (32) differ
c
by only small amounts. For larger values however, radiation is
more favorable than the individioal shot; f.i.:
for ~ =» 0.1 becomes r — irr-rr » 1.11 and e^'^^ = 1,105
c 1 - 0,1
for — = 0.5 becomes :: — i— r = 2.0 and e^^^ « i^^
c "^ 1 - 0.5
for ~ " 0.9 becomes -^ ^ q q « 10,0, and e°*9 « 2.14-6
for — = 1.0 becomes . ^ « oo and e''-'^ = 2,72
c 1-1
To determine the duration of free flight, let us ignore the
small influence of the Earth rotation and let us assume further
that T2 " ri. The flight then consists of two sections:
59
I. ty from the end of tlie accelerating stage at r^ = 8,^90 km
e beginning of the rettxrn ellipse at ro = 800,000 fon;
to the
II. The tinye tjj to cover the return ellipse from the apogee
at ro = 800,000 km to the largest perigee at rg^ =» 6,^55 km.
The time tj equals the free fall of a body with zero initial
velocity from a height r^ = 800,000 km dcwn to r-j_ = 8,^^90 km. Here
at an arbitrary point r the velocity v according to (27) is:
- ^ •
dt
dt "J ro */ r
/2g^o2 f 77 dr
PBo-^o^
t « - 7r(ro - r) + r^ arc sin / — + C;
for r = r^: = + r^ • 2^ + C;
thus generally:
Sgo^o"
^3
and for r = r^^:
t - /r(r3 - r) + r^ij^ - arc sin 1^),
*I "j^^^ Z.7ri(r3 - r^J * r^{f - arc sin J7^)J>
for large values of r^ with respect to r^ -- is the case — one
may put
AT Fi.
arc sip / —
so that
'i-'^js^
OAW^ . ,\) . r^q . ^)J;
60
also
tj = yf§§^ Z7S>90 (800,000 - 8^90) + 800,000 (4^ -
J^lololo U " 1 • Z?i>900 + 1,171^^02^ = 1,256,300 sec =
^349 hours.
The time tjx to cover half the elliptic orbit follows frcaa
(18a):
t o ^^^
whereby
Tq + r
a «
ilia ^ 800,000 . 6,1^^^ ^ ^^03,227 km
and
Vor-
. °-Q? • QQQ^QQ^ - 72,1.00 ion,
/800.000 ^
y 800; 000 - °-°^
therefore
*II - 'llof-VoO^T'^ ■ ^'^^3,000 sec - -35* h.„s.
The entire diiration of free flight is therefore
tj + tji - 3^9 + 35^ = 703 liours = /^29 1/3 days
and the entire trip, including laimching and landing, takes
703 + 22,6 » 725.6 hours = /^^30 1/5 days,
(roughly one month).
Our investigations so far permit a more accurate estimate of
the vehicle weight G-^, so far asstuoed as 2 tons. It has to include:
a) the passengers and personal equipment,
b) liquid and solid food.
61
c) fuel for heating,
d) oxygen for breathing and combustion,
e) containers for the supplies mentioned,
f ) equipment for heating, ventilation, waste
removal, measurements and observations,
g) weight of the parachutes far the glide,
consisting of breaking area, wing, elevation con-
trol, nose cone and alx frame,
h) the weight of the fuselage wall,
i) the cannon and ammunition for the direc-
tional shots.
kg
For a; Two men with clothing and personal
belongings to weigh at most 2 • 100 = 200
For b) The daily consumption per man of a suitable
diet and water is about k kg; thus for 2 men/month:
2 • 30 • ii- = 240
For c) Since the vehicle radiates heat to space
(no heat conduction), heat losses will not be larger
than those of a Dewar of the same shape and dimension,
i.e., very small for a smooth surface. If in addi-
tion the part of the shell facing the sun is partially
or entirely painted black to take any more of the sun's
heat, then the internal temperature will remain tolera-
ble without further artifices. To calculate unfavora-
bly, lieat transfer is to be established for conduction
and not radiation. The rate of heat loss/hour is then
V = 2lt • f • ip, where At represents the difference be-
tween internal and external temperature, f the separa-
tion area and f the heat conduction/hour for Im^ area/
degree in heat units, which depends on the nature of
the wall (1 heat Tin it = 1 k cal). By lagging the wall
with a suitable insulator — which must be as light as
possible (say peatmoss) — the heat conduction may be
taken as ^ = 0.5. The vehicle surface f is to be kept
as small as possible; the sphere has max. volume for
min. surface, but since for other reasons the min.
dimension is to be I.5 m and since the space (see Fig-
ure 13) for 2 persons and supplies should have a min.
volume of i4-.5 m3, an ellipsoid of revolution, having _^^__
carry kkO
62
kg
transfer khO
1.6 m dia. and 3«^ ^ length with a voluioe of ^.55 ^
and surface f =■ livA5 nr^ nay be xised. !I!he internal
temperature may be + 10*-* C and it is further assumed
that the wall, exposed to the sun, has a teniperature
of + 70^ C, that facing space about - 270^ C, i.e.,
the mean external temperature is - 100° C and the
difference At = 110*^ C. The rate of heat loss is
V = 110 • li^.i^5 • 0.5 « 800 k cal/hr, the daily loss
2k • 800 = 19,000 k cal. This loss must be compen-
sated by burning a suitable fuel. The optimum fuel
is petroleum with 11,000 k cal/kg, requiring a daily
fuel consumption of at least 19,000 and with
11,000 =" -L-f ^
regard to the requirements under d), we assume a fuel
consumption of 2 kg/day, in 30 days therefore 30 • 2 = 60
For d) Since 1 kg of petroleum requires 2.7 kg
of oxygen, a daily provision of 2 • 2.7 = 5.ii- kg
oxygen is required. Further, one man requires per
day about 0.6 kg oxygen, thus 2 men 1.2 kg, making
a total consumption of 5.^4- + 1.2 « 6.6 kg daily and
a grand total of 30 • 6.6 « 200
The oxygen is to be taken along in the liquid
state and contained in vacuum containers, since the
containers for compressed oxygen are too heavy, due
to the wall thickness required to withstand the high
pressures. Liquid oxygen however has a temperature
of - 190° C; and for the transformation from the
gaseous to the liquid state a latent heat of 5 00 k
cal/kg is assumed. For the heating of the gaseous
oxygen with a specific heat of O.27 from - 190° to
+ 10° C, a further 0.27 • 200 = ^k k cal are neces-
sary. This gives a total requirement of 6.6 • 55^ "
3,560 k cal/day, to make available the necessary
6.6 kg of oxygen. This in t\arn requires 3»^60 ^ ^ ^
11,000 ^'-^
kg of petroleum, thus increasing fuel consumption
under c) from I.7 kg by O.3 kg to 2.0 kg, so that the
quantity quoted under c) is enough.
For e) The vessels for storing liquid oxygen
will be Dewar flasks with an assumed ratio of weight
to that of the constants of O.k, while the remaining
supplies will be assumed contained in vessels with a
carry 7OO
65
transfer 7 00
weight ratio of 0.2, so that we get 200 • O.k +
i2k0 + 60) • 0.2 - li^O
For f ) Fdr an efficient petroleum stove, for
ventilation and garbage removal, for chronometers,
for protractors, distance meters and so on we assiime 200
For g) For the wings F^ =» 59 m and breaking
area F =* 6 m^, elevation (and preferably also
lateral steering) control ^ 3 nr. For the nose
cone, which, to decrease the weight and heat trans-
fer, is to be separated from the vehicle proper, a
cone of about 1.6 m base dia. and h m side; 1,6 K •
k.O 3 n Q ^ ; together 6+59+5 + 10 « 80 n^ per
6 Isg/m^ « 2I1-O
For h) The surface of the rump, according to c),
is Ik.k^ m^, the weight, including heat insulation,
may be taken as 50 kg/nr, a total of Ik.k^ • 50 - 78O
For i) The cannon 200
Therefore the total weight without ammunition: 2,26o
If the total loss of weight during the journey be-
cause of consumption of supply is ignored, and one as-
sumes three directing shots of l/lO of the total mass
each, then there res\ilts, as the initial weight after
acceleration ceases, G^ » 2,260 • l.l3 « g ^ 000
thus the ammunition to be taken along is 3,000 - 2,260 « jkO
At the beginning of the glide all supplies of aimminition, food,
fuel and oxygen are used up and the remaining weight is
Gi* = 3,000 - yifo - 2ii.o - 60 - 200 - 3,000 - i,2ivo » 1,760 kg.
