Question

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

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

Answer 1

The correct option is C $1:2\sqrt{2}$

Let the radius and mean density of the earth is $R_e$ and $\rho$ respectively.

The mass of the earth, $M_e=\frac{4}{3}\pi (R_e{)}^3\rho$

Then the radius of the planet, $R_p=2R_e$

Density of the planet, ${\rho }_p=2\rho$

Mass of the planet $M_p=\frac{4}{3}\pi (R_p{)}^3\times 2\rho$
$\Rightarrow M_p=\frac{4}{3}\pi (2R{)}^3\times 2\rho$

The escape velocity from the surface of a planet having mass $M$ and radius $R$ is given by
$v_e=\sqrt{\frac{2GM}{R}}$
$\Rightarrow v_e=\sqrt{\frac{2G\times \left(\frac{4}{3}\right)\pi R^3\rho }{R}}$
$\Rightarrow v_e=R\sqrt{\frac{8G\pi \rho }{3}}$

We can say that
So, $v_e\propto R\sqrt{\rho }$

Now
$\frac{\text{Escape velocity from surface of earth}}{\text{Escape velocity from surface of planet}}=\frac{v_{e.earth}}{v_{e.planet}}$

$\Rightarrow \frac{v_{e.earth}}{v_{e.planet}}=\frac{R_e\sqrt{\rho }}{R_p{\sqrt{\rho }}_p}$

$\Rightarrow \frac{v_{e.earth}}{v_{e.planet}}=\frac{R_e\sqrt{\rho }}{2R_e\sqrt{2\rho }}$

$\Rightarrow \frac{v_{e.earth}}{v_{e.planet}}=\frac{1}{2\sqrt{2}}$

Hence, option (c) is correct.