A satellite of mass M is launched vertically upwards with initial speed u from the surface of the earth. After it reaches height $R\left(R=\text{radius of the earth}\right)$ , it ejects a rocket of mass  $\frac{M}{10}$ so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is ( G is the gravitational constant; $M_e$is the mass of the earth):

detailed solution

Correct option is C

Applying energy conservation , to find velocity of satellite when it reaches at height R.​−GMeMR+12Mu2=−GMeM2R+12Mv2​⇒−GMeR+12u2=−GMe2R+12v2⇒−GMeR+GMe2R+12u2=+12v2​⇒−2GMe+GMe2R+12u2=+12v2⇒−GMeR+u2=v2​⇒ v=u2−GMeR    As the satellite reach at height R,  it has only radial velocity v, no tangential velocity.After ejecting rocket, satellite has only tangential velocity and  moves with orbital speed = GMe2RLet the rocket has velocity Vr  in radial direction and VT ​in tangential direction.Applying momentum conservation in tangential direction, ​M10VT=9M10GMe2R⇒Tangential velocity,VT =9GMe2R=81GMe2R−−−−−−1  ​Now applying momentum conservation in radial direction, M10Vr=Mu2−GMeR​⇒Vr = 10 u2−GMeR−−−−−−2 Total Kinetic energy of rocket=12M10VT2+Vr2=M2081GMe2R+100u2−100GMeRFrom 1 &2​                                                     =M20100u2−119GMe2R  =100M20u2−119GMe200R​                                                     =5Mu2−119GMe200R