Question

The acceleration due to gravity g and mean density of earth ρ are related by which of the following relations? [G = gravitational constant and R = radius of earth].
(A) \[\rho = \dfrac{{4\pi g{R^2}}}{{3G}}\]
(B) \[\rho = \dfrac{{4\pi g{R^3}}}{{3G}}\]
(C) \[\rho = \dfrac{{3G}}{{4\pi gR}}\]
(D) \[\rho = \dfrac{{3G}}{{4\pi g{R^3}}}\]

Answer 1

Hint: The acceleration of an object due to the gravitational force is called the acceleration due to gravity (g). The relation that relates the acceleration due to the earth's gravity and mass of earth combined with the gravitational constant is used in this problem to get the density of earth. By knowing the value of g and R, the Gravitational constant can be measured by Cavendish’s experiment.

Complete step by step solution:
We know that the acceleration due to gravity of earth is,
\[g = \dfrac{{G{M_E}}}{{{R^2}}}\]where,
 G is the gravitational constant
R is the radius of the earth
G is the acceleration due to gravity
\[{M_E}\]is the mass of the earth
From the above equation, the mass of the earth can be written as,
  \[{M_E} = \dfrac{{g{R^2}}}{G}\]
Density is a measure of mass per volume.
   \[\rho = \dfrac{{{M_E}}}{{\dfrac{4}{3}\pi {R^3}}}\]
  \[{M_E} = \rho \dfrac{4}{3}\pi {R^3}\]
   \[\dfrac{{g{R^2}}}{G} = \rho \dfrac{4}{3}\pi {R^3}\]
   \[\rho = \dfrac{{3g}}{{4\pi GR}}\]
The above expression gives the relation between the acceleration due to gravity and mean density of the earth.

Hence the correct option is C.

Note: The value of acceleration due to gravity varies due with the altitude and depth from the surface of the earth. For an object, the value of acceleration due to gravity is less at an altitude h than at the surface of the earth. Also the value of g decreases with increase in the depth.