MHT CET · Physics · Gravitation
The height above the surface of the earth where acceleration due to gravity becomes \(\left(\frac{g}{9}\right)\) is ( \(R\) is radius of the earth, \(g\) is acceleration due to gravity)
- A 2R
- B \(\frac{R}{3}\)
- C \(\frac{R}{\sqrt{2}}\)
- D \(\sqrt{2} R\)
Answer & Solution
Correct Answer
(A) 2R
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & g^{\prime}=\frac{g}{\left(1+\frac{h}{R}\right)^2} \\ & \frac{g}{9}=\frac{g}{\left(1+\frac{h}{R}\right)^2} \\ & h=2 R\end{aligned}\)
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