Question

The density of earth in terms of acceleration due to gravity(g), radius of the earth (R) and universal gravitational constant (G) is:
$\begin{aligned} & \text{a) }\dfrac{\text{4 }\\!\\!\pi\\!\\!\text{ RG}}{\text{3g}} \\\ & \text{b)}\dfrac{\text{3 }\\!\\!\pi\\!\\!\text{ RG}}{\text{4g}} \\\ & \text{c)}\dfrac{\text{4G}}{\text{3 }\\!\\!\pi\\!\\!\text{ RG}} \\\ & \text{d)}\dfrac{\text{3g}}{\text{4 }\\!\\!\pi\\!\\!\text{ RG}} \\\ \end{aligned}$

Answer 1

The density of any substance is given by the ratio of mass of the body to its volume. To solve the above question we have to consider earth be sphere of uniform mass M. Further we can substitute the relation between acceleration due to gravity near the surface of the earth, in terms the average radius of the earth and further substitute the result and replace the possible terms in the definition of density and obtain the required answer.

Complete step-by-step answer:
Let us consider Earth to be a sphere of radius R, of uniform mass M and average density of Earth be denoted by $\text{ }\\!\\!\rho\\!\\!\text{ }$.
The acceleration due to gravity near the surface of the Earth is equal to i.e. $g=\dfrac{GM}{{{R}^{2}}}m{{s}^{-2}}....(1)$ where M is the mass of the earth only, R is the average radius of the Earth, and G is the gravitational constant.
From equation 1 we can write the mass of the earth as $M=\dfrac{g{{R}^{2}}}{G}kg...(2)$.The volume of the Earth is equal to $\text{V=}\dfrac{\text{4}}{\text{3}}\text{ }\\!\\!\pi\\!\\!\text{ }{{\text{R}}^{\text{3}}}\text{ }{{\text{m}}^{\text{3}}}...(3)$ .
The average density of the earth is the ratio of mass of the earth to its volume. Hence from equation 1 and 2 we get the density of the Earth as,
$\text{ }\\!\\!\rho\\!\\!\text{ }=\dfrac{\text{Mass of the Earth}}{\text{Volume of the Earth}}\text{kg}{{\text{m}}^{\text{-3}}}$
$\text{ }\\!\\!\rho\\!\\!\text{ }=\dfrac{\dfrac{g{{R}^{2}}}{G}}{\dfrac{\text{4}}{\text{3}}\text{ }\\!\\!\pi\\!\\!\text{ }{{\text{R}}^{\text{3}}}}$
$\text{ }\\!\\!\rho\\!\\!\text{ }=\dfrac{3g}{4G\text{ }\\!\\!\pi\\!\\!\text{ R}}kg{{m}^{-3}}$

So, the correct answer is “Option d”.

Note: We have assumed that the mass of the Earth to be uniform and at the same time the radius of the earth to be constant. Hence we cannot say the above density of the earth is the actual density of the Earth. The actual density of earth will be somewhat the same as the result we obtained.