COMEDK · Physics · 9. Gravitation
The radius of earth is \(R\) and acceleration due to gravity on its surface is \(g\). The height at which the acceleration due to gravity becomes \(\dfrac{g}{8}\) is:
- A \((2 \sqrt{2}-1) R\)
- B \(2 \sqrt{2} R\)
- C \(\sqrt{2} R\)
- D \(2 R\)
Answer & Solution
Correct Answer
(A) \((2 \sqrt{2}-1) R\)
Step-by-step Solution
Detailed explanation
The acceleration due to gravity at a height \(h\) above the surface of the Earth is given by the formula \(g' = g \left( \dfrac{R}{R+h} \right)^{2}\), where \(g\) is the acceleration due to gravity at the surface and \(R\) is the radius of the Earth. Given that…
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