Question

Derive an expression for the value of acceleration due to gravity decreases with the increase in height h

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Acceleration}\mathrm{due}\mathrm{to}\mathrm{gravity}\mathrm{on}\mathrm{the}\mathrm{surface}\mathrm{of}\mathrm{earth}: \\ \mathrm{g}=\frac{\mathrm{GM}}{{\mathrm{R}}^2} \\ \mathrm{Acceleration}\mathrm{due}\mathrm{to}\mathrm{gravity}\mathrm{at}\mathrm{some}\mathrm{altitude}\text{'}\mathrm{h}\text{'}\mathrm{from}\mathrm{the}\mathrm{surface}\mathrm{of}\mathrm{earth}: \\ \mathrm{g}\text{'}=\frac{\mathrm{GM}}{{\left(\mathrm{R}+\mathrm{h}\right)}^2} \\ \frac{\mathrm{g}\text{'}}{\mathrm{g}}=\frac{\frac{\mathrm{GM}}{{\left(\mathrm{R}+\mathrm{h}\right)}^2}}{\frac{\mathrm{GM}}{{\mathrm{R}}^2}} \\ ={\left(\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}\right)}^2 \\ ={\left(\frac{\mathrm{R}+\mathrm{h}}{\mathrm{R}}\right)}^{-2} \\ ={\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)}^{-2} \\ =1-\frac{2\mathrm{h}}{\mathrm{R}}(\mathrm{by}\mathrm{neglecting}\mathrm{higher}\mathrm{terms}\mathrm{in}\mathrm{binomial}\mathrm{expansion}\mathrm{when}\mathrm{h}<<\mathrm{R}) \\ \mathrm{Hence}, \\ \mathrm{g}\text{'}=\mathrm{g}\left(1-\frac{2\mathrm{h}}{\mathrm{R}}\right) \\ \mathbf{Case}\mathbf{}\mathbf{I}\mathbf{:}\mathbf{}\mathbf{When}\mathbf{}\mathbf{h}\mathbf{}\mathbf{is}\mathbf{}\mathbf{considerable}\mathbf{}\mathbf{compared}\mathbf{}\mathbf{to}\mathbf{}\mathbf{R} \\ \mathrm{g}\text{'}=\mathrm{g}{\left(\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}\right)}^2 \\ \mathbf{Case}\mathbf{}\mathbf{II}\mathbf{:}\mathbf{}\mathbf{When}\mathbf{}\mathbf{h}\mathbf{}\mathbf{is}\mathbf{}\mathbf{not}\mathbf{}\mathbf{considerable}\mathbf{}\mathbf{compared}\mathbf{}\mathbf{to}\mathbf{}\mathbf{R} \\ \mathrm{g}\text{'}=\mathrm{g}\left(1-\frac{2\mathrm{h}}{\mathrm{R}}\right) \end{gathered}$$
This is the required relation between g and h which indicates that the acceleration due to gravity decreases with increasing altitudes.

Regards.