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MHT CET · Physics · Gravitation

When the value of acceleration due to gravity ' \(\mathrm{g}\) ' become \(\frac{\mathrm{g}}{3}\) above surface of earth at height ' \(h\) ' then relation between ' \(h\) ' and ' \(\mathrm{R}\) ' is \((\mathrm{R}=\text { radius of earth) }\)

  1. A \(\mathrm{h}=\frac{\mathrm{R}}{\sqrt{3}-1}\)
  2. B \(\mathrm{h}=\frac{\sqrt{3}}{\mathrm{R}}\)
  3. C \(\mathrm{h}=(\sqrt{2}-1) \mathrm{R}\)
  4. D \(\mathrm{h}=(\sqrt{3}-1) \mathrm{R}\)
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Answer & Solution

Correct Answer

(D) \(\mathrm{h}=(\sqrt{3}-1) \mathrm{R}\)

Step-by-step Solution

Detailed explanation

\( \begin{aligned} & \mathrm{g}^{\prime}=\mathrm{g} \frac{\mathrm{R}^2}{(\mathrm{R}+\mathrm{h})^2}=\frac{\mathrm{g}}{3} \\ & \frac{1}{3}=\frac{\mathrm{R}^2}{(\mathrm{R}+\mathrm{h})^2} \\ & \frac{1}{\sqrt{3}}=\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}} \\ & \mathrm{h}=(\sqrt{3}-1) \mathrm{R} \end{aligned} \)
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