The resulting final weight at landing is therefore a little
less than the 2 tons assumed in section II, on the other hand, the
initial weight is about I.5 of that given in section I. Therefore,
according to section I, the mass radiated during acceleration, has
to increase also by the factor of I.5, i.e., the length dimensions
of Figure k would increase, other things remaining egioal by
(1.5)^/3. If at the same time air resistance on the way up is
considered also, which according to the end of section I would
6k
require an increase in Initial mass dIq in the ratio 9 g3 then the
825
required linear increase, according to Figure h, is given by
71-5 • §5" = ViT^ = 1.192,
so that for c = 2,000 m/sec and o^c « 30 m/sec
the height 27 • I.I92 « /w/32 la
the lower diameter 18,7 • I.I92 = ^22 m
the upper diameter O.65 • I.I92 « /^ O.77 m
and the weight at the beginning of the ascent will be
^o " ^1 • IT " 3 • 933 « 2,799 tons.
The use of a single gun presupposes that the vehicle can be
rotated arbitrarily, which is possible, if a part of the nass in
the vehicle is txirned in the opposite direction, f .i, by having
the passengers cliirib about the wall of the vehicle by means of
rungs, put there for this purpose. If the living masses m^ here
move with an angular velocity co^ at an average distance x^ from
the vehicle center of mass, while the dead weight m^ moves with
an opposed angular velocity a>^ at a distance x^ from the center
of nass, then, since the total angular momentum must remain zero.
or
therefore
2mvx = 0, or, da v » x
'CO,
Zmwx^ =
; • H * ^t^ " ™1 * '^l *
Xn^.
*^1 mt . x^.i=^'
(33)
i.e., the angular velocities are inversely as the moments of inertia
of the masses. If a total weight of l40 kg is assumed for the pas-
sengers, leaving a dead weight for the vehicle in the most unfavora-
ble case (i.e., at the beginning of free travel) 3,000 - 1^0 =
2,860 kg, then there follows with the average distances to the cen-
ter of gravity, given by Figure I7:
65
f,2^
Figirre I7.
2
^1 2,860 • 1.22
/v
120
Thus, to effect a 360^ rotation, the Ininates have to clinib
120 times around the wall, for 180<^ 60 times and 30 times for 90^.
Since this will give them the illusion of gravity under liands and
feet, this climbing exercise will be a welcome change in the other-
wise gravitation -less existence. If they move their mass centers
with a velocity of O.5 m/s^c, then they will need approximately
I.Otl a 5 sec ^^ ^ 90*^ rotation, i.e., 30 • 6 = I80 sec. Since
0.5
at a distance r^ = 40,000 km from the center of the Earth, where
the first shot Is to be fired, the trajectory velocity is about
4A6 lau/sec, a distance of k.h6 • I80 = 800 km is covered, before
the laterally positioned vehicle is brought into the position re-
quired to change the velocity by -^V2 (t corresponding to the sign
of V2 with respect to the gun position). With respect to the
lfO,000 Ian, this difference of 8OO km is insignificant.
The correct positioning for the wings at the beginning of the
glide, a rotation about the main axis of the ellipsoid may be simi-
larly effected, but will be a little faster, since the dead weight
has less distance from the main axis.
At the end of this section, we want to derive in brief the
laws of motion in a gravitational field, which have been and will
be used repeatedly.
1.
orbits.
Experimental fact: ^he planets describe nearly circular
2. If a body, mass m, describes a circular orbit r with
velocity v, then the centripetal acceleration ^^r , according to
dt
Figure I8, is given as follows :
66
Figure 18.
After a short tiioe At, the path covered is given by
^x » V • At or At « 4i ^
V
and
dv.
^y
dt 2 dt 2v^
and because of the similarity of the right-angled triangles with
angle A^:
^y
2 * r " 2r
Coniparing the two expressions for Ay
dt " r '
or, if a central force P produces this centripetal acceleration:
v2
m
(31^)
F
k
(ttie - sign, because P is opposed to r).
3. Experimental fact: The aqudres of the periods Tt and T2
of two planets are proportional to the cubes of their distances r-^
and r2 (Figure 19) from the sun, or
Ti2 ri3
■ ■■II ta ■ ■ iMi
67
X
thus
Figure I9.
If v-j^ and V2 are the respective velocities, then
2ri7t 2r2 7L
Ti - -7^ and T2 - -t;;^ ,
223
ri V2 r^-*
V3_^ r^^ T^
or
VgS n
vi2 r2
l^-. From equations (3U) and (35):
ni]_V3_2
T2
(35)
and therefore
mi
1 ^ ri2
(negative, because P is
directed toward and r
away from the center);
or generally the law:
m
(36)
where /c is a ratio depending on the center of attraction <
68
5. For the sun as a center, /c follows from the Earth's mean
distance r^ = lit 9, 000, 000 km, moving with a period of Tg = 365 days,
i.e., with a mean velocity of
^^e'^ 2 • 149.000,000 "JL ,
"t; 365 ' 86>00 ' 29.T Wsec,
so that by (3^^) and (36):
"- ^e
Ve^ me
or
^= Ve^ • ^e ° (27.7 Ian/sec )2 • 114-9,000,000 km,
IX. = 132,000,000,000 ^2__. . (37)
sec'=^
6. For the Earth as center, /<- follows from the distance v-^ »
392,000 Ian and period of 28 days, i.e., with a mean velocity of
of the moon, so that
^^in^ 2 ' 392. 000 "JL. , _, , /
^= "t; 28-8^,1,00 ' 1-°^ ^^/^^='
/^ - Vm^ • rm = 1.01^ • 392,000 = 1^00,000 i22£-. . (38)
km3
sec
2
7. At the Earth svirface (ro = 6,380 km) the central force hy
(36) should be:
r> 1^' I" « 400.000 ^
or in acceleration
So = ;^ - X3^ ' °-°°^^ ^-^'^'^ " ^'^ '°^'^'^'
which would follow innned lately^ from observations of free fall on
Earth, giving
^ = g^^2 ^ 0.0098 • 6,380^ = 400,000 i2£_ .
sec
8. Tlie area theorem: For central motion of a mass point under
the influence of a force P, directed toward a fixed center, the fol-
lowing is true:
69
At the distance r^ the velocity Vj^ changes rjHgnitude and direc«
tian in consequence of P^, which causes central acceleration. The
new velocity V2 nay be regarded as the diagonal of a parallelogram
of velocities. The area covered by the radius vector is in unit
time according to Figure 20:
»>-^^^ y
for a velocity Vn:
Figure 20.
dF^ r]_ • V3_ sin f^.
dt 2
for a velocity V2:
dFg rn • Vn sin a^-y
dt
Similarly the velocity vo at T2 p and given by ^2 ^^^ -^2^ ^^
to be regarded as a diagonal m such a parallelogram. The rate,
at which the radius vector covers the area, is then:
for a velocity V2:
djF2 ^2 • "^2 ^-^ ^
W ^ 2
for a velocity v^;
dFo r2 • V2 sin ^2
dt ^ 2
70
This gives
dP-L dPg <iFQ
dT = dt" = dt ' constant (39)
i.e., eqvial areas are swept in equal times.
9. Energy conservation: At each point, according to Figure 1,
the force consists of two components with fixed directions X and Y,
so that:
dv^ dv^
5^ and Y « m • ^^ i
where
• m n s V and, — * ■ " » V !
from this X • dx = mvj^dVjj and Y • dy = mvydvyj
(Xd. - ^ - !!h! ; (Ydy =. ^ - ^ ;
/ 2 2/2 2
2 2^2
^ = ^x -^ ^y '
between two points with velocities Vo and v:
^Xdx + CYdy-^
iv2 mva'
2
2
Further by Figiire 21:
Figure
21.
