Question

The escape velocity from the earth's surface is $11km/sec$. If the radius of a planet is double to that of the earth but the average density is same as that of the earth, then the escape velocity from the planet would be:

• A. $22km/sec$,
• B. $11km/sec$,
• C. $5.5km/sec$,
• D. $15.5km/sec$,

Answer 1

The correct option is A $22km/sec$,

Answer is A.

The minimum velocity with which a body must be projected up so as to enable it to just overcome the gravitational pull, is known as escape velocity.

If $v_e$ is the required escape velocity, then kinetic energy which should be given to the body is $\frac{1}{2}m{v_e}^2$.

$\frac{1}{2}m{v_e}^2=\frac{GMm}{R}$

$⟹v_e=\sqrt{\frac{2GM}{R}}$.

We know, Mass M = Density d * Volume V.

Volume of the earth is given as $\frac{4}{3}\pi R^3$.

In this case, the radius of a planet is double to that of the earth but the average density is same as that of the earth.

So, $v_e=\sqrt{\frac{8}{3}\pi dG{R}^2}$.

Therefore, escape velocity is proportional to R if density d is constant. Since the planet having double radius in comparison to earth, the escape velocity becomes twice.

That is, $11km/sec\times 2=22km/sec.$

Hence, the escape velocity is $22km/sec.$