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.) $4G{{M}_{p}}m{{D}_{p}}^{2}$
B.) $\dfrac{4G{{M}_{p}}}{{{D}_{p}}^{2}}$
C.) $\dfrac{G{{M}_{p}}m}{{{D}_{p}}^{2}}$
D.) $\dfrac{G{{M}_{p}}}{{{D}_{p}}^{2}}$

Answer 1

Hint: An object near the surface of the earth will have the force of gravity acting on it. The force of gravity acting on it will produce an acceleration on the body. The force of gravity on an object depends on the mass of the planet and the mass of the body and the radius of the planet.

Complete step by step answer:
We know that the gravitational acting between two masses is given by,

${{F}_{g}}=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{R}^{2}}}$

Where,

${{m}_{1}}$ is the mass of the first body.
${{m}_{2}}$ is the mass of the second body.
R is the distance between the masses.

For a planet of mass, ${{M}_{p}}$ and diameter ${{D}_{p}}$ the force of gravitation with an object of mass m near the surface of the planet is given by,

${{F}_{g}}=\dfrac{G{{M}_{p}}m}{{{\left( \dfrac{{{D}_{p}}}{2} \right)}^{2}}}$

So, the above equation can be written as,

${{F}_{g}}=\dfrac{4G{{M}_{p}}m}{{{D}_{p}}^{2}}$

This gravitational force acting on the mass m, will accelerate the mass towards the planet, so that can be written as,

${{F}_{g}}=m{{a}_{m}}$

Where,

${{a}_{m}}$ is the acceleration of the mass m.

Substituting the value of ${{F}_{g}}$ in the equation above, we get,

$\dfrac{4G{{M}_{p}}m}{{{D}_{p}}^{2}}=m{{a}_{m}}$
$\therefore {{a}_{m}}=\dfrac{4G{{M}_{p}}}{{{D}_{p}}^{2}}$

So the object of mass m will experience an acceleration of ${{a}_{m}}=\dfrac{4G{{M}_{p}}}{{{D}_{p}}^{2}}$ when it is near the surface of the earth.

So, the answer to the question is option (B).

Note: The acceleration due to the gravity of a planet of mass M and radius R is given by,
$g=\dfrac{GM}{{{R}^{2}}}$, where g is the acceleration due to gravity, G is the gravitational constant whose value is $6.674\times {{10}^{-}}^{11}~{{m}^{3}}k{{g}^{-}}^{1}{{s}^{-}}^{2}$.

The escape velocity of the above planet is, ${{v}_{e}}=\sqrt{\dfrac{2GM}{R}}$
The value of acceleration due to gravity decreases as we go from the surface of the earth to higher altitudes or deeper depths.