X
= P •
cos
?=
ijc
= ds •
cos
ds
= dr
Y
= P •
sin
S;
dy
=" ds •
sin
^■-
cos
f
71
and
(p(cos ^ COS ^ + sin ^ sin ^)
dr mv^ ^^a
2
cos w 2 2 ^
or, since cos c, cos ^ + sin ? sin C^ = cos (^ - ^) = cos cp:
jPdr = ^^ ^ . (1.0)
10. Application to arbitrary mot'ion under gravity: In Fig-
ure 22, Z is the center of attraction, Vg^ the orbital velocity of
a body at its perigee r^^, v the orbital velocity at an arbitrary
distance r with conrponents along and at right angles to the radius
vector, given by dr and dg respectively, we have by the law of
dt * dt
equation (36):
r'
according to theorem equation (^O):
2 '
I
/ dr inv^ ^a^
Pdr = -^ j •-^- -|- - -^~
or
for r = re
thus
+ it^ C = ^1 ?r— :
r 2 2^
-^ + C =
ro
r " r^ 2 " 2
or
v2 ^^^ +2^.2^; (1,1)
a r r^
72
Figure 22,
by the theorem (39):
Va • -At • r^
■ (r . a^ • ^t) ■ I • If • At;
dt
2 dt
from this
dt 2 dr ^
or for At = dt = 0:
dt" " ^2 '
(i^2)
by Pythagoras :
(v/\t)2 - (^ • ^t)2 + (r ^ • At)2
dt dt
or
v^ = (dt) + ^^(dt) = (dt) + ":r~ J
by comparison with (^l-l):
v2r,2
(^)2 „ „ 2 ^. 2^ . 2^ . ^a ra- ,
further by (U2):
r r^ r'
2^ 2
■d£\2 ^a ^a~
r^
Mt^
therefore
<if Va^ra2 *■ a Tq r r2^
or
73
dr
ddi
,2 _
2^
.2 + 2/^
Va^r^S
(i^3)
11. The equation of an ellipse (see Figure 23):
-
4
^
"T"
d
Figu
V
re 23.
- b^ or a2 - e^
= b2 .
a + e cos f '
dr b^ • e • sin ^ .
d<p " (a + e cos y>)^ ^
here we rmy put
b2
■ b2
(a + e cos <p)'^
and
e sin ^ = /e
2 - e^ cos^ ^ ,
further
•
2 2 ^^^ \2 -. b^ 2ab2
e^ cos^ f "" ^? ^^ 2 ' r
-a2,
therefore
^k
r? p . 2ab^ b^ / ,.p , 2ab'2' b^
e Bxuf = /e^ - a^ + -^ - -J = /- b^ + -^ - — ;
we have
dr i£ A 2 + 2ab2 b^
or
12. By coioparisoD (k^) and (ifJ+) for d£ it follcfws timt the
d^
motion of a body under gravity (36) is an ellipse^ for wliich
and
therefore
further
thus
further
1__ ^ J- Ta
b2 va^ra^
2a ^ 2/>c
b2 va^ra^ '
a=-^
r a
^a
^2 ^a^^a2 v^^ra^
b"^ = a
^ 2^. v2 '
(U5)
r a
•^a
'ra a
75
e2 = a2 - b2 = a2 - a "1^^
A '
by adding
there follows
= t 2ara - 2arg^
.2
3^a^ * a 2^
e2 = a'^ - 2ara + —T^ (— - ^^T),
ov, since
-. (--v^S) --is:
^a
e2 a a2 - 2ara + ra^ « (a - Ta) ;
thus
e « t (a - r^);
i.e., the center of attraction Z is the focus of the ellipse (see
Figure 22 and 23).
13. As long as 2^^ 2 n ^ ^ renains positive, b real,
^a
i.e., the orbit is an ellipse.
If a^ . y 2 « ^ *^^° ^ ^^ infinite, b infinite, the orbit
is now a parabola.
If 2_ 2 n ^ ^^^° ^ •'•^ negative, b iiaaginary and the
„ va ^ ^
orbit a hyperbola.
If a is to be equal to ra^ we must havie
^a
or
thus
2/^- ^a^^a "/^
76
^a
the orbit in this case is a circle.
lif. The period far an elliptic orbit is given by the area
theorem (39):
dP
dt
■ constant
Va^a
' 2
P =
Vai"a
2
• t =
' ab'ft j
thus
and if, according to (^6), we substitute the value
then we get:
t ^2ait^=2uy^. m
IV
CIRCUMNAVIGaTIQN OF OTHER HEAVENLY BODIES
A circumnavigation of the Moon, f,i.^to find the nature of its
untoown side, will not substantially differ from free space travel,
as long as one does not approach it close enough, so that its at-
traction as well as that of the Earth (which at the saioe distance
is 80 times as effective) becocies significant. Since diiring the
30 days of the journey the Moon will also orbit the Earth once, we
cannot speak of a circumnavigation, but rather a crossing of paths,
which may take the form shown in Figure 2^^, where E, M and F are
Earth, Moon and rocket respectively, while the numbers indicate
simultaneous positions. The largest Moon perigee is therefore
about half of the largest Earth apogee, the relative largest at-
traction by the Moon therefore about h _ 1_ of the simultaneous
5o " 20
Earth attraction. Its influence will not further be investigated
here.
77
1i^
Figure 2k,
In our consideration only the Earth attraction was considered,
while that of the Sun was ignored, because the vehicle follows the
30 km/sec, which the Eai'th covers in its orbit about the Sun. This
is only strictly true at the instance, when the vehicle is at rest
relative to lilarth, i.e., at its highest point ro, and even then
only if this point has the same distance from the Sun as the Earth.
Assuming that the vehicle leaves the Earth tangentially to the Earth
orbit, then its velocity, if 10 km/sec relative to Earth is with
respect to the Sun 30 + 10 = i^-0 or 30 - 10 = 20 km/sec, according
to a positive or negative direction. In the latter case, its in-
stantaneous trajectory has a higher curvature, in the first case,
less ciirvatiire than the Earth orbit, because of the solar attrac-
tion. Since the*vehicle *s velocity relative to Earth however
quickly decreases because of the Earth's gravitational effect,
and the total time of the ascent is only 15 days, i.e., l/2i^ of
an Earth orbit, the trajectory in the range considered will not
perceptably deviate from the Earth orbit. If on the other hand
the ascent is radial with respect to the Earth orbit, then at the
apogee ro the velocity of the vehicle relative to the Sun equals
that of the Earth, but the distance from the Sun is larger or
smaller tlian the Earth-Sun distance, depending on the ascent being
from or away from the Sun. In the former case, the tirajectory
again has a higher curvature, in the latter a lower curvature.
But since an apogee of 800,000 km is negligible compared to the
distance of 150,000,000 km, the deviation here too is hardly no-
ticeable. The direction of ascent is therefore so far arbitrary.
It is advisable however to direct it toward the Sun, so that the
Earth can be seen in its entirety and brightly, which is necessary
for the distance and velocity measurements. The distance ro =
800,000 laa to be attained will therefore always be assumed m that
direction during our further considerations, even if ro is ignored
with respect to the distance from the Sun.
li' at that point the tangential velocity vo is chosen 3 km/sec
instead of O.O9 km/sec (see Figure lU ) as in section III, then under
the influence of the Earth's attraction alone the trajectory is now
a flat layperbola rather than an ellipse, since
78
aff _ ^ 2 - 2 ; 400 ,000 2 - a
r^ 3 " 800,000 ^
and the- vehicle vill pursue a path directed away from the effective
range of the Ekrth gravitation with nearly uniform velocity, until
it finally -- so to say as an independent comet — it is subject
only to the attraction of the Sun. Initially the tangential veloc-
ity relative to the Sun is vj = 29.7 t 3,0 = 32.7 or 26^7 Ion/sec,
according to whether Vi> is along or against the Earth velocity of
29.7 lon/s^c. In either case the vehicle describes an ellipse about
the Sun in the first case outside, in the second inside of the
Earth orbit.
If the vehicle *s path is to touch at a distance rjj from the
Sun, the orbit of a planet other than the Earth, distant r j from
the Sun (see Figiare 25), then the major ajcis of the ellipse is
ri + rii
1
and by (^5 )
a =
JL
ri -L
thus
from this
or
2al 2 2AA.
ri 1 rj + rii '
2 2U. rii
I rj- + rii r-t
/ 2>t ^^
{^9)
The Earth's mean distance from tbs Sun is ri = 1^9,000,000 km,
that of Venus f.i. ru = 108,000,000 km. Since further by (37)
^ = 132,000,000,000 km3/sec^, we have for a journey close to Venus
/2 614-.000 10^ ^^ ^ , I
VI °/ 237 • lH? = ^^'^ km/sec.
