During thrusting, Sr is varying according to d v≡a F 日= d 1dr 2ar1/2 (3) and so d(a2) d(89)3 2ar1/2 3 (4) For integration, we will regard r=roas a constant(small variations) d(89) 3a =--t+ constant Starting from t=0, 69=00(89)20 e obtain t69=3at(t<t,) After(t=ti), we continue to drift at a constant rate dt and, since we start from (t,)2To the 59 during the coasting phase is 3 a t2-t1(t-t1)=- t At the end of coasting(t=At-t), we have then 16.522, Space P rtinez- sanchez Lecture 4 Prof. Manuel martinez3 Ω δ r d (δϑ) δΩ = - = (1) 2 r dt During thrusting, δ r is varying according to d ⎛ µ ⎞ F r ⎝ a= M ⎠ ⎟ (2) dt ⎜ ⎝ - 2r ⎟ ⎠ ≅ M v ≅ a µ ⎛ F ⎞ ⎜ or µ dr ≅ a µ → 1dr = 2ar 12 (3) 2r2 dt r rdt µ and so d (δΩ) d2 (δϑ) 3 2ar12 3a = = - Ω = − (4) dt dt2 2 µ r For integration, we will regard r ≅ r0 as a constant (small variations): d (δϑ) 3a = - t constant + (5) dt r0 Starting from t= 0, = 0, δϑ d (δϑ) = 0, dt we obtain d (δϑ) = - 3a t ; δϑ = - 3 a t2 (t < t1 ) (6) dt r0 2 r0 After (t=t1), we continue to drift at a constant rate d (δϑ) 3a = - t1 dt r0 and, since we start from δϑ ( ) 3 a 2 t= - t1 , 1 2 r0 the δϑ during the coasting phase is t1 ⎞ δϑ = - 3 a t1 2 - 3a t1 (t - t1 ) = - 3at1 ⎜ ⎛ t - ⎟ (7) 2 r0 r r0 0 ⎝ 2 ⎠ At the end of coasting (t= ∆t t - ) , we have then 1 16.522, Space Propulsion Lecture 4 Prof. Manuel Martinez-Sanchez Page 2 of 10