Question

The escape velocity for the earth is $v_e$. The escape velocity for a planet whose radius is $\frac{1}{4}th$, the radius of the earth and mass half that of the earth is :

• A. $\frac{V_e}{\sqrt{2}}$
• B. $\sqrt{2}{V}_e$
• C. $2V_e$
• D. $\frac{V_e}{2}$

Answer 1

Answer: (B)

$\sqrt{2}{V}_e$

Escape velocity of body from the surface of the earth is given by
$v_e=\sqrt{\frac{2G{M}_e}{R_e}}$ ....(i)
So, according to the question,
$M_P=\frac{M_e}{2}$ and $R_P=\frac{R_e}{4}$
Now, escape velocity of a body from the surface of the planet is given by
$v_P=\sqrt{\frac{2G{M}_P}{R_P}}$
or $v_P=\sqrt{\frac{2G\times M_e\times 4}{2R_e}}=\sqrt{2}\sqrt{\frac{2G{M}_e}{R_e}}$
$\sqrt{2}{v}_e$ [from Eq. (i)]