79
F
k
k
Figure 25.
Kow the Earth velocity is v^ = 29.7 km/sec and accordingly the
velocity to be imparted to the vehicle, after it reaches its apogee,
has to be
AVj « vj - Ve = 27,3 - 29.7 = - 2,i|- km/sec
and could be the result of a tangential shot of mass
^m = ID •
Avj
"c" >
where m is the mass of the vehicle before the shot and c is the
velocity of the projectile. In this case, one can no longer use
the value of c = 1 km/sec, assumed in section III, and also a single
shot of the required strength would endanger the vehicle and its
passengers. A system of continuous mass radiation as in section I
must be used with a min. velocity of c « 2 km/sec. We have now for
the ratio of total mass before and after the action by (32)
mo ,Av.
— - a^ e ^c )
mi
But since during the initial parallel paths of planet and ve-
hicle_orbit interference is unavoidable, an additional safety fac-
tor /see Kote/"/ say V = 1.1, must be added, which necessitates:
Er
(^)i
= = 1.1
.2.0
1.1
.1.20
3.65,
where the loasa has to be thrown forward in the direction of the
Earth's motion.
(/.Note/ Such interferences may be obviated by radiating the mass
dm ^ (see (ic)), directed against the distiorbing planet and
dt
80
equal to the gravitational effect g, so that at a distance x from
the planet by (la) and (2)
dv ^o ^ ^o cxt
5^ = co< = g = g^ ^ and — == e .
At ttie assumed initial point x = 800,000 km from the Skrth with
Sq = 9,8 m/sec^ and Tq = 6,380 km:
and after one day = 86^^00 sec, when c = 2,000 m/sec :
J- Co< . _ 86,^00 ^ r\ no^n.
'^^ T' ^ 16,000 . 2,000 '^ °-°2T0^
at the distance x = 800,000 km from Venus with g^ = 8.7 and r^ =
6,090;
Q r, 6.090^ 1 / 2
^^ " 8-^ * 500,000^ = 2o;:ooo "'/"^^
and
4- - 86,1^00 ^ ^ ^.x;.
at a distance x =« 800,000 km from Mars with gp = 3-7 and Vq = 3,392:
ccr = ^ 7 • .3;2?ii ■■ = .^ — i
^^ '^•^ 800,000^ 150,000
and
^* - i;o,oo&''?°2,ooo - °-°°=88.
With each succeeding day, x becomes larger, i.e., the daily
increase -xt smaller. By plotting planet and vehicle positions,
one gets for the first 5 days the folloving distances x and the
corresponding daily values cx,t:
Earth Venus l^rs
3 X km <a^ X km o^ x^ km. <^t
800,000 0.0270 800,000 0.0216 800,000 0.0029
1 850,000 o.oei^o 850,000 0.0191 900,000 0.0023
81
Barth
Days X km o<,t x km a^
2 900,000 0.0213 900,000 0.0170 1,000,000 0.0018
3 1,000,000 0.0173 1,000,000 0.0138 1,200,000 0.0013
k 1,100,000 0.0143 1,200,000 0.0096 i,it-oo,ooo 0.0009
5 1,200,000 0.0120 1,^00,000 O.OCTJO 1,700,000 O.OOO6
Sum Hait = 0.1159 2-<t = 0.0881 2!o<t = 0.0098
Accordingly after the first 5 days y = -2. = e :
m
for iihrth: >/ = ^^.116 « 1.123; for Yenus: v = e^'^^^ = I.O93;
for Mars: y = e^'^-"- = 1.01.
Tbe above safety factor y = 1.1 is thus only a rough Diean,
which njust be corrected for each planet. The interference cor-
rection need not be done in one step^ it will be sufficient to
do it daily once or several times with corresponding intensity.
/End of xiotej) The time required to cover half the ellipse by (^8),
using ^I ^ ^11
a = —2 ' 128,500,000 km:
Tj = ^yT: ^ '^/l32, 000, 000, 000 " 12,600,000 sec - lk6 days.
The Earth moves in its orbit with an angular velocity of
36QQ ^ nQrrO/A ^enus With 360^ ^ o/ During-^
3^5 days == O-^ST^/day, 22fdays = 1.6cr77day.
the time of 1U6 days, the Earth, thus covers an arc of lk6 • O.987 =»
Ikh^^ Venus an arc of IkS • I.607 « 23^.5^. In order to have the
vehicle approach Venus in fact (say at a distance of 800,000 km
from the center of Veniis and on the side closest to the Sun), the
launching has to take place at a time when Venus is 23^.5 - I80 =
55.5^ behind Earth in the sense of the motion of the planets (points
Vi and El in Figure 25). If the vehicle would continue its joijrney
unchanged, then it would return after a further lh6 days to its
initial point in space via the dotted half of the ellipse, the Earth
however would be retarded with respect to the vehicle by a further
36° or a total of 72^ (point Eo in Figure 25). To make it possible
for the orbits to intersect, the time for the return trip must some-
how be increased. Two possibilities present themselves:
1st possibility (see Figure 25). If the dotted branch of the
ellipse is to lead back to ihrth, then the Earth at the time of
82
departure at ¥9 would have to be 36 in front of, rather than be-
hind, Venus, i.e., at Ep*, not Ep. The vehicle would have to be
kept near Venua until the correct position of the two planets oc-
currs, i.e., until Venus has almost caught up again with Earth,
except for the 36°. Because of its faster travel, Venus gains
a daily angle of I.6O7 - O.987 = 0.62^ and so it would require
k6M- E^rth days for it to advance from its initial 36^ advantage
over Earth, the remaining 288^ to arrive 36^ behind the ikrth.
During tiiis time, the vehicle can be made to circle Venus arbi-
trarily often. To achieve this, it must first of all be suitably
decelerated, say by ^Vjj and thus subjected to the permanent in-
fluence of the gravitational field of that planet, just as it was
previously taken out of the Earth's gravitational field by /^vj.
The Venus near point V2 (Figure 25) is attained with a velocity
r
I 111
^11 '^I • ^^^^^^
];oa ** 37-6 km/sec.
while the orbital velocity of Venus is
^v = ^ '22f'^8^;?Q0 '^^ " 35.1 km/sec.
To attain velocity zero with respect to Venus, a decrease of
3T»6 - 35.1 = 2.5 km/sec must be made. If the orbit abcut Venus
is to be a circle of radius a, then the period by (kQ) is t ■»
o^y^ . With regard to the correct vehicle position with respect
to the later departure, the following has to be observed concern-
ing the choice of t: During the h&^ Earth days, Venus covers its
orbit k6k , i.e., when the orbitting has
22I;- = 2.07 = 2 + 0.07 times '
to stop, Venus will be further in its orbit by 0.C7 rotations
about the Sun than at the beginning (see Figure 25a). Since the
vehicle's velocity, when entering the field of attraction of Venus
(vjj), as well as at the exit from this field (vjj'), must be di-
rected at right angles to the Sun-Venus radius vector, there are,
according to Figure 25a, at the moment, when the vehicle leaves
the orbit, O.O7 parts of an orbit missing. The total number of
orbits may therefore be 3.93 or U.93 or 5.93 and so on, so that
f.i. for 5.93;
t = ^—^ = 78.2 days = 6,750,000 sec.
If mass conditions pertaining to Earth are assumed to apply,
for simplicity's sake, also in the case of Venus, which is of the
same size or nearly so (exact observations of the trajectory devia-
tions of comets indicate, that Venus has a mass only of 0.82 that
85
of E^rth)^ thee we may again put fK = ifOO^OOO km^/sec^ and this leads
far a to:
^ " y^2i) " 7^^00,000 {--^-^tt ) " 773,000 km,
and fOQT a trajectory velocities during the orbits of
V, = 2a2 ^ 2 - 772; QQQ ' %- « q 72 km/sec
V3 ^ 6,750/000 ^-'"^ ion/sec.
