Question

The average density of the earth in terms of $g,G$ and $R$ is:
A. $\dfrac{{4\pi GR}}{{3g}}$
B. $\dfrac{{3g}}{{4\pi GR}}$
C. $\dfrac{{3g}}{{4\pi G{R^2}}}$
D. $\dfrac{{4\pi G{R^2}}}{{3g}}$

Answer 1

Hint: In this question we will use the newton’s second law i.e. F=ma , or force is equal to mass times acceleration and law of gravitation on the surface of earth i.e. equal to $9.8m/{s^2}$ .
Formula used: By newton’s second law and universal law of gravitation on the surface of earth, i.e. $mg = \dfrac{{GMm}}{{{R^2}}}$ $ \Rightarrow g = \dfrac{{GM}}{{{R^2}}}$

Complete Step-by-Step solution:
According to the question we have to find the average density of the earth in terms of $g,G$ and $R$.
Hence, it is given that $g$ is gravitational constant :
$g = \dfrac{{GM}}{{{R^2}}}\;or\;M = \dfrac{{g \times {R^2}}}{G}\; \Rightarrow \;Density\;D\; = \dfrac{{mass}}{{volume}}\; = \dfrac{{g \times {R^2}}}{{G \times {V_e}}}$
Where ${V_e}$ is the volume of the earth
$D = \dfrac{{g \times {R^2}}}{{G \times \dfrac{4}{3}\pi {R^3}}} = \dfrac{{3g}}{{4\pi GR}}$
Hence, the average density of the earth in terms of $g,G$ and $R$ is $\dfrac{{3g}}{{4\pi GR}}$.

Note: Acceleration due to gravity is the acceleration of a body falling freely under the influence of the Earth’s gravitational pull at sea level. It is approximately equal to $9.8m/{s^2}$. Its measured value varies slightly with latitude (due to rotation of earth) and longitude (due to non-spherical shape of earth) and also with depth and height from the earth’s surface.