Question

A rocket is launched straight up from the surface of the earth. When its altitude is $\frac{1}{3}$ of the radius of the earth, its fuel runs out and therefore it coasts. If the rocket has to escape from the gravitational pull of the earth, the minimum velocity with which it should coast is (Escape velocity on the surface of the earth is $11.2km{s}^{-1}.$ )

• A. $11.2km{s}^{-1}$
• B. $10.7km{s}^{-1}$
• C. $9.7km{s}^{-1}$
• D. $8.7km{s}^{-1}$

Answer 1

Answer: (C)

Given, escape velocity on the surface of the earth.
$v_e=\sqrt{2g{R}_e}=11.2km/s$ ......(i)
where, $R$ is the radius of earth.
Height of the rocket from the surface of earth,
$h=\frac{R_e}{3}.$ If the rocket has to escape from the
gravitational pull of the earth from $h=\frac{R_e}{3}$.
Then, the potential energy at the altitude,
$\frac{R_e}{3}=$ kinetic energy
$\frac{G{M}_{e}m}{R_e+h}=\frac{1}{2}m{v}_{e_1}^2\Rightarrow \frac{G{M}_e}{R_e+\frac{R_e}{3}}=\frac{1}{2}{v}_{e_1}^2$
$[v_{e_1}=$ escape velocity of rocket from $\frac{R_e}{3}]$
$\left(\begin{array}{c}{M}_e=\text{ mass of the earth and }\\ m=\text{ mass of the rocket }\end{array}\right)$
$\frac{g{R}_e^2}{4R_e}=\frac{1}{2}{v}_{e_1}^2$
$\Rightarrow \frac{3}{4}g{R}_e=\frac{v_{e_1}^2}{2}\left[\because g=\frac{G{M}_e}{R_e^2}\right]$
$\frac{3}{4}\cdot 2g{R}_e=v_{e_1}^2[\text{ From Eq. (i) }]$
$9.7=v_{e_1}$
$\Rightarrow v_{e_1}=9.7km/s$