The desired orbit will automatically occur, if at the time
of transit at V2 (Figure 22) the relative velocity is not zero,
but 0.72 km/sec, i.e., the retardation not equal to 2.5, but
-Avjj « 37.6 - 35-1 - 0.72 « ^1.8 Wsec.
This again requires a radiation of mass of
1.8
m,
"o c
2.0
1.1 • e =1.1
0.9
e = 2.65
in the direction of the motion, i.e., toward the front.
Figure 25a.
After the \Q\ days, necessary for the 5.93 orbits, an equal
radiation of I^O\ ^ ^^ must be used to withdraw the vehicle
in opposite direction from the gravitational field of Venus and to
return it into its own elliptic orbit, in which after a further lk6
days it retxirns to the neighborhood of Earth. At the instant of
the crossing, again assumed to occur ro » 800,000 km from the
Earth center, the relative velocity with respect to Earth has to
be reduced by further mass radiation to the value of vo = O.09 km/sec,
established in section II. Since at this instant the velocity of
the vehifcle is Vj « 27.3 km/sec and the speed of the Earth in its
trajectory is v^ = 29*7 km/sec, the necessary increase in velocity
is
^vj' = 29.7 - 27.3 - 0.09 « --^2.3 km/sec
8!+
and must now be effected, so as to accelerate the vehicle, i.e.,
toward the rear:
2.3
(^) . = V. e2.0 = i.i . el-15 =3.1,7.
The entire journey in this case lasts -- including the 30 days
required for ascent and launching:
30 + 146 -^ k6\- -^ llf6 = 786 days =» 2,15 years.
If m^^ denotes the mass of the returning vehicle, m^ the total
masG at the beginning of the ascent, including fuel, then -- not
taking into account the change of loass due to supplies being used
up -- approximately;
?• = 933 • 3-65 • 2.65^ • 3.^7 = 83,000.
^1
2nd possibility (see Figure 26): From the point V2 ttie vehicle
is to return to Earth E^^ not directly, but by detour via F^. The
coincidence with Iferth can happen at best I.5 likrth years after
separation at E-^, The Sun distance rjjj of point F^ is to be chosen
therefore such, that the entire travel time from E-j^ via Vg and Fo
to Ei^ takes 5^1-7.5 days. The total tiioe T is composed out of the
times T-j^ and T2 and T^, required to cover the 3 half ellipses I,
II and III with the major semi -axes
r "t* r
ai = ~ ^ = 123,500,000 Ian;
a2 2 * ^3 2 •
From these two expressions :
^3 . a2 = ^^i^ - 11^9,000,000-108,000,000 ^ 20,500,000 ion.
Further
T3 + T2 = T - T^ = 5I+7.5 - 11^6 = i<-01.5 days,
or by (ij-3) -- for half an elliptic orbit --
-.ml + 71: ^2- = 1^01.5 days = 31^,700,000 sec,
or
85
J^.JZ^- 2^70^0,000 . ^- ^ ^4,700,000 y ,3,^ 000, 000, 000;
therefore
>/ap" + Jb^ = 1+, 010, 000, 000, 000/
and
a^ - ag = 20,500,000.
Tliese two equations are satisfied by:
a2 =» 169,000,000 km and a2 = 148,500,000 km.
Therefore from ^11 '*' ^III
^2 ." 2 •
rjjj = 2a2 - rjj = 297,000,000 - 108,000,000 « 189,000,000 laii.
Departure at E-^^ occurred with a velocity Vj = 27.3 km/sec
and arrival at V2 with a velocity:
^11 - ^I • ?^ = 27.3 • j^ = 37.6 Wsec.
The velocity required at Vg, to reach Fo, is by {h$)'.
I 2/^ ^III 1 26k, OOP 189 , /
""II V ^11 * ^III ' ^ ^""2^ • loj = 39.i^ Wsec;
which determines the velocity of arrival at F^:
^III = ^II' • 1^ = 39.4 • i§ = 22.5 Wsec.
IThe departiore velocity necessary at F.., to attain Ej^., is
/ ^A rr /2 64. OOP iW 01, n 1 /
^III
and finally the velocity, arriving at Ej^
^III
2I+.8 • JJ^ " 31.5 Wsec
86
compared to the Earth velocity
Vg " 29-7 to/sec.
Accord iiJglj^ the f ll.wiog v.lo:l.y c] ajges are jecessary:
Vg.
at departure Ej
at passixig
at passing
at arrival
■.,k Iou/e
/iVj = liTo ^9-7 ^ " ^'^ lau/sec.
F3; Avjjj- « 24.8 - 2c. 5 ^ + 2.3 Wsec,
^h^* ^^£-,1 ^ 2.S.7 " 3- '5 * O.C> ^••- - 1.7 km/sec
{i;}ltiatii3£ iaudicg).
Tile iiaeses Pccssary fco atteiu ti.ese velocity cljauges, txsing
c » 2o0 to'/'3ec^ ere giver) ic seqfaence Id^
tlie?e 5re to be di-
rected forward at E..
«nd Ei,^ and baclcferarda
at Yp ^3Dd F5
-:2.0 . i„3
i.8
. ,].::o
i
)/
.2.0 .3.-1
. ^0.^.
2 71 (
m^
2a
'if 'HI '
V •
.2.0 . i,i .
. ei.15
- 3-V(
nirs
la
y
e2-0 -^ 1.1 '
■ eO-^'5
- 2.57
With the saiTje rjotatiori as tcirr
^ 933 " 3'>65 • 2.71 ^ 3^i^7 2-57 - 82,000.
Tlie entire Journey takes
30. 5 + 547^5 ==• 578 dsys « 1.58 yearS;,
iBcludiug landing and ascent.
Of these poseibilitles^ the second cne for the same fuel con-
sumpticr has the advaritage of a shorter travel time^ while the
first periflits a longer stay in tte neighborhood of Venus.
A visit to Mars would take a similar form. Here however its
position at the instant of proximity would have to be more ac-
curately calculated, since its orbit is more eccentric than tbat
of iikirth or Venus, its Sun distance varying between 248,000,000 km
87
and 205,000, CX)0 km. Now the detoui' via F^, according to Figiire 26,
at its apogee rju equals 189,000,000 km, is nearly equal to the
smallest distance of Mars to the Sun, leaving only 16,000,000 km.
With a suitable choice of the time of ascent during a mutiial con-
stellation of Earth, Venus and Jfexs and with a suitable adjustment
of rjj and rjjj, a passage at relatively small distance (about
^ - 8 million km each time) °^ ^" ^^ ^^^^ ^^ ^^°^ ^^ ^^
achieved in a single journey of about 1 1/2 yeaxs' dxiration.
I^n
Figure 26.
This 580 day journey would not quite take 20 times as much
as the 30 (lay journey into space, described in section III. For
a rough estimate of the vehicle mass now required, we multiply by
20 the parts of the weight, which depend on the duration and are
denoted on pages 67-68 under b), c), d), e), while those inde-
pendent of time, i#e., listed under a), f ), g) and i), are left
as is. With regard to the larger weight h), required by the larger
storage space, we allow 3 times its previous weight. Since with
the storage space also the heat -transferring surface increases,
we imply here a better insulation. These assumptions lead to a
weight less fuel
(21^0 + 60 + 200 + li^O) • 20 = 12,800 kg
+ 200 + 200 + 2ij-0 + 200 + T^^-O « 1,580 kg
+ 780-3 « 2,3^0 kg
total of 16,720 kg = 16.72 t.
Between E-^ and V2, T^ » ±h6 days pass, between V^ and F2
88
between F^ and Ei^.
'3 . . /169.03
To - Ti • h^ = 146 r^^'\. = 220 days .
Of the 12.8 tons of supplies, consumed were:
during the 15 days ascent up to E]_: 12.8 • r^ = O.33 t,
between ^2. Si^ ^2' 12.8 • rr| - 3-20 t,
between Vg and F^: 12.8 • ^ - 3-95 t,
between F3 and Ei^: 12.8 • =^ =« iv.80 t,
between departure and Ej^ therefore 12.28 tons.
After arrival at Ei,., the weight of the vehicle remains 16.72 -
12.28 - k.hh tons.
