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Unformatted text preview: ﬁlm—M
. Pinyin: I: i Ergnlﬁm 11:1; For the direct method shown in Figure 11—9, one single
3 maneuver injects the spacecraft into the desired hyperbolic orbit.
3 _Energy conservation in the hyperbolic orbit requires that the speed at .1 perigee: vp be related to the speed at infinity, v“, by _1_3_E__12__E_ _1_2
va a0 Zvn uZvu Solving for the perigee speed in the hyperbole. we have v I v2 + ._2E_
p u no
The speed before the maneuver is, of course, circular orbital velocity
at a radius e0 nvdm as vp — vein 0 O The Olberth method requires two maneuvers. The first of these is _
identical to the final maneuver in a Hohmsmn transfer. The speed i
before the first. maneuver is circular speed. v1 . The transfer 6 I‘  * ellipse has a semimajor axis of at  ( a0 + rp)/2, so the energy of
E the transfer ellipse is t 28 a+r : O p
Eguating this to the ellipse's energy at apogee, vi/Z  u/ao, where v
a
"I .
# 1
v. i
h n
+ "luau MM” 14; _. Hams/é #6 50416;», ( Pg; any 63) is the speed in the ellipse at apogee, the speed after the first _
Olberth maneuver is v a ” _____J£___
a so so + rp The first Olberth maneuver is then / .E.  ._Ji_  .____J£___
no so so + rp The speed before the second maneuver is found similarly. If v
is now the speed in the ellipse at perigee, P H II ._+_
p rp “o rp Replacing the radius aowith the radius r in the hyperbolic orbit P
energy equation above, the speed after the second maneuver, vh is
h n rp Finally, this second maneuver is n a ‘1' 3+1: The first Olberth maneuver slows the spacecraft down, while the second
speeds it up. rankmas mmqw
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:Pr;é&m Z: Eggnlgn 11:5; The parking orbit radius ro is likely to be well known,
while errors in the rocket burn will produce errors in the burnout
speed v0. These will, in turn, produce errors in the spacecraft’s
speed at the activity sphere crossing, v". So, taking differentials
of the energy equation, and holding ro constant, we have ' vm dvIn a v0 dvo This easily yields v
o dvmI _. v dv;
D Since Va is substantially larger than v , the error in the crossing
speed is larger than the error in the burnout velocity. This gets still worse. The aphelion distance is R2  2 at + Re'
so the error in aphelion distance is dR2  2 dat, since the radius of
the earth's orbit is known. The differential of the semimajor axis of the transfer ellipse is
da =n{\ar2 211/12}
t e 1 e a How, v1 = va + vm by velocity addition at the activity sphere, so dV1  dvk. Replacing dat and (W1 with the equivalent expressions in terms
of an and dv , respectively, we find
2 on dR= ‘31 2av 2 [ Vi / 2 ' "e / Rs ] " Inserting the error expression for the departure hyperbole links this all the way back to burnout in the parking orbit.
This is not a complete analysis, since the injection velocity can also have the wrong direction, as well. arable]: 11:64, The required departure speed at infinity for a Hers
mission is V"  2.98 km/sec, as given in problem 114. The necessary
departure speed vo from a low altitude parking orbit (say r0 3 6500
km) is given by equation 11.19 2 2". 1/2
yo  { v“ + r0 }  11.46 km/sec MM 14/ Hommré #5 50M».  (P12 and So, the error growth in the departure hyperbola is l 7 7 v0 , 7 ,
dvm — Vm dv; = 3.34 de However, things get much worse during the heliocentric portion of the flight. Using “0 = 1.33 x 1011 km3/sec2 and Re‘ 1.5 x 10' km for the earth’s distance from the sun, the earth’s orbital velocity is
iabout , , * , 1/2 7 , V© = { “o / Re } = 29.77 km/sec
iand the heliocentric injection speed for the spacecraft is V1 = Ve +
v0° = 32.75 km/sec. Evaluating the error growth in the heliocentric
phase, we have , ’u V , dR 7: '____—_.__9___1__—__——_ dv 2 co [ Vi / 2  "o / Re )2 12
= we... av = 3.54x107dv. 1.227 x 105 ” Combining these two error propagation equations, the aphelion
error at Mars arrival is related to the burnout velocity error by , 7— 7 8* dR2 — 1.36 x 10 dvo
iif dR2 is in kilometers and dvo is in km/sec. If the error at Mars
arrival is 100 km, we need to achieve the correct burnout speed to
within
100 0 1.36 x 108 '= 7.3 x 10"7 km/sec ’dv This is an error of only 0.073 cm/sec, far more than can be expected
of booster hardware. In order to actually fly an interplanetary trajectory, provision
must be made for midcourse correction maneuvers. 'Since these are
usually performed by very small rocket engines attached to
(relatively) massive spacecraft, they can be done to high precision.
Of course, the actual trajectory must first be determined to rdgh
accuracy before the error can be eliminated. mg I I _ Hana6f! #J sic/22m
jaglltﬂi :5: anhlsl 11:1. The injection from an incoming hyperbole to a (nearly) parabolic orbit about the planet requires dropping tron periapse speed
in an hyperbole (see probles 113, direct nethod) to (just below) escape speed, v_a = /Eu 7 r1, . So, the required
maneuver is _ / 5 5H _ / in
AV v; + rp rp Now, this is a function of v;, the speed at which we enter the activity sphere of the planet, and rp, the periapse radius in the approach hyperbole. The first quantity is not under our control, since it will be mandated by the heliocentric orbit on which the spacecraft approaches the planet. However, as shown in the sketch,
small changes in this heliocentric orbit can produce major changes in
r . It is or interest to ask if there is an optimum r by minimizing P P
the maneuver Av with respect to rp. Taking the derivative or the maneuver with respect to r , we have For a minimum maneuver, we want this derivative to be zero. However, it is never zero. If rp is verylarge (undesirable,
since the spacecraft is supposed to approach the planet), the
derivative does approach zero. In another fashion, it 'approaches'
zero as rp e U, as well. In this case, the expression above reduces
to d Aw/ d r = a  n. Reexamining the original maneuver expression, P
however, shows that as rp 9 0, the term in v: becomes negligible, and the required maneuver approaches ﬂ ﬂ
Av x / E r o 0 P P since rp cannot really go to zero, the best we can do is approach as
close to the planet as is safe. This was actually done in the case of
Mariner 9, the first spacecraft to orbit another planet. The orbit of the probe about Mars was highly elliptical, with a low perimartem
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14000 '12000 ’10000"
First case: A V along the y—axis ’> 7  — — Second case: A V along the x—axis
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 Winter '08
 Villac

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