Question

A rocket is launched normal to the surface of 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 $v _{e}=11.2\, km\, s ^{-1}$. The minimum initial ( $v _{ s }$ ) 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. $v_{s} = 22\,km\,s^{-1}$
• B. $v_{s} = 42\,km\,s^{-1}$
• C. $v_{s} = 62\,km\,s^{-1}$
• D. $v_{s} = 72\,km\,s^{-1}$

Answer 1

Answer: (B)

$\frac{1}{2} mv _{ es }^{2}-\frac{ GMm }{ R }-\frac{ G \left(3 \times 10^{5} M \right) \times m }{2.5 \times 10^{4} R }=0$
$\Rightarrow V _{ es }=\sqrt{13}\,\,v _{e}=40.3\, km / s$