Question

The acceleration due to gravity near the surface of a planet of radius R and density d is proportional to
$
  {\text{A}}{\text{. }}\dfrac{d}{{{R^2}}} \\
  {\text{B}}{\text{. }}d{R^2} \\
  {\text{C}}{\text{. }}dR \\
  {\text{D}}{\text{. }}\dfrac{d}{R} \\
 $

Answer 1

Hint: We will first calculate the expression for acceleration due to gravity from Newton's law of gravitation and Newton's second law of motion. Then we can use the relation that density is equal to mass per unit volume. From the obtained relation, we can check the dependence of acceleration due to gravity on the density and radius of a given planet.

Complete step by step answer:
In order to solve this question, we first need to obtain the expression for acceleration due to gravity. We know that according to Newton’s law of gravitation, the gravitational force between planet of mass M and an object of mass m is given as
$F = \dfrac{{GMm}}{{{r^2}}}$
Here r is the distance from the centre of the planet to the centre of the object. If an object is near the surface of the planet then r is equal to the radius of the planet which is denoted by R.
 $F = \dfrac{{GMm}}{{{R^2}}}$
We also know that according to Newton’s second law of motion, for an object moving near the surface of planet, the force is given as
$F = mg$
Here g is the value of acceleration due to gravity near the surface of the planet. Equating the expression for forces, we can get an expression for g.
$
  mg = \dfrac{{GMm}}{{{R^2}}} \\
  g = \dfrac{{GM}}{{{R^2}}} \\
 $
Now we know that the density of a planet can be given as mass of the planet per unit volume of the planet. If d is the density of the planet and if planet is spherical in shape then we have
$
  d = \dfrac{M}{{\dfrac{4}{3}\pi {R^3}}} \\
   \Rightarrow M = \dfrac{4}{3}\pi {R^3}d \\
 $
Now we can insert this in expression for acceleration due to gravity. Doing so we get
$
  g = \dfrac{{GM}}{{{R^2}}} = \dfrac{{G \times \dfrac{4}{3}\pi {R^3}d}}{{{R^2}}} = \dfrac{4}{3}\pi GRd \\
   \Rightarrow g \propto Rd \\
 $

Therefore, the correct dependence of g on density and radius of the planet is given by option C.

Note:
1. It should be noted that the value of acceleration due to gravity changes with distance from the surface of earth. Due to this the dependence of g will also change according.
2. The value of g is maximum at the surface of earth and decreases as we go away from the surface of earth. At the centre of earth as well as at a very large distance from the surface the planet, the value of g is equal to zero.