2015 USA Physics Olympiad Exam Part B 16 RM RE i.Derive an expression for the velocity boost Av to change the orbit from circular to elliptical.Express your answer in terms of ve and a. Solution Consider a circular orbit of radius Re.Considering centripetal forces, GMs v2 Re2 Re Consider an elliptical orbit that has closet approach RI and farthest point R2.Consid- ering energy, 2-6=6 Considering angular momentum at the closest and farthest points, U1R1=2B2. Combining,and focusing on the closest point, -2 B)2 GMs R2 R2 which can be solved for v, (-() R2-R1 2 GMs-R1R2 or,if we let R1/R2=a, GMs (1-a) 2n21-a)= which simplifies into 2 V1=UEV1+a This is necessarily greater than vE,so the boost is △wU1=VE 2 - 1+a Copyright C2015 American Association of Physics Teachers2015 USA Physics Olympiad Exam Part B 16 RE RM i. Derive an expression for the velocity boost ∆v1 to change the orbit from circular to elliptical. Express your answer in terms of vE and α. Solution Consider a circular orbit of radius Rc. Considering centripetal forces, GMS Rc 2 = vc 2 Rc Consider an elliptical orbit that has closet approach R1 and farthest point R2. Consid￾ering energy, 1 2 v 2 − GMS r = E. Considering angular momentum at the closest and farthest points, v1R1 = v2R2. Combining, and focusing on the closest point, 1 2 v1 2 − GMS R1 = 1 2 v1 2  R1 R2 2 − GMS R2 which can be solved for v1, 1 2 v1 2 1 −  R1 R2 2 ! = GMS R2 − R1 R1R2 or, if we let R1/R2 = α, 1 2 v1 2 ￾ 1 − α 2
= GMS R1 (1 − α) which simplifies into v1 = vE r 2 1 + α This is necessarily greater than vE, so the boost is ∆v1 = vE r 2 1 + α − 1 ! Copyright c 2015 American Association of Physics Teachers