Question

The escape velocity from the Earth is about $11km{{s}^{-1}}$. The escape velocity from a planet having twice the radius and the same mean density as the earth is
A. $22km{{s}^{-1}}$
B. $11km{{s}^{-1}}$
C. $5.5km{{s}^{-1}}$
D. $15.5km{{s}^{-1}}$

Answer 1

Hint: Escape velocity is the amount of energy an object needs to escape the gravitational clutches with the body. The escape velocity of an object depends only on the mass and size of the body from which it is trying to escape.

Formula used:
Escape velocity $v=\sqrt{\dfrac{2GM}{R}}$

Complete step by step answer:
Escape velocity is expressed as the speed that an object requires to be traveling to break free of a planet’s (or say a body) gravitational pull and leave it without further propulsion. The object should not fall back to the surface of the body or its orbit. Escape velocity is the function of the mass of the body and the distance to the center of mass of the body. Once we achieve escape velocity, no further impulse is required to be applied on the body for it to continue in its escape. We can say that in given escape velocity, the object will move away from the body, continuing getting slow, and will asymptotically approach zero speed as the distance between object and body reaches infinity, and the object will never return back to the body. In case of speed higher than the escape velocity, the object will have positive velocity at infinity.

Expression for escape velocity:
$v=\sqrt{\dfrac{2GM}{R}}$
Where $G$ is the gravitational constant
$M$ is the mass of body from which object is trying to escape
$R$ is the distance between center of body and center of mass
We are given that escape velocity from Earth is ${{v}_{e}}=11\dfrac{km}{s}$
Radius of planet is twice the radius of Earth ${{R}_{p}}=2{{R}_{e}}$
Mean density of planet and earth is equal ${{\rho }_{p}}={{\rho }_{e}}$
Escape velocity $v=\sqrt{\dfrac{2GM}{R}}=\sqrt{\dfrac{2G\times \dfrac{4}{3}\Pi {{R}^{3}}\rho }{R}}=\sqrt{\dfrac{8}{3}\Pi G{{R}^{2}}\rho }$
We get, $v\propto \sqrt{\rho {{R}^{2}}}$
Therefore, $\dfrac{{{v}_{p}}}{{{v}_{e}}}=\sqrt{\dfrac{{{\rho }_{p}}{{R}_{p}}^{2}}{{{\rho }_{e}}{{R}_{e}}^{2}}}$
Given that ${{v}_{e}}=11\dfrac{km}{s}$and ${{R}_{p}}=2{{R}_{e}}$
$\dfrac{{{v}_{p}}}{11}=\sqrt{\dfrac{{{\rho }_{p}}{{(2{{R}_{e}})}^{2}}}{{{\rho }_{e}}{{R}_{e}}^{2}}}=2$
${{v}_{p}}=2{{v}_{e}}$
Therefore, ${{v}_{p}}=22\dfrac{km}{s}$
Escape velocity from the planet will be $22km{{s}^{-1}}$
Hence, the correct option is A.

Note: Students should note the point that the minimum escape velocity assumes that there is no friction, such as atmospheric drag, which would increase the needed instantaneous velocity to escape the gravitational pull. Also, there will not be any acceleration or deceleration, which would change the value of required instantaneous velocity, the escape velocity.