lecture3

# lecture3 - 16.522 Space Propulsion Prof Manuel...

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16.522, Space Propulsion Lecture 3 Prof. Manuel Martinez-Sanchez Page 1 of 9 16.522, Space Propulsion Prof. Manuel Martinez-Sanchez Lecture 3: Approximate V for Low-Thrust Spiral Climb Assume initial circular orbit, at co 0 v = v = r µ . Thrust is applied tangentially. Call F a= M . By conservation of energy, assuming the orbit remains near-circular v r µ ± , d - a dt 2r r µ µ ± 2 dr a dt r 2r µ µ ± -3 2 1 2 r dr a dt 2 µ ± When we integrate, b t 0 a dt = V, and so b t 1 -1 2 2 0 - r = V µ ( ) 0 b V = - r r t µ µ or final co c V=v -v (1) The result appears to be trivial, but it is not. Notice that the “velocity increment” V is actually equal to the decrease in orbital velocity. The rocket is pushing forward, but the velocity is decreasing. This is because in a r -2 force field, the kinetic energy is equal in magnitude but of the opposite sign as the total energy (potential = - 2 × kinetic). If thrust were applied opposite the velocity (negative a), the definition of V would be b t 0 (-a) dt , so the result in general is c V= v (2)

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16.522, Space Propulsion Lecture 3 Prof. Manuel Martinez-Sanchez Page 2 of 9 For simplicity, assume now F a = M = constant, which is actually optimal for many situations. Equation (1) can be recast as 0 - = at r r µ µ and solving for r, 0 0 2 2 co 0 r r r = = at at 1- 1- v r µ (3) This shows how the radial distance “spirals out” in time. In principle, this says r at co v t = a , a crude indication of “escape”. But of course, the orbit is no longer “near- circular” when approaching escape, so this result is not precise. One can get some improvement for the estimation of escape V as follows. The radial velocity r i can be calculated from (3) by differentiation. Notice that this is in the nature of an iteration, since r i was implicitly ignored in the energy balance which led to (3). We obtain 0 co 3 co 2a r v r = at 1- v i (4) The tangential component
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