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Unformatted text preview: 16.522, Space Propulsion Prof. Manuel MartinezSanchez Lecture 4: Repositioning in Orbits Suppose we want to move a satellite in a circular orbit to a position apart in the same orbit, in a time t (assumed to be several orbital times at least). The general approach is to transfer to a lower (for positive ) or higher (for < 0 ) nearby orbit, then drift in this faster (or slower) orbit for a certain time, then return to the original orbit. The analysis is similar for low and high thrust, because we have radius ratios very close to 1, so that, in either case the satellite is nearly in orbit even during thrusting periods, and V's for orbit transfer amount (in magnitude) to the difference of the beginning and ending orbital speeds. In detail, of course, if done at high thrust the maneuver involves a twoimpulse Hohmann transfer to the drift orbit and one other twoimpulse Hohmann transfer back to the original orbit. For the lowthrust case, continuous thrusting is used during both legs, with some guidance required to JG remove the very slight radial component of v picked up during spiral flight (although ignored here). We will do the analysis for the lowthrust case only, then adapt the result for high thrust. Let be the advance angle relative to a hypothetical satellite remaining in the original orbit and left undisturbed. The general shape of the maneuver is sketched below: Since the orbital angular velocity is = 3 , r its variation with orbit radius is 16.522, Space Propulsion Lecture 4 Prof. Manuel MartinezSanchez Page 1 of 10 3 r d ( ) =  = (1) 2 r d t During thrusting, r is varying according to d F r a = M (2) dt  2 r M v a F or dr a 1 dr = 2 ar 1 2 (3) 2 r 2 dt r rdt and so d ( ) d 2 ( ) 3 2 ar 1 2 3a = =  = (4) dt dt 2 2 r For integration, we will regard r r 0 as a constant (small variations): d ( ) 3a + =  t constant (5) dt r 0 Starting from t = 0, = 0, d ( ) = , dt we obtain d ( ) =  3 a t ; =  3 a t 2 ( t < t 1 ) (6) dt r 0 2 r 0 After ( t = t 1 ), we continue to drift at a constant rate d ( ) 3 a =  t 1 dt r and, since we start from ( ) 3 a 2 t = t 1 , 1 2 r 0 the during the coasting phase is t 1 =  3 a t 1 2  3 a t 1 ( t  t 1 ) =  3 at 1 t  (7) 2 r 0 r 0 2 r 0  At the end of coasting ( t = t t ) , we have then 1 16.522, Space Propulsion Lecture 4 Prof. Manuel MartinezSanchez Page 2 of 10 3 ( t t ) =  3 at 1 t...
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 Fall '05
 ManuelMartinezSanchez
 Propulsion

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