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)
- A \(2.5 \mathrm{~m} \mathrm{~s}^{-2}\)
- B \(7.5 \mathrm{~m} \mathrm{~s}^{-2}\)
- C \(5 \mathrm{~m} \mathrm{~s}^{-2}\)
- D \(10 \mathrm{~m} \mathrm{~s}^{-2}\)
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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