MHT CET · Physics · Gravitation
At a height ' \(R\) ' above the earth's surface the gravitational acceleration is ( \(\mathrm{R}\) = radius of earth, \(\mathrm{g}\) = acceleration due to gravity on earth's surface)
- A g
- B \(\frac{\mathrm{g}}{8}\)
- C \(\frac{\mathrm{g}}{4}\)
- D \(\frac{\mathrm{g}}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{g}}{4}\)
Step-by-step Solution
Detailed explanation
Acceleration due to gravity is given by
\(
\begin{aligned}
& \mathrm{g}^{\prime}=\frac{\mathrm{GM}}{\mathrm{r}^2}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^2}=\frac{\mathrm{GM}}{(2 \mathrm{R})^2} \\
& =\frac{\mathrm{GM}}{4 \mathrm{R}^2}=\frac{\mathrm{g}}{4}
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{g}^{\prime}=\frac{\mathrm{GM}}{\mathrm{r}^2}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^2}=\frac{\mathrm{GM}}{(2 \mathrm{R})^2} \\
& =\frac{\mathrm{GM}}{4 \mathrm{R}^2}=\frac{\mathrm{g}}{4}
\end{aligned}
\)
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