Question

A spherical planet has a mass $ {M_p} $ and diameter $ {D_p} $ . A particle of mass $ m $ falling freely near the surface of this planet will experience an acceleration due to gravity of
(A) $ 4G{M_p}m{D_p}^2 $
(B) $ 4G{M_p}/{D_p}^2 $
(C) $ G{M_p}m/{D_p}^2 $
(D) $ G{M_p}/{D_p}^2 $

Answer 1

Hint : To solve this question we need to use the formula of the Newton’s law of gravitation to find the force acting on the freely falling particle on the planet. Then, we can easily obtain the acceleration of the particle.

Formula used: The formula used in solving this question is given by
 $\Rightarrow F = \dfrac{{GMm}}{{{d^2}}} $ , here $ F $ is the force exerted on a particle of mass $ m $ by another particle of mass $ M $ located at a distance of $ d $ from it. $ G $ is the universal constant of gravitation.

Complete step by step answer
The particle falls freely on the surface of the given planet due to the gravitational force acting on the particle by the planet. We know that the gravitational force is given by the Newton’s law of gravitation as
 $\Rightarrow F = \dfrac{{GMm}}{{{d^2}}} $ .............................(1)
According to the question, the particle of mass $ m $ is near the surface of the planet. So the distance between the particle and the centre of mass of the planet is equal to the radius of the planet. We know that the radius is equal to half the diameter. So the distance becomes
 $\Rightarrow d = \dfrac{{{D_p}}}{2} $ .............................(2)
Also, the mass of the planet is given as
 $\Rightarrow M = {M_p} $ .............................(3)
Substituting (2) and (3) in (1) we get
 $\Rightarrow F = \dfrac{{4G{M_p}m}}{{{D_p}}} $ .............................(4)
From Newton’s second law of motion we know that
 $\Rightarrow F = ma $
 $ \Rightarrow a = \dfrac{F}{m} $
Substituting (4) above we have
 $\Rightarrow a = \dfrac{{4G{M_p}m}}{{m{D_p}^2}} $
 $ \Rightarrow a = \dfrac{{4G{M_p}}}{{{D_p}^2}} $
Thus, we get the acceleration due to gravity of the particle equal to $ \dfrac{{4G{M_p}}}{{{D_p}^2}} $ .
Hence, the correct answer is option B.

Note
The distance of the particle from the centre of mass of the planet was not mentioned directly in the question. But the phrase “near the surface of this planet” means that the particle can be approximately assumed to be on the surface of the planet.