Immediately before arrival at %,
the total mass is i^.i^l^ • 2.57 » 11.1^0 t
after arrival at P3 ll.it'O +1^.80 = 16.20 t
immediately before arrival at F^ 16.20 • 3.^7 = 56.3O t
after arrival at V2 56.30 + 3.95 ■ 6O.25 t
immediately before arrival at Vg 6O.25 ' 2.7I - I63.OO t
after arrival at E^ I63.OO + 3.20 « 166.20 t
immediately before arrival at E]_ 166.20 ' 3.65 = 606.67 t
after acceleration has ended 606.67 + O.33 - 6C7 t
at departure Gq » 6C7 • 933 - 567,000 t
or abbreviated:
Go » Cl/J^-^ ' 2.57 + ^.8) • 3A7 + 3.957 • 2.71 + 3.2J • 3.65 +
0.3^ • 933 - 567,000 t.
Q9
F
k
k
The inain part of the ammunition, to be taken along, is taken
up by the fuel required for the initial acceleration, but such fuel
is also necessary (say 607 - I7 =* 59O tons), to change the velocity
during the journey and such a mass will present difficulties as re-
gards maneuverability. How much Gq depends upon the velocity, with
which the mass is radiated, c, is clarified by the following list
of values for Gq, resulting for dii'ferent values of c for a constant
acceleration ofo(c = 30 m/sec^:
c = 2 km/sec: G^ =« C{il}^^^^ ' 2.57 + ^.8) * 3*^7 + 3.9^ • 2.71 +
3-2} • 3.65 + 0.327 • 933 = 567,000 t
c =» 2.5 km/sec: G^ - /^j/JkM * 2.I7 + 1^,8) • 2.77 + 3*9^ •
2.27 + 3.2 i ' 2.87 + 0.3^7 * 235 = 69,500 t
c = 3 km/sec: G^ « C{lJ^^^ * 1*95 + ^^8) • 2.38 + 3*9^ • 2.00 +
95 =• 17,600 t
M • 1.69 + h^Q)
30 = 3,150 t
M • 1.55 + ^.8)
15 = 1,130 t.
V
lANDING CK OTHER CELESTmL OBJECTS
3.2| • 2.i^5 + 0.3^7
k km/sec: G^ = C{lJ^
3.2^ • 2.00 + 0.3^7
5 km/sec: G^ = C\lJ^
3.2 j • 1.78 + 0.3^7
1.98 + 3-957 • 1-73 +
1-75 + 3.9^ • 1.57 +
Of the planets, Venus seems to be particularly suited for a
landing, because, presumably, it has an atmosphere similar to that
of Earth. This and the further assumption of similar gravitation
conditions would accordingly permit a landing exactly as described
in sections II and III for Earth. It could begin by imparting to
the vehicle at the distance ro = 800,000 km from the center of
Venus a tangential velocity Vo = O.O9 km/sec (see Figure 1^).
(Compare what has been said about Mars and Venus on pages 88-89.
Since furthermore the atmosphere of Venus is very high and dense,
landing should be siiapler than on Earth. ) The preceding journey
proceeds exactly as determined for the segment E-|^ - V2, following
Figure 25, i.e., V2 is passed with a velocity vjj = 37»6 km/sec,
compared with a velocity of Venus v^ » 35.1 km/sec; the relative
velocity at that instant being 37*6 - 35*1 = 2.5 km/sec. To re-
duce it to 0.09 km/sec, a reduction by approximately ^vjj = 2.i^-
km/sec is necessary, requiring a mass
90
^v
(^)ll » V • e = = 1.1 ' e^-O = 1.1 • e^-2 = 3.65,
while at Eq as before
(^)l=3.65.
The travel time is now:
ascent at E^^ 15 days
Sun -centered orbit E^ * V2 1^6 days
landing at V2 15 days
total 176 days,
i.e., about 6 times the 30 days, discussed in section III. In de-
termining the iBass, the parts of the weight, b), c), d), e) may be
multiplied by 6, while a), f ), g), i) are taken as is and the ve-
hicle weight h) is about doubled, giving a total weight (less fuel)
of
(21^0 + 60 + 200 + llvO) • 6 = 3,860
+ 200 + 200 + 2k0 + 200 + 7^0 = 1,580
+ 780-2 = 1,560
total =« 7,000 kg « 7.0 t.
Of the supplies, used up, were, as before:
between departure and E-j^ O.3 t
between Ej^ and V2 3*2 t
therefore between departure and V2 3-5 "t,
therefore after arrival at V2 a weight of: 7.O -3.5 =3.5 t re-
mains. The total weight at the ascent from Earth is therefore as
follows :
f or c « 2 km/sec: Gq =» /T3-5 * 3.65 + 3.2) • 3.65 + 0.^ • 933 =
54,800 t
for c = 2.5 km/sec: G^ - /X3.5 • 2.&J + 3-2) • 2.87 + 0.^7 ' 235 »
8,800 t
91
for c » 3 km/sec: G^^ « ZJ3^5 ' ^-^^^ + 3^^) * ^^^3 + 0.;^ • 95 «
2,800 t
f or c = it Wseci G^. « £13-5 ' 2*00 4 3.2) ' 2.00 + 0,^7 • 30 »
620 t
f or c » 5 Wsec; Gq « /TS-^? ' 1^78 -^ 3-2) * I.78 + 0.^ • I5 «
260 te
For a self -motivated retiirn to Earth from Veniis, tbe saine
ascent weight is neceseaxy, if however the mass necessary for the
return trip has to be taken along immediately^ then the following
weight at departiire would be necessary:
for c = 2 km/sec: 5i|-,800
f or c = 2 5 -km/eeLt 8,800
for c « 3 km/sec: 2-800
for c = ^ km/sec: 620
for c ^ 5 km/s£c: 260
A landing on Venus therefore presupposes that tl:e fuel neces-
sary for a return^ may be nanufaotujed by simple means of raw ma-
terials available there.
A landing on Mars cannot be undertaken after the fashion de-
scribed for Venus or Earth because presumably the atmosphere there
is lac kit) g. Retardation must now be effected in a manner inverse
to acceleration, described in section I, Mars has a radius of
Tp = 3^373 loD, the gravitational acceleration — as determined
frcm the motion of its 2 moons - is g^ » 3.7 m/sec^ » O.OO37
kffi/^sec^
If again an acceleration of coc= 0^03 km/sec^ and an exhaust
velocity c = 2 km/sec is assumed, so that ^ „ c^ a 0*0^ » Q>01^ ,
c 2.0 sec
then the distance r^^ at which the acceleration must commence,
is by (7):
ri - rod - ^) - 3,392 (1 - ^^SgX) . 3,800 km
3.65^ •
933 -
670,000,000 t
2.8J^ '
235 =
17,000,000 t
2.1f5^ •
55 "
1, 600, 000 t
2.002 •
30 -
71^,000 t
1.78^ •
15 -
1,21+0 t.
92
and the velocity at arrival in r^ from a large distance is by (8);
/ggpTo^ I2 . 0.0037 • 3.392^ 1, r^^ , /
vi -y-7^ -J 37800 ^^ = ^'^^ ^^^'''
further the average retardation by (9) J
a ^o / ^o^v 0.0037 , 3.392^ . ^ / 2
^ = co< - 3- (2 + ^) = 0.03 - - 3 • ^ ' (2 + ^'^g^ ) = 0.02655 Wsec^,
the approximate time therefore for retardation by (lO):
and the imss ratio of the radiatioa by (ll):
^ „ e^^l « eO-015 • 177 „ e2-66 , ii^^3.
If Tj = 1^9^000,000 km denotes the Barth-Sion distance, and if
Mars is to be reached at its perigee Vjj » 205,000,000 km, then the
vehicle, after leaving Earth, inust receive a tangential velocity,
according to (kS), of
'l-/lF7^.32.0Wse=
compared to the Earth velocity of 29.7 km/sec, while Mars is passed
with a velocity of
vii - 32.0 • 2^ = 23.2 km/sec
compared to a velocity in its orbit around the Sun of 26.5 km/sec.
The necessary changes in velocity are therefore after leaving Earth
2ivj =» 32.0 - 29.7 =» 2,3 km/sec
with
24
• e^^ = 1.1 • el*^5 = 3,1^.7;
■^0^
(^)l " "
for landing on Mars :
AVjj = 26.5 - 23.2 -3.3 Wsec
93
with
(—■)ll " >' * e^'O - 1.1 • e^*o5 - 5.73.
