Question

The ratio of escape velocity at earth$\left(v_e\right)$ to the escape velocity at a planet$\left(v_p\right)$ whose radius and mean density are twice as that of earth is:

• A. $1:2$
• B. $1:2\sqrt{2}$
• C. $1:4$
• D. $1:\sqrt{2}$

Answer 1

Answer: (B)

$1:2\sqrt{2}$

We know that: $v_e=\sqrt{\frac{2GM}{R}}$

Given that: $R^{\prime }=2R$ and ${\rho }^{\prime }=2\rho$

$\Rightarrow M=\frac{4}{3}\pi R^3\times \rho$

$\Rightarrow M^{\prime }=\frac{4}{3}\pi (2R{)}^3\times 2\rho =16M$

$\Rightarrow v_p=\sqrt{\frac{2G\left(16M\right)}{2R}}=\sqrt{\frac{16GM}{R}}$

$\Rightarrow \frac{v_e}{v_p}=\sqrt{\frac{1}{8}}=\frac{1}{2\sqrt{2}}$