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

Hint: We can solve this problem with the help of Newton's law of universal gravitation. Force can be calculated from this law and then we can get the relation between the force and the acceleration through Newton's 2nd law of motion.

Complete step by step answer:
Newton’s law of universal gravitation states that each particle attracts each other particle in the universe with a force directly proportional to the product of their masses and inversely proportional to the square of their distance.
According to the question the mass of the planet is ${M_p}$ and the mass of the particle is $m$ and the distance between them is $\dfrac{{{D_p}}}{2}$ (because the diameter is ${D_p}$ ) then the gravitational force acting between them is
$F = G\dfrac{{{M_p}m}}{{{{(\dfrac{{{D_p}}}{2})}^2}}}$ where $G$ is the gravitational constant.
    $ = \dfrac{{4G{M_p}m}}{{{D_p}^2}}$
Newton’s 2nd law of motion states that the acceleration of an object is directly related to the net force and inversely related to its mass.
Now applying Newton’s 2nd law of motion:
If the acceleration due to the gravity is $g$ then according to the question we call tell that
$g = \dfrac{F}{m}$
Putting the value of $F$ in this equation we get
$g = \dfrac{{4G{M_p}m}}{{{D_p}^2m}}$
   $ = \dfrac{{4G{M_p}}}{{{D_p}^2}}$
So the particle falling near the surface of the planet will experience the acceleration due to the gravity is $\dfrac{{4G{M_p}}}{{{D_p}^2}}$ .

The correct answer is option B.

Note: We can solve this problem directly by applying this formula $g = \dfrac{{GM}}{{{R^2}}}$ where $g$ is the acceleration due to the gravity, $M$ and $R$ are respectively the mass and the radius of the planet and $G$ is the gravitational constant.