Question 327: Gravitation

Question

The depth at which acceleration due to gravity becomes $$\frac{\mathrm{g}}{2 \mathrm{n}}$$ is $$(\mathrm{R}=$$ radius of earth, $$\mathrm{g}=$$ acceleration due to gravity on earth's surface, $$\mathrm{n}$$ is integer)

Accepted Answer

The gravitational acceleration at depth is given as $$g_d=g\left[1-\frac{d}{R}ight]$$
Given $$\mathrm{g}_{\mathrm{d}}=\frac{\mathrm{g}}{2 \mathrm{n}}$$
$$\begin{aligned}
	herefore \quad \frac{g}{2 n} & =g\left[1-\frac{d}{R}ight] \
& \frac{d}{R}=1-\frac{1}{2 n} \
d & =\left[\frac{2 n-1}{2 n}ight] R
\end{aligned}$$

Answer

$$\frac{\mathrm{R}(1-2 \mathrm{n})}{\mathrm{n}}$$

Answer

$$\frac{\mathrm{R}(1-\mathrm{n})}{2 \mathrm{n}}$$

Answer

$$\frac{\mathrm{R}(\mathrm{n}-1)}{\mathrm{n}}$$

Answer

$$\frac{\mathrm{R}(2 \mathrm{n}-1)}{2 \mathrm{n}}$$

Gravitation

The depth at which acceleration due to gravity becomes $\frac{\mathrm{g}}{2 \mathrm{n}} is (\mathrm{R}= radius of earth, \mathrm{g}= acceleration due to gravity on earth's surface, \mathrm{n}$ is integer)

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