The travel time is composed as follows:
ascent from Eeixth about 15 days
Sun-centered journey Earth -^fe,rs :
a = -=~ — =^ = 177,000,000 km and
h km3
J; /I « 132,000,000,000 2 therefore
^y isIioO^OOo'oOO = 20,350,000 sec - 235 days
landing on Mars approximately 1$ days
total: 265 days
i.e., about 9 times the 30 day journey of section III* Similar to
the journey around Venus, the initial weight of the vehicle, less
fuel, may be found by:
I • 3,860 + 1,580 + 1,560 = 5,790 + 3,1^0 » 8,930 kg - ^9 t.
Of the approximately 5.8 t supplies consumed were:
during the ascent from Earth =4— • 5»8 ■ ^0.3 t
255
during the Sun -centered journey Earth-^krs 265 * 5»8 » 5 •2 t
dxiring the landing on Melts ^0.3 t
At arrival on Mars there are left 9.0 - 5.8 « 3.2 t, and the
total weight at the beginning of the ascent is
f or c =« 2 km/sec: Gq =- JZT3.2 • lk.3 + 0.3) • 5-73 + 5*27 • 3*^7 +
0-3| • 933 « 875,000 t
9h
for
c = 2.5 km/sec: G^ = JZT3.2 • 8.3 + 0.3) • I+.I3 + 5.27 • 2.77 +
0.3] • 235 = 76,500 t
■
for
c = 3 Wsec: G^ = ^/J3.2 • 5-9 + 0.3) • 3-32 + 5.27" • 2.38 +
0.3} • 95 = 15,600 t
for
c = i^ km/sec: Gq = JZT3.2 • 3.8 + 0.3) • 2.5I + 5 -g/" * 1-98 +
0.3^ ' 30 » 2,200 t
for
c = 5 km/sec: G^ - {/X3.2 * 2.9 + 0.3) * 2.11+ + 5.27 • 1.75 +
0.3} • 15 - 690 t
thus
i very much less favorable than in the case of Venus, which has
an atmosphere. Much, more favorable however is the self-propelled
retxirn to Earth from Mars — again of course under the assumption.
that fuel can be found and processed there -- in this case the fac-
»
tor
933, etc., is eliminated because of the E&rth atmosphere and
the
remaining factors in the inverse sequence, corresponding to a
return trip, are
«
for
c = 2 km/sec; G^ = f/Js .2 + O.3) * 3-^7 + 5 -g/ ' 5-73 +
0.3 \ ' 11+.3 = l,i^30 t
for
c = 2.5 km/sec: G^ = JZJ3-2 + O.3) • 2.77 + 5-27" " ^.13 +
0.3} • 8.3 = 515 t
for
c = 3 km/sec; G^ = JZT3.2 + 0.3) * 2.38 + 5 -g/ * 3-32 +
0.3 j • 5.9 =- 265 t
for
c ° k km/sec: G^ =• JZX3.2 + O.3) • I.98 + 5-27 * 2.51 +
0.3J • 3.8 =« 118 t
for
c = 5 km/sec: Gq = fZT3.2 + 0.3) • 1.75 + 5.2/ • 2.1^+ +
0.3 j • 2.9 - 71 t.
,
A landing on the Moon would be similar to one on ^rs. Here,
using the same notation, we have:
95
r = l^T^O km; g-, = 0.0016 km/sec^ (since the density of the Moon
1 7i»-0
is lower than Earth's, we have gQ < O.OO98 • ^3^); <KC = O.03
/ ? / 0.015 , , 0.0016 ,
km/sec'^; c = 2 .0 km/sec; °< = "gg^ ; r^ = l,lh-0 (l + q^q^ )
/2 • 0.0016 . 1.-;
1
0.0016
/2 • 0.0016 ' I.7H02
,830 km; v-j^y J-830 " 2*30 km/sec; fi = ^0.03
0016 , 1.7^0^ . ' o, / P "^1 2.30 „,
3 ^^ "^ 183^^ " ^'^^ Ion/sec^; t^ = -^ = o.028lt " ^^ ^^='
!?o = gocti = eO.015 • 81 =. el'22 = 3,1^0.
mi
Since the duration of the journey in this case is at most tvice
as long as tliat required for the extended space journey of section
^ ITI, requiring a correspondingly snaller stock of supplies, one may
J now assume a weight, less fuel, averaging 2.6 t in place of 3 t,
^ giving an initial weight for the one-way trip Efeirth-Moon:
f cxr c = 2 Ion/sec: G^ =» 2,6 • 3.^ * 933 " 3,250 t
for c =■ 2.5 km/sec: Gq = 2.6 • 2.6U • 235 " l,6lO t
f or c = 3 Weec: G^ - 2.6 • 2.25 • 95 - 555 *
for c = 4 km/sec: G^ » 2.6 • I.85 • 30 - iWf t
for c = 5 km/sec: Gq = 2.6 • 1»&* * I3 ' Oi t
and the initial weight for the return trip Moon -Earth:
f or c = 2 km/sec: Gq = 2.6 ♦ 3.^ » 8.9 t
for c = 2.5 km/sec: Gq » 2.6 • 2.6^ = 6.9 t
f or c = 3 kra/sec: Gq •» 2.6 • 2.25 = 5.9 t
for c " k km/sec: Gq = 2.6 • I.85 » k.Q t
f or c = 5 km/sec: Gq = 2.6 • 1.61^ - k,3 t.
If on the other hand the return to Earth is to be assured, then
the weight at departure from Earth must be:
f or c = 2 km/sec: Gq " 2.6 • 3.^^ • 933 =• 28,000 t
for c = 2.5 km/sec: Gq « 2.6 ♦ 2.61^^ • 235 " ^,^50 t
96
f or c = 3 km/sec: Gq « 2.6 • 2.25^ • 95 == 1,250 t
for c = U km/sec; Gq « 2.6 • 1.85^ . 30 « 890 t
f or c =» 5 km/sec: Gq » 2.6 • 1.6k^ • 15 = TOO t.
The relative case of the attainability of the Mood and the low
mass ratio ^ „ ii n ^* departure from the l^on leads to the concept
mi "
of using the Itoon as an intermediate station for further undertakings.
A condition here is that the necessary fuel may be gained from the
Moon itself, i.e., an explosives factory can be constructed there.
To investigate this possibility, an investigative 2 -way trip with
assured return, f .1. one, involving G^ = 38,000 t at c = 2 km/sec,
would have to be carried out, which after all is not coinpletely out-
side the range of possibility. In case of positive findings, every
further l-Ioon journey would require only 8,250 t and each return from
Moon to Earth only 8.9 t, and for each trip to the planets, initiated
from Moon rather than from Earth, the mass ratio required would be
^o ) 4_ > etc., in place of the Earth -mass ratio ^ _ , etc.,
idt -^ * m-i y^~f
whereby however the return trip would be directly to Earth, because
of the more favorable landing conditions.
