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 equal to
A) $\dfrac{{4G{M_p}}}{{{D_p}^2}}$
B) $\dfrac{{G{M_p}m}}{{{D_p}^2}}$
C) $\dfrac{{G{M_p}}}{{{D_p}^2}}$
D) $\dfrac{{4G{M_p}m}}{{{D_p}^2}}$

Answer 1

Hint: The spherical planet exerts a gravitational pull on the particle that is falling freely near its surface. This gravitational force is proportional to the product of the two masses (planet and particle) and is inversely proportional to the square of the distance between the two masses.

Formulas used:
-The gravitational force exerted by a mass on another mass is given by, ${F_g} = \dfrac{{GMm}}{{{R^2}}}$ where $G$ is the gravitational constant, $M$ and $m$ are the two masses and $R$ is the distance between these two masses.
-The acceleration due to gravity of a particle is given by, $g = \dfrac{{{F_g}}}{m}$ where ${F_g}$ is the gravitational force acting on the particle by another object and $m$ is the mass of the particle.

Complete step by step solution:
Step 1: List the known parameters of the problem at hand.
The mass of the planet is represented as ${M_p}$ .
The diameter of the planet is represented as ${D_p}$ .
The mass of the freely falling particle is represented as $m$ .

Step 2: Express the gravitational force exerted by the spherical planet on the particle.
The gravitational force exerted by the spherical planet on the freely falling particle can be expressed as ${F_g} = \dfrac{{GMm}}{{{{\left( {\dfrac{{{D_p}}}{2}} \right)}^2}}}$
$ \Rightarrow {F_g} = \dfrac{{4G{M_p}m}}{{{D_p}^2}}$ ----------- (1)
Equation (1) represents the gravitational force exerted by the planet on the particle.

Step 3: Express the relation for the acceleration due to gravity of the particle falling freely near the surface of the spherical planet.
The acceleration due to gravity of the given particle can be expressed as $g =
\dfrac{{{F_g}}}{m}$ -------- (2)
Substituting equation (1) in (2) we get, $g = \dfrac{{4G{M_p}m}}{{{D_p}^2m}} = \dfrac{{4G{M_p}}}{{{D_p}^2}}$
$\therefore $ the acceleration due to gravity of the freely falling particle is obtained to be $g = \dfrac{{4G{M_p}}}{{{D_p}^2}}$ .

Hence, the correct option is A.

Note: As the particle is mentioned to be falling freely near the surface of the planet, the radius of the spherical planet is considered to be the distance of separation between the planet and the particle. Since the radius of the spherical planet will be half of its diameter, we take $R = \dfrac{{{D_p}}}{2}$ as the distance of separation between the two masses.