Question

The average density of the earth in terms of g (acceleration due to gravity), G (Universal gravitational constant) and R (Radius of earth) is

• A. $\frac{3g}{4\pi GR}$
• B. $\frac{3g}{5\pi GR}$
• C. $\frac{3G}{4\pi gR}$
• D. $\frac{3G}{5\pi gR}$

Answer 1

The correct option is A

$\frac{3g}{4\pi GR}$

We know $g=\frac{GM}{R^2}$ where g (acceleration due to gravity), M (mass of earth), G (Universal gravitational constant) and R (Radius of earth)

$\Rightarrow M=\frac{g\times R^2}{G}\Rightarrow$ Density D = $\frac{\text{mass of earth}}{\text{volume of earth}}=\frac{g\times R^2}{G\times V_e}$
(Where $V_e$ is the volume of the earth)

$V_e$ = $\frac{4}{3}\pi R^3$
$\Rightarrow D=\frac{g\times R^2}{G\times \frac{4}{3}\pi R^3}=\frac{3g}{4\pi GR}$