A satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to that of the earth's radius $R_e$. By firing rockets attached to it, its speed is instantaneously increased in the direction of its motion so that it becomes $\sqrt{\dfrac32}$ times larger. Due to this the farthest distance from the centre of the earth that the satellite reaches is $R$. Value of $R$ is:
Answer: (C) $3R_e$
Orbital speed at $R_e$: $v_0^{2} = \dfrac{GM}{R_e}$. New speed $v^{2} = \dfrac32 v_0^{2}$, at perigee $R_e$.
Vis-viva: $v^{2} = GM\left(\dfrac2{R_e} - \dfrac1a\right) \Rightarrow \dfrac{3}{2R_e} = \dfrac{2}{R_e} - \dfrac1a \Rightarrow a = 2R_e$.
Farthest distance $= 2a - R_e = 3R_e$.
(Equivalently, use conservation of angular momentum $vR_e = v'R$ and of energy between perigee and apogee.)
Solution by Sreeraj P, M.Sc Physics