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 $v_e=11.2km{s}^{–1}$. 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)

• A. $v_s=62km{s}^{–1}$
• B. $v_s=22km{s}^{–1}$
• C. $v_s=72km{s}^{–1}$
• D. $v_s=42km{s}^{–1}$

Answer 1

The correct option is D $v_s=42km{s}^{–1}$


loss in KE = Gain in PE

$\Rightarrow \frac{1}{2}m{v}_s^2=\frac{G{M}_{1}m}{R}+\frac{G{M}_{2}m}{r}$

$\Rightarrow \frac{1}{2}{v}_s^2=\frac{GM}{R}+\frac{G\times \text{3}\times \text{1}{\text{0}}^{5}M}{2.5\times \text{1}{\text{0}}^{4}R}$

$\Rightarrow v_s=\sqrt{2\times \frac{GM}{R}\times \text{13}}$

= $11.2\times \sqrt{13}=40.4km/s$

$\simeq 42km/s$