Question

Acceleration due to gravity $g$ and the mean density $\rho$ of the Earth are related by the relation:
$[G$ is the gravitational constant and $R_{e}$ is the radius of the Earth]

• A. $\frac{g}{G} \frac{4 \pi}{3} R_{e}^{3}=\rho$
• B. $\frac{\frac{g}{G}}{\frac{4 \pi}{3} R_{e}}=\rho$
• C. $\frac{g}{G} \frac{4 \pi}{3} R_{e}^{2}=\rho$
• D. $\frac{\frac{g}{G}}{\frac{4 \pi}{3} R_{e}^{3}}=\rho$

Answer 1

Answer: (B)

$g=\frac{G M}{R_{e}^{2}}=\frac{G}{R_{e}^{2}}\left(\frac{4}{3} \pi R_{e}^{3} \cdot \rho\right)$
$g=\frac{4}{3} \pi R_{e} \cdot G \cdot \rho \Rightarrow \rho=\frac{\frac{g}{G}}{\frac{4 \pi}{3} \cdot R_{e}}$