Question

A rocket is launched normal to the surface of the Earth, away from the Sun, along the line, joining the Sun and the Earth. The Sun is $3\times {10}^5$ times heavier than the Earth and is at a distance $2.5\times {10}^4$ times larger than the radius of the Earth. The escape velocity from Earth's gravitational field is $v_e=11.2\text{km/s}$. The minimum initial velocity $\left(v_s\right)$ required for the rocket to be able to leave the Sun-Earth system is closest to $($Ignore the rotation and revolution of the Earth and the presence of any other planet$)$

Answer 1

For earth, the escape velocity will be,

$v_e=\sqrt{\frac{2GM}{R}}=11.2\text{km/s}$

And for the system (Sun +Earth), potential energy will be,

$U_{system}=(-\frac{GM}{R}-\frac{3\times {10}^5}{2.5\times {10}^4}\frac{GM}{R})m$

$U_{system}=-\frac{GMm}{R}(1+12)=-\frac{13GMm}{R}$

For minimum velocity of satellite,

$\frac{1}{2}m{v}_s^2=\frac{13GMm}{R}$

$v_s=\sqrt{\frac{2GM}{R}}\cdot \sqrt{13}=11.2\times \sqrt{13}$

$\therefore v_s=40.38\text{km/s}\approx 42\text{km/s}$

Hence, option $\left(\text{B}\right)$ is correct.