Question

The escape velocity from the earth is about $11km{s}^{-1}$. The escape velocity from a planet having twice the radius and the same mean density as the earth, is

• A. $22km{s}^{-1}$
• B. $11km{s}^{-1}$
• C. $5.5km{s}^{-1}$
• D. $15.5km{s}^{-1}$

Answer 1

The correct option is A

$22km{s}^{-1}$

Let, $v_e$ be the escape velocity, $G$ be the universal gravitational constant, $M$ be the mass of the planet, $R$ be the radius of the earth and $\rho$ be the density of the earth.
We know, $v_e=\sqrt{\frac{2Gm}{R}}$
We know mass of the planet, $m=volume\times density=\frac{4}{3}\pi R^3\times \rho$
Therefore, $v_e=\sqrt{\frac{8}{3}\pi \rho G{R}^2}$
$\therefore v_e\propto Rif\rho$ = constant.
Given escape velocity from the earth is about $11km{s}^{-1}$
Since the planet is having double the radius in comparison to earth, therefore the escape velocity becomes twice i.e. $2\times 11=22km{s}^{-1}$.