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AP EAMCET · PHYSICS · Gravitation

The acceleration due to gravity at a height of \((\sqrt{2}-1) \mathrm{R}\) from the surface of the earth is
(Acceleration due to gravity on the surface of the earth \(=10 \mathrm{~m} \mathrm{~s}^{-2}\) and R is radius of the earth)

  1. A \(2.5 \mathrm{~m} \mathrm{~s}^{-2}\)
  2. B \(7.5 \mathrm{~m} \mathrm{~s}^{-2}\)
  3. C \(5 \mathrm{~m} \mathrm{~s}^{-2}\)
  4. D \(10 \mathrm{~m} \mathrm{~s}^{-2}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(5 \mathrm{~m} \mathrm{~s}^{-2}\)

Step-by-step Solution

Detailed explanation

\(R+h = R+(\sqrt{2}-1)R = R\sqrt{2}\) \(g_h = g \left( \frac{R}{R+h} \right)^2 = 10 \left( \frac{R}{R\sqrt{2}} \right)^2\) \(g_h = 10 \left( \frac{1}{\sqrt{2}} \right)^2 = 10 \left( \frac{1}{2} \right)\) \(g_h = 5 \, \mathrm{m} \, \mathrm{s}^{-2}\)
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