Question

How the value of g varies with height.

Answer 1

Let, $A$ be the point on the surface of the Earth and $B$ be the point at un ultitude $h$, let, $M$ be the man of the Earth and $R$ be the radius of the Earth.

Consider the Earth as a spherical body. The acceleration on due to gravity at point $A$ on the surface of the Earth is

$y=\frac{CM}{R^2}⟶\left(1\right)$

let, the body be placed at $B$ at a height $h$ from the surface of the Earth.

The acceleration due to gravity at $B$ is,

$g^1=\frac{CM}{{(R+h)}^2}⟶\left(2\right)$

Dividing equation $\left(1\right)$ by $\left(2\right)$ we get,

$\frac{g}{g^1}=\frac{{CM/R}^2}{\left[CM/{(R+h)}^2\right]}$

$\frac{g}{g^1}=\frac{CM\times {(R+h)}^2}{CM\times R^2}$

$\frac{g}{g^1}=\frac{{(R+h)}^2}{R^2}$

$\frac{g}{g^1}={(\frac{R}{R}+\frac{h}{R})}^2$

or, $\frac{g}{g^1}={(1+\frac{h}{R})}^2$

$g^1=g{(1+\frac{h}{R})}^{-2}$

Now, expanding by using binomial theorem, we get

$g^1=g(1-\frac{2h}{R})$

The value of acceleration due to gravity decreases with increase in height above the surface of the Earth.