The height above the earth's surface at which the acceleration due to gravity becomes…

The height above the earth's surface at which the acceleration due to gravity becomes $\left(\frac{1}{n}\right)$ times the value at the surface is ( $R=$ radius of earth)
  1. $\frac{R}{\sqrt{n}}$
  2. $\mathrm{R} \cdot \sqrt{\mathrm{n}}$
  3. $(\sqrt{\mathrm{n}}+1) \mathrm{R}$
  4. $(\sqrt{\mathrm{n}}-1) \mathrm{R}$

Solution

$\begin{array}{ll} & \text { Given: } \frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{1}{\mathrm{n}} \\ & \frac{1}{\mathrm{n}}=\frac{\mathrm{R}^2}{(\mathrm{R}+\mathrm{h})^2} \Rightarrow \frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}=\frac{1}{\sqrt{\mathrm{n}}} \\ \therefore \quad & \sqrt{\mathrm{n}} \mathrm{R}=\mathrm{R}+\mathrm{h} \\ \therefore \quad & \mathrm{R}(\sqrt{\mathrm{n}}-1)=\mathrm{h}\end{array}$

Asked in: MHT CET 2024 (02 May Shift 1)

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