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Unformatted text preview: W5 11;. Homwérg #5 some»... ?%bé&mn I; (9” ifaﬁﬁahbwaﬁ:‘”ga§‘
I h m Erghlem 3:1; VThe period of the orbit is given as T 23 56 4.098 a
86164.09 seconds. The radius of a circular orbit with this period can be found using Kepler’s third law (2.37) a = [ ~———~— T ] = 42164.18 km 2 H A Hohmann transfer from a low altitude 280 parking orbit to geosynchronous equatorial orbit is perhaps the most commonly performed transfer. Using a1 = 6578 km as the radius of the parking orbit, the semimajor axis a of the Hohmann transfer ellipse is a = ——~———~—— = 24371.09 km Then, using the notation of section 3.2 and Figure 3.1, the speed of
the spacecraft in the parking orbit is given by the circular orbital speed at r = a1 _ ﬂ =
vcl — a1 7.7843 km/sec The required speed in the transfer ellipse at perigee is given by equation (3.2) evaluated at r = a1 1/2
= 2“  .2. =
v1 { a1 a } 10.2389 km/sec The difference between these two Speeds gives the first Hohmann maneuver Avl = v1  vcl = ‘2.4546 km/sec The calculation of the second maneuver is somewhat similar, but we must perform the inclination change at apogee. Section 3.3 shows that inclination maneuvers are much more expensive when performed at
points where the satellite already possess a high velocity. M195 1%. HomWrg # { Soﬁa/Em 2/;
all..l__l__llll_ll_l_ _ _ . The required final speed is just the circular speed at geosynchroneus distance, r = a2 = L:
v02 3.0746 km/sec 32 At apogee on the transfer ellipse, the spacecraft will have speed 1/2
_ 2a _ U 
v2 ~> { a 5— }  1.5973 km/sec As shown in the sketch, these two velocity vectors differ by the 280 difference in the planes of the orbits. The cheapest way to (i)
circularize the orbit, and (ii) change the orbital plane is to do both
as one maneuver by drawing the Av vector between the tips of the initial and final velocities. The required maneuver is then found
from the cosine law for plane triangles as 1/2
_ 2 2 ._ o
sz  { vc2 + v2 2 vczv2 cos 28 } H 1.8254 km/sec Any other strategy would be more expensive. For example,
Circularizing the orbit first, and then performing the plane change is ‘~; 32, The ascent trajectory flown by an Apollo Lunar Medule
(L3) is the moral equivalent of a Hohmann transfer, where there is no lower circular orbit. Instead, the powered flight trajectory
terminates directly on the ascent ellipse. MM 145. Hamemrk #4: acclaim
a_N*a___**“*__“*_ﬂ_~“u~_~_wiiuu_; The eeaimajor axis a of the ascent ellipse is a = (1740+30) + (1740+250 = .lZZQ_:ll229_ = 1330 km 2 2
where a1 = 1770 km and a2 = 1990 km. We will need the value of ”m for
the moon. Since u = GM, and the mass of the moon is 0.01213 the mass
of the earth, “m = 0.01213 uQ = 4.83503 x 103 km3/sec2. The burnout speed of the LM at perilune is the required velocity
in the ellipse when r = a 1
2pm ”m 1/2
v1 = { — ——~ } = 1.7004 km/sec
a1 a The situation at rendezvous is the usual Hohmann transfer
condition, since the LM must make an actual maneuver at this point. The speed of the Command Module (CM) in its circular orbit at r = a
is 2 ch = ——E = 1.5587 km/sec while the speed of the LM in the ascent ellipse at apolune will be 2am um 1/2
v = { — ~—— } = 1.5124 km/sec
2 a2 a
_ The required second maneuver is then Av = 1.5587 ~ 1.5124 2 0.0462 2
46.2 m/sec. This is just slightly over 100 mph. In order to have the LM and CM come together directly after the
second Hohmann burn, it is necessary to time the liftoff correctly.
