Question

The escape velocity of a body from earth is about $11.2\text{km/s}$. Assuming the mass and radius of the earth to be about $81$ and $4$ times the mass and radius of the moon. The escape velocity in $\text{km/s}$ from the surface of the moon will be

• A. $0.54$
• B. $2.48$
• C. $11$
• D. $49.5$

Answer 1

The correct option is B $2.48$

Given that,
Escape velocity from surface of earth = $11.2\text{m/s}$
Let the mass and radius of the earth is $M_e$ and $R_e$ respectively.
The mass and radius of the moon is $M_m$ and $R_m$ respectively.
So,
$\frac{M_e}{M_m}=81$
$\frac{R_e}{R_m}=4$

The escape velocity $v_e$ from surface of a planet/spherical body of mass $M$ and radius $R$ is given by
$v_e=\sqrt{\frac{2GM}{R}}$
We can say that
$v_e\propto \sqrt{\frac{M}{R}}$
Now
$\frac{\text{Escape velocity from surface of moon}}{\text{Escape velocity from surface of earth}}=\frac{v_{e.moon}}{v_{e.earth}}$
$\Rightarrow \frac{v_{e.moon}}{v_{e.earth}}=\sqrt{\frac{M_m}{M_e}\times \frac{R_e}{R_m}}$
$\Rightarrow \frac{v_{e.moon}}{v_{e.earth}}=\sqrt{\frac{1}{81}\times 4}=\frac{2}{9}$
$\Rightarrow v_{e.moon}=\frac{2}{9}\times 11.2\text{km/s}$
$\Rightarrow v_{e.moon}=2.48\text{km/s}$
Hence, option (b) is correct.

Why this question? To make students understand the dependence of escape velocity on mass and radius of the planet.