Question

The escape velocity from the earth's surface is $11km/sec$. A certain planet has a radius twice that of the earth but its mean density is the same as that of the earth. The value of the escape velocity from this planet would be

• A. $11km/sec$
• B. $22km/sec$
• C. $33km/sec$
• D. $16.5km/sec$

Answer 1

The correct option is B $22km/sec$

Escape velocity, $v_e=\sqrt{\frac{2GM}{R}}$
where, $M=\frac{4}{3}\pi R^{3}d$

$\therefore v_e=\sqrt{\frac{2G\times \frac{4}{3}\pi R^{3}d}{R}}$

$=\sqrt{\frac{8G\pi R^{2}d}{3}}$

$\Rightarrow v_e\propto R$
Now, $\frac{(v_e{)}_{earth}}{(v_e{)}_{planet}}=\frac{R_{earth}}{R_{planet}}$

$\Rightarrow \frac{11}{v_e}=\frac{R}{2R}$
$\Rightarrow v_e=22km/sec$