The analysis of section 3.4 does not apply, since the moon rotates
very slowly ( once per 29 days ), and corrections for rotation of the
moon would have been made by the CH. The ascent trajectory is one
half of an elliptical orbit, so the time which elapses from burnout to rendezvous is given by Kepler’s third law divided by 2: t = _i_ T = —_.E~—— a”: = 3682.8 sec 2
V “m For comparison, the period of the CM in its circular orbit is given by “ km/sec T = —33——— a“2 = 8021.5 sec CM r—— 2
“m 3/? MM (a Ham/we # new»... The ascent trajectory thus lasts 3682.8 ”555175‘ 0.45911 of one orbit of the CM. If the CM is to be near the rendezvous point
when the LM arrives there, it must be located 0.45911 x 360° = 165928 back from the rendezvous point at LM burnout. In other words, it must
be 1800  165928 = 14972 ahead of the LM burnout point at LM burnout.
As a time difference, the CM must pass the burnout point
14°72 m 5
At = ———L— T = 327.9 sec = 5 28
360° CM before LM burnout. This is not the difference between the LM liftoff time and the passage of the CM over the landing site, since the actual
dynamics of the powered flight trajectory must be known before this
time can be calculated. ¥%b£6wn .3 ; Cb éuw cﬁaa . gzgplgm 35, The circular orbital speeds are determined solely by the
radii of the initial and final orbits ...___ , c1 a ’ c
2 a2 Similarly, the perigee and apogee speeds in the transfer ellipse, v
and va respectively, are given by V = { Zu  H }1/2 _ para r a
for rp.— a1 and ra = a2. p,a Since these are fixed, the only question that remains is where is
the best place to do the inclination change. The sketch shows the two velocity triangles, assuming that inclination change Ai performed at the first maneuver,
second maneuver. 115 and A12 = i  Ail is performed at the ch ”£1 sz M35 l5 {fammrg #{ 934265», Applying the cosine law of plane trigonometry to
each triangle, the total velocity change is given by A A +15 2 2 " 1/2
v — v1 v2 ~ { vc1 + vp  2 vC1 vp cos A11 }
2 2 . . 1/2
+ { ch + Va  2 vC2 Va cos( 1  A11 ) } Since vp, vcl' va, and vC2 are all fixed, this is a function of the
one variable which is at our disposal: Ail. To minimize Av with respect to Ail, take itsderivative and set it
equal to zero 0 = d Av z vc1 v Sln A11
d A11 2 2 1/2
{ vcl + vp  2 vcl vp cos A11 } v v sin( i  Ai1 ) 2 1/2
{ v02 + Va  2 vC2 Va cos( 1 — A11 ) } This is a very nonlinear function of Ail, and its zeros, if any, give
the optimal inclination changes at the first and second maneuvers.
Although solving the above equation for the optimal Ai1 is not a trivial task, it is simple to show that Ai1 = O (as considered in problem 31) is not the optimal solution. Evaluating the derivative
at Ail = 0, we have d Av  vC2 va sin( i )
d A1 2 2 .
l { vc2 + va — 2 vC2 va cos( 1 ) } This is only zero if i 0, in which case there is no need to perform
any inclination change at all. So Ai1 0 is not optimal, although it
is nearly so for the case considered in problem 31. H W MM (45, Mama2 #4 swan, we 4: 0,, KW megs, EKQQLQE aé. During one orbit of the earth, taking time T given by;
Kepler's third law, the earth rotates AA == ma T, where we is the? rotation rate of the earth. Using the expression for Kepler's third
law, this becomes 3/2
2 n mg a ﬂ As shown in the sketch, divide the swath of the earth visible from the
satellite in two, and notice that each AA = side is a right triangle. Since, from either triangle, cos 6/2 a
RQ/a, the swath visible from the satellite has angular width R
_ 1 e
6 — 2 cos { a } The entire surface of the earth will be visible from any orbits
where 9 <: AA. Both of these angles are functions of the orbital
semimajor axis a, and they are shown in the plot. For satellites
below about 116 km altitude, the swath visible from the satellite is
saaller than the spacing between ground traces, and the spacecraft
will no longer cover the entire earth within one day. Actually, the
areas near the edge of the earth (the ’limb’ of the earth) are highly compressed and far away, both of which limit the usefulness of images of the edge of the earth.
Radio communication is not so badly affected long slant distances. MA; 14; HMWE # 6 50%;“. deg
0 10 20 3040 50 60 ...
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 Spring '08
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