Question

Find the acceleration due to gravity 'g' in terms of mean density of the Earth 'd'. Assume it to be a perfect sphere.

• A. $4\pi r^{2}dG$​
• B. $\frac{4\pi r^2}{G}$
• C. $\frac{3}{4\pi RdG}$
• D. $\frac{4\pi RdG}{3}$

Answer 1

The correct option is D $\frac{4\pi RdG}{3}$

Acceleration due to gravity is given by:
$g=\frac{GM}{R^2}--\left(1\right)$
where
$M:$ Mass of Earth
$R:$ Radius of Earth
$G:$ Gravitational constant

Density is the ratio of mass and volume.
$d=\frac{M}{V}$
$M=dV$

Volume of sphere,
$V=\frac{4}{3}\pi R^3$

$⟹M=d\times \frac{4}{3}\pi R^3-\left(2\right)$

Substituting equation (2) in (1):
$g=\frac{GM}{R^2}$
$g=G\left(\frac{d\times 4\pi R^3/3}{R^2}\right)$
$g=\frac{4\pi RdG}{3}$