Question

The radius of a planet is $1/4^{th}$ of $R_e$ and its acc, due to gravity is $2g$. What would be the value of escape velocity on the planet, if escape velocity on earth is $v_e$

• A. $\frac{v_e}{\sqrt{2}}$
• B. $v_e\sqrt{2}$
• C. $2v_e$
• D. $\frac{v_e}{2}$

Answer 1

Answer: (A)

$\frac{v_e}{\sqrt{2}}$

The escape velocity on the earth is defined as
$v_e=\sqrt{2g_e{R}_e}$
Where $R_e$ & $g_e$ are the radius & acceleration due to gravity of earth
Now for planet $g_p=2g_e,R_p=R_e/4$
So $v_p=\sqrt{2g_p{R}_p}=\sqrt{2\times 2g_e\times R_e/4}=\frac{v_e}{\sqrt{2}}$