2015 USA Physics Olympiad Exam Part B 17 ii.Derive an expression for the velocity boost Av2 to change the orbit from elliptical to circular.Express your answer in terms of ve and a. Solution Most of the previous work still applies,except we want to focus on the second circular orbit at R2.Then 2%21-(1ya)2)=9 1 GMs (1-(1/)) which simplifies to 2 02 =UM 1+1/a Tis is less than vM,so the rocket must receive a second positive boost: △2=UM 2 -V1+1/a Note,however,that we want the answer in terms of vE.So there is some more math to do.First,by Kepler's law, UE2RE =UM2RM: which implies UM =VEVa Then 2 △2=vEVa 1+1/a iii.What is the angular separation between Earth and Mars,as measured from the Sun,at the time of launch so that the rocket will start from Earth and arrive at Mars when it reaches the orbit of Mars?Express your answer in terms of a. Solution Kepler's Third law gives the time for the orbital transfer 3/2 (RE+RM) M=2 RM During this time Mars has moved an angular distance OM=2%TM a+13/2 while the rocket moves an angular distance m,so the angular separation from the launch point,which is the position of Earth,will be 0= -() Copyright C2015 American Association of Physics Teachers2015 USA Physics Olympiad Exam Part B 17 ii. Derive an expression for the velocity boost ∆v2 to change the orbit from elliptical to circular. Express your answer in terms of vE and α. Solution Most of the previous work still applies, except we want to focus on the second circular orbit at R2. Then 1 2 v2 2 ￾ 1 − (1/α) 2
= GMS R2 (1 − (1/α)) which simplifies to v2 = vM s 2 1 + 1/α. Tis is less than vM, so the rocket must receive a second positive boost: ∆v2 = vM 1 − s 2 1 + 1/α! Note, however, that we want the answer in terms of vE. So there is some more math to do. First, by Kepler’s law, vE 2RE = vM 2RM, which implies vM = vE √ α Then ∆v2 = vE √ α 1 − s 2 1 + 1/α! iii. What is the angular separation between Earth and Mars, as measured from the Sun, at the time of launch so that the rocket will start from Earth and arrive at Mars when it reaches the orbit of Mars? Express your answer in terms of α. Solution Kepler’s Third law gives the time for the orbital transfer T TM = 1 2 1 2 (RE + RM) RM !3/2 = 1 2  α + 1 2 3/2 During this time Mars has moved an angular distance θM = 2π T TM = π  α + 1 2 3/2 while the rocket moves an angular distance π, so the angular separation from the launch point, which is the position of Earth, will be θ = π 1 −  α + 1 2 3/2 ! Copyright c 2015 American Association of Physics Teachers