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MHT CET · Physics · Gravitation

The height ' \(h\) ' above the earth's surface at which the value of acceleration due to gravity ( g ) becomes \(\left(\frac{\mathrm{g}}{3}\right)\) is ( \(\mathrm{R}=\) radius of the earth)

  1. A \((\sqrt{3}+1) \mathrm{R}\)
  2. B \((\sqrt{3}-1) R\)
  3. C \(\sqrt{3} R\)
  4. D \(3 \sqrt{R}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \((\sqrt{3}-1) R\)

Step-by-step Solution

Detailed explanation

We know that,
\(\begin{aligned}
& \mathrm{g}_{\mathrm{h}}=\mathrm{g}\left(\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}\right)^2 \\
& \text { For } g_h=\frac{g}{3} \\
& \frac{g}{3}=g\left(\frac{R}{R+h}\right)^2 \\
& \frac{1}{\sqrt{3}}=\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}} \\
& \sqrt{3} \mathrm{R}=\mathrm{R}+\mathrm{h} \\
& \therefore \quad h=(\sqrt{3}-1) \mathrm{R}
\end{aligned}\)