The following initial weights would be necessary;
a) round -trip Moon -Venus -Mars -Earth (no intermediate landing on
Venus or Mars ) :
f or c = 2 km/sec: Gq - ^ • 567,000 » 2,Cr70 t
933
for c =« 2.5 km/sec: G^ - ^Ji^ • 69,500 = 78O t
f or c = 3 km/sec: G-. - ^i^ • 1T,600 = U17 t
"95
f or c == if km/sec: G^ =» ii^ • 3A50 » 19^ t
f or c = 5 km/sec: G^ « i^^ • 1,130 ' i2k t
b) for a trip Moon-Mars, including landing, but without insuring
a return:
97
f or c = 2 km/sec: Gq - 2A • 875^000 = 3,190 t
for c = 2.5 km/sec: G„ = ^.6^ . 76,500 = 86O t
o 235
f or c = 3 km/sec: G„ = 2^^ . 15 6OO = 37O t
95
for c =■ !<• Ian/sec: G- - il§l * 2,200 = I36 t
30
for c = 5 km/sec: Gq = IiSL • 69O = 76 t
c) for a trip Moon-Venus, including landing, but without insuring
a return:
fcxr c = 2 km/sec: G„ = lA * 5^»800 = 200 t
° 933
for c = 2.5 km/sec: Gq =• ^^ ' 8,800 - 99 t
235
f or c = 3 km/sec: Gq " ij^ • 2,800 » 67 t
95
f or c = Iv km/sec: Gq ■ ii^ • 620 - 38 t
for c = 5 km/sec: Gq = ij^ * 260 « 29 t
d) for a landing on Mars, safeguarding tbe return trip (safe for
an Initial investigation), using ^ ^ ^^^ , > etc., and taking into
consideration a further 5.8 t of supplies for the return trip:
f or c = 2 km/sec: Gq = 3,190 • l^+.S * ? * ^■-- - 75,000 t
for c = 2.5 km/sec: Gq = 86O • 8.3 ' ^ *J'^ = 11,800 t
f or c = 3 km/sec: Gq = 370 • 5-9 • ^ *J'^ ' 3,600 t
f or c = 1+ km/sec: Gq » I36 • 3.8 • ? * ^'^ •» 85O t
98
for c = 5 km/sec: G^ =« 76 • 2.9 • ^ ^^" = 36O t
e) for a landing on Venus, seifeguarding a return in a suitable way:
f or c = 2 km/sec: G^ = 200 • 933 • T "^3'? = 290,000 t
for c = 2.5 km/sec: Gq « 99 • 235 • T .t 3-9 = 36,300 t
f or c = 3 km/sec: G^ =» 6? • 95 • ^ "^3'? = 9,900 t
f or c = 4 km/sec: Gq = 38 • 30 • ^ '*' 3'? = 1,780 t
f or c = 5 km/sec: G^ = 29 • I5 • T '*' 3*? = 68O t.
Safeguarding the return in case e) is more difficult than in
case d), nevertheless and in spite of the fact that a self-propelled
return from Venus (with nearly the same values G^ as at departure
from Sarth) could only be accomplished with a high exhaust velocity
c; the probability of finding there an atmosphere and consequently
conditions for life, similar to those on Earth, is so great, and
the difficulties of tlie one-way trip there — having once estab-
lished the station on Moon -- so small that presumably Venus must
be primarily considered as a goal for immigration, Mars on the other
hand a goal for scientific investigation trips*
During departure from Moon, one should, strictly speaking,
take into account the velocity of the Moon around the Earth, as
was done (Figure I6) with the rotation of the Earth; its influence
will however not here be investigated.
For simplicity, we have up to now only discussed those connec-
ting elliptic segments between planets, which touch the 2 planets,
which are to be connected, i.e., those requiring changes in veloc-
ity, but no changes in direction. It is not obvious that these
tangential ellipses constitute the most favorable connection.
Rather it is conceivable that other ellipses, inters ectij:ig plane-
tary orbits, would be more expeditious, since without doubt they
would provide shorter connections. For this reason we examine
first the other limit of a change in direction at constant veloc-
ity.
99
The connecting ellipse to be foxmd roust cross both planetary
orbits with velocities equal to the velocity of those planets;
using the notation of Figure 27 ve have for the connecting el-
lipses, according to (kl):
2 Sfi 2 §iff
1- ^a - ^ == ^1 - 7^ >
Ya r 2
a ^ ^2
and for the circular orbits r^ and r2> by (37):
'1^=^'
'^'■^^
therefore we would require:
1. v^2 .g^=.^
-n-
2. v^2 . 2^« j^
^ ^a ^2
■t--
or
1. 2^ . V 2 = Ji ;
2. 2ii_ 2 „^.
ra a rg
The two equations are contradictory. It follows that the con-
dition of crossing both orbits with corresponding velocities cannot
be met.
If now only the condition is required, that one of the plane-
taiy orbits, say radius r2, has velocity equal to that of the el-
lipse, then only one of the equations remains :
iit. v2 = ^ .
^a ^ ^2 '
100
and choosing r arbitrarily:
JFurther by (k^):
and from (^4-6):
^ Pa r2 '
2/i- _ Va2 ^ ^
^a ^2
Va^a
v=r
a-^a
A^ - Va2
■a/rp
1 ;
i.e., each ellipse, whose semi-iaajor axis a equals the radius r2
on a circular orbit, will cross this orbit with the corresponding
velocity.
Figure 27.
The angle, subtended at the crossing by the two trajectories,
which at the same time gives the tangential inclination of the
trajectory, is by Figure 28:
dr
dr
^^ '< rgdfp rg ' d^ '
i.e., according to (k^) with r = rg:
101
2 2^
'a ~ ra 2 2/*
tan o< = / ^— r- • rg + "U - " g * rg
1 ;
or, since in this case
2 _ 2A^ _ ^
we have:
tan a(
Mr2 2ac~2
va^ra^ va^ra^
1 =
/ /^2
1 .
Figiire 28.
Of the many possible connecting ellipses with semi-najor axis
a = T2t ve now wish to examine the one which touches the planetary-
orbit with radius r]_, i.e., the one which requires a change of di-
rection only near one planet and a change in velocity only near the
other one. For this we must choose
so that
ri r2
J^'
^2 - ^1
rir2
and
102
tan ^
sl ^1 ^ rir2 ^
or
/r22 " 2rir2 + ri2 / ( r2 - ri)2
To produce tbe change of direction without change in velocity
Vp at the intersection, a velocity coniponent has to be added at
right angles to the bisector of the angle o<, having a nagnitude
ziv a 2 • V2 • sin ^ (see Figure 28).
F.i.jfor an ellipse, touching the Earth orbit and intersecting
that of Venus in the desired way, i.i. for
r]_ =» 1^1-9,000,000 km
rg =* 108,000,000 km
Vg « 35*1 km/sec:
/ (108 - 1I^S)2 ^1 » n ki •
o< = --22 lA°; ^iv » 2 • 35.1 • sin 11 l/8° •» 13. 5' km/sec;
for the ellipse, touching the orbit of Venus and cutting the orbit
of Earth, i.e., for
r-L = 108,000,000 km
rg =• 1^4-9,000,000 km
Vg ■ 29.7 km/sec:
*^° '^ ■ /loS I29S - 108) 7 108 . 190 °*^^^'
190
e^ = ^^16°; AS = 2 " 29.T • sin 8° = 8.3 km/sec;
103
for the ellipse^ touching the Earth orbit and intersecting that
of l^krs^ i.e., for
r-j_ = li;9,000,000 km
rg = 205,000,000 Ian (circular orbit assuioed)
V2 "* 26.5 km/sec:
711^9 (1^10 - 11.9) y ii ^g Tg^i
^a ^Igo. ^v = 2 • 26.5 • sin 8^ = 7A km/sec;
for the ellipse, touching the Mars orbit and intersecting that of
Earth, i.e., for
r = 205,000,000 km
rg = 1^9,000,000 km
Vg '-^ 29.7 km/sec:
o< « ^ 22^; Av =^ 2 • 29.7 • sin 11^ = 11.4 km/sec.
One can see that in all cases the velocity change -Av is much
larger tlaan that required for an orbit touching both planetary-
orbits. F.i., in the most favorable case (touching the Earth orbit
and crossing that of Mars), ve have a Av = 7.^4- km/sec (in place of
Avjj ==3.3 km/sec as on page 98), requiring a mass expenditure
~ = V • e ^ of the following values:
f or c = 2 km/sec: ^ = 1.1 • e^-O - hk.^ in place of 5.73
m
far c = 2.5 " — = 1.1 • e^'^ - 21. i<- in place of U.13
mr
for c « 3 km/sec: — = 1.1 • e^^^ » lU.l in place of 3.32
10i+
f or c = i^ km/sec: ^ « 1.1 • e^*^ = ?•
05 in place of 2.5I
ID
lA
for c = 5 km/sec: ~ = 1.1 • e^*^ = If. 85 in place of 2.1^.
It is to be added that also during the transit from the plane-
tary orbit^ touched tangentially into the connecting ellipse, cut-
tirjg tiie other planetary orbit, one requires a larger velocity
change Avjj than in the case of an orbit, touching both planetary
orbits, since in tlie latter case the changing curvature has a
miniinuni*
;?^om these results it loay be concluded that the ellipse,
touching both planetary orbits, does in fact represent the most
favorable trajectory.
Translated by U. S. Joint Publications Research Service,
205 East ^2nd Street, Suite 300,
New York 17, N. Y.
NASA - Langley Field, Va. p _ 1^1^
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