Gravitation - Acceleration due to Gravity

At what height $h$ above the Earth's surface will the acceleration due to gravity be $25\%$ of its value on the surface? (Let $R$ be the radius of Earth)

Given $g' = \frac{g}{4}$. Using the formula $g' = g \left( \frac{R}{R+h} \right)^2$, we have: $\frac{g}{4} = g \left( \frac{R}{R+h} \right)^2$ $\frac{1}{4} = \left( \frac{R}{R+h} \right)^2$ Taking the square root on both sides: $\frac{1}{2} = \frac{R}{R+h}$ $R + h = 2R \implies h = R$

Since the value of $g$ decreases with the square of the distance from the center, to reduce $g$ to a quarter, the distance must be doubled. Thus, $h$ must equal the radius of the Earth.

Calculate the depth $d$ below the Earth's surface where the weight of a body becomes $\frac{1}{3}$ of its weight on the surface.

Weight $W = mg$. If weight becomes $\frac{1}{3}$, then $g' = \frac{g}{3}$. Using the depth formula: $g' = g \left( 1 - \frac{d}{R} \right)$ $\frac{g}{3} = g \left( 1 - \frac{d}{R} \right)$ $\frac{1}{3} = 1 - \frac{d}{R}$ $\frac{d}{R} = 1 - \frac{1}{3} = \frac{2}{3}$ $d = \frac{2R}{3}$

Acceleration due to gravity decreases linearly with depth. At a depth of two-thirds the radius, $g$ is reduced by two-thirds, leaving only one-third of the surface value.