**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 :

Correct Option: , 2

**Solution:**

$\mathrm{V}_{0}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}_{\mathrm{e}}}}$

$\frac{-\mathrm{GMm}}{\mathrm{R}_{e}}+\frac{1}{2} \mathrm{mv}^{2}=\frac{-\mathrm{GMm}}{\mathrm{R}_{\max }}+\frac{1}{2} \mathrm{mv}^{\prime 2}$.$\ldots(i)$

$\mathrm{VR}_{\mathrm{e}}=\mathrm{V}^{\prime} \mathrm{R}_{\max }$....(ii)

Solving (i) & (ii)

$\mathrm{R}_{\max }=3 \mathrm{R}_{\mathrm{e}}$