Question

How does the value of ‘g’ changes with height?

Answer 1

Value of ‘g’ depending on the change of height:

Let A be a point on the Earth's surface and B be a point at an altitude of h, M be the Earth's man, and R be the Earth's radius.

Consider the Earth to be a sphere. At point A on the Earth's surface, the acceleration due to gravity is:

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

Place the body at B, at a height of h above the earth's surface:

Acceleration due to gravity at B is,

${\mathrm{g}}^{\text{'}}=\frac{GM}{{(\mathrm{R}+\mathrm{h})}^2}$ ---------------- (2)

Dividing equations (1) and (2) we get:

$\frac{\mathrm{g}}{\mathrm{g}\text{'}}=\frac{{\displaystyle \raisebox{1ex}{$GM$}\!\left/ \!\raisebox{-1ex}{${\mathrm{R}}^2$}\right.}}{\left[{\displaystyle \raisebox{1ex}{$GM$}\!\left/ \!\raisebox{-1ex}{${(\mathrm{R}+\mathrm{h})}^2$}\right.}\right]}$

$=\frac{GM\times {(\mathrm{R}+\mathrm{h})}^2}{GM\times {\mathrm{R}}^2}$

$=\frac{{\left(\mathrm{R}+\mathrm{h}\right)}^2}{{\mathrm{R}}^2}$

$={\left(\frac{\mathrm{R}}{\mathrm{R}}+\frac{\mathrm{h}}{\mathrm{R}}\right)}^2$

$\mathrm{g}\text{'}$ $=$$\mathrm{g}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)}^{-2}$

Hence, with an increasing height above the Earth's surface, the value of acceleration due to gravity decreases.