Question

Average density of the earth
(a) Does not depend on g
(b) Is a complex function of g
(c) Is directly proportional to g
(d) Is inversely proportional to g

Answer 1

Hint: in this question use the direct formula of the acceleration due to gravity that is ${g_e} = \dfrac{{GM}}{{{R_e}^2}}$, where G is the gravitational constant, M is the mass of the earth and ${R_e}$ is the radius of the earth. Use the relationship between the mass, density and volume that is $M = V \times \rho $. This will help establish the required relationship between density of the earth and the acceleration due to gravity.

Complete step-by-step answer:
As we know (g) is known as acceleration of gravity and the value of (g) is constant at the earth's surface or at sea level.
g = 9.8 ${m^2}$/sec (at earth surface).
But as we know g is dependent on the radius of the earth which is given as (i.e. gravitational force of the earth)
${g_e}$ = (GM/$R_e^2$)……………….. (1)
 where G = universal gravitational constant.
             M = mass of earth.
          ${R_e}$ = radius of the earth
Now as we know that the mass of the earth is the volume of the earth multiplied by the density of the earth ($\rho $).
$ \Rightarrow M = V \times \rho $
As we know earth is in the shape of a sphere so the volume (V) of the earth is $\dfrac{4}{3}\pi R_e^3$.
So the mass (M) of the earth is 
M = $\dfrac{4}{3}\pi R_e^3 \times \rho $
Now substitute this value in equation (1) we have,
${g_e}$ = (GM/$R_e^2$) = $\dfrac{{G \times \dfrac{4}{3}\pi R_e^3 \times \rho }}{{R_e^2}} = G \times \dfrac{4}{3}\pi {R_e} \times \rho $
Now calculate the density of the earth from the above equation we have,
$ \Rightarrow \rho = \dfrac{{{g_e}}}{{G \times \dfrac{4}{3}\pi {R_e}}}$
So as we see from the above equation that density of the earth is directly proportional to the earth gravitation (${g_e}$)

So this is the required answer.
Hence option (C) is the correct answer.

Note – There is often a confusion between g and G. g is the acceleration due to gravity whose value is 9.8 at the surface of the earth however G is the proportionality constant and has a default value of $6.674 \times {10^{ - 11}}{m^3}K{g^{ - 1}}{s^{ - 2}}$. It is advised to remember the direct formula for the force of gravitation between two masses that is ${F'_g} = G\dfrac{{{m_3}{m_4}}}{{{d^2}}}$ as it is very helpful while dealing with forces between two bodies and has involvement of G in it as well.