Question 296: Gravitation

Question

The depth 'd' at which the value of acceleration due to gravity becomes $\frac{1}{n-1