Question

A spherical planet has a mass ${M_p}$ and diameter ${D_p}$ . Find the acceleration due to gravity experienced by a particle of mass $m$ falling freely near the surface of this planet.
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

The given spherical planet will exert a gravitational pull on the particle which is falling freely near its surface. This gravitational pull will be proportional to the product of the masses of the spherical planet and the particle and inversely proportional to the square of the distance of separation between the two masses. The acceleration due to gravity of the particle can then be determined.

Formulas used:
-The gravitational force exerted by one mass on another mass is given by, ${F_g} = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$ where $G$ is the gravitational constant, ${m_1}$ and ${m_2}$ are the two masses and $r$ is the distance of separation between the two masses.
-The acceleration due to gravity of a body is given by $g = \dfrac{{{F_g}}}{m}$ where ${F_g}$ is the gravitational force exerted on the body by another body and $m$ is the mass of the given body.

Complete step by step answer.
Step 1: List the given parameters of the planet and the particle.
The mass of the spherical planet is given to be ${M_p}$.
The diameter of the spherical planet is given to be ${D_p}$.
The mass of the particle falling freely is given to be $m$.
Step 2: Express the gravitational pull exerted by the planet on the particle.
The gravitational force exerted by the spherical planet on the particle can be expressed as ${F_g} = \dfrac{{G{M_p}m}}{{{{\left( {\dfrac{{{D_p}}}{2}} \right)}^2}}}$ -------- (A)
On simplifying, the gravitational pull experienced by the particle becomes
${F_g} = \dfrac{{4G{M_p}m}}{{{D_P}^2}}$ --------- (1)
Step 3: Express the relation between the acceleration due to gravity of the particle and the gravitational pull experienced by it.
The acceleration due to gravity of the particle can be expressed as $g = \dfrac{{{F_g}}}{m}$ ------- (2)
Substituting equation (1) in (2) we get, $g = \dfrac{{4G{M_p}m}}{{{D_P}^2m}}$
On simplifying, we get the acceleration due to gravity of the particle as $g = \dfrac{{4G{M_p}}}{{{D_P}^2}}$ .

So the correct option is A.

Note: Here the distance of separation between the spherical planet and the particle is considered as the radius of the spherical planet. This is because the given particle is mentioned to be falling freely near the surface of the planet. The diameter of the planet is given as ${D_p}$ so the radius of the planet will be ${R_p} = \dfrac{{{D_p}}}{2}$ . This is substituted as the distance of separation in equation (A).