Question

A robot has fallen from a spaceship, as a result he is orbiting a far away planet all by himself. The robot is in an elliptical orbit with min = rR and (max = 25rR/7, IR is the rescue orbit radius. If the robot changes his orbit to a circle of radius rr_ then he will be rescued. The robot has loose papers with him which he can throw to change the orbit. M. = mass of planet, m = mass of robot, m/3 = mass of papers. 1) Before the robot throw the papers, the papers and him are still travelling together. Find Vm, speed at perigee. Express answer in terms of G, M and rr. 2) Calculate VR, the speed of the object in a circular orbit, radius rr, around the planet. 3) At perigee, the robot throws the papers in the same direction he is moving to slow himself down. Find Vnd, the speed of the papers relative to the robot right after the papers are thrown. 4) Calculate the eccentricity of the orbit of the notes.

A robot has fallen from a spaceship, as a result he is orbiting a far away planet all by himself. The robot is in an elliptical orbit with rmin = rR and rmax = 25rR /7, rR is the rescue orbit radius. If the robot changes his orbit to a circle of radius rR then he will be rescued. The robot has loose papers with him which he can throw to change the orbit. M = mass of planet, m = mass of robot, m/3 = mass of papers.

1) Before the robot throw the papers, the papers and him are still travelling together. Find vm, speed at perigee. Express answer in terms of G, M and rR.

2) Calculate vR, the speed of the object in a circular orbit, radius rR, around the planet.

3) At perigee, the robot throws the papers in the same direction he is moving to slow himself down. Find vnd, the speed of the papers relative to the robot right after the papers are thrown.

4) Calculate the eccentricity of the orbit of the notes.

Answer 1

(0) We object of mars m around a planet of mass m im know that relocity of elliptical orbit IGM (22 h is V = Where G= gravitational constant r= instantanious position of the object w.rit. the planet a= half of the semi major axis. proof Kinetic GM m total GMm (²1) 2 energy E= energy = 12 mv², potential energy Where 'r' is the position of the object at time t Zet from conservation of energy, mv² - GMm GMm (e²-1) 2et ea eccentricity of the elliptical orbit VEM (2 + 22-1 Hency et Now an e-1 Hences Va √GM (2 2 2 Hence at perigee r=rmin VR We have as rmin 4 rmax rR+ 250R/7 16rR 7 2 2 Hence speed at perigee Vm = min Hence, Vm = GM ( 2 / 7765R) Hence, Vm= 25 GM 167R Hency Vm = GM YR

(2) We have VR = GM (2 á rmax Hence, VRE GM 14 Pora) 250R Hence, VR= 49GM 400rr Hence, 7 20 GM NR = VR (3) When the robot is at the circular orbit of radius VR we have GMm the total 2rR energy If to be the velocity of the robot after loosing the paper at perigee total energy of the robot 1 / 1 m vo ² GMm rmin From the we have, conservation of energy GM m GMm Ź mv.? 27R Emin Hence, GMm 2TR 11 / 2 m v. ²- GMm YR Hences Gm Vo YR Now if vo be the relocity of the paper then the kinetic energy Ź m ovo of the paper muck 2

from the the sum of total energy of the conservation of energy robot and the paper we have + m vo = ² (m + 3) vm² Hence, tvo? = 2 / 3 25 G Hence, voa 오 6 12 mvo 25 GM Gm 2 NR TR 13 GM 48R Hence the velocity of the paper relative to robot is, Vnd = Volvo 13 GM GM 414 YR Hence, Und - (-1) VR

(4) We know the velocity at perigee, Vond jam (ite) Now as lore 7 Hences GM YR VB-1) 7.6M 7 GM lite (te) 16" Hence, ite l-e = 0.022 Hence Ite 0.022 0.02.2e 2 Hency lel=0.9565