6.522, Space Propulsion Prof. manuel martinez-sanchez Lecture 3: Approximate Av for Low-Thrust Spiral climb Assume initial circular orbit atV=V Thrust is applied tangentially. By conservation of energy, assuming the orbit remains near-circular v 2r2 dt dr= a dt When we integrate =△V The result appears to be trivial, but it is not. Notice that the"velocity increment"Av 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 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 Av would be l(-a)dt, so the result in general is 16.522, Space P pessan Lecture 3 Prof. Manuel martinez Page 1 of 916.522, 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 2r dt r µ µ  -3 1 2 2 r dr a dt 2 µ  When we integrate, bt 0 a dt = V, ∆ ∫ and so 1 -1 tb 2 2 0 - r =V µ ∆ ( ) 0 b V= - r rt µ µ ∆ or final ∆V=v -v co c (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 bt 0 (-a) dt ∫ , so the result in general is ∆ ∆ V= vc (2)