Question

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 is become $\sqrt{\frac{3}{2}}$ times larger. Due to this the farthest distance from the centre of the earth that the satellite reaches is $R,$ value of $R$ is :

• A. $4R_e$
• B. $3R_e$
• C. $2R_e$
• D. $2.5R_e$

Answer 1

Answer: (B)

$V_0=\sqrt{\frac{GM}{R_e}}$
$\frac{-GMm}{R_e}+\frac{1}{2}m{v}^2=\frac{-GMm}{R_{\max }}+\frac{1}{2}m{v}^{\prime 2}$... (i)
$V{R}_e=V^{\prime }{R}_{\max }$...(ii)
Solving (i) $\&$ (ii)
$R_{\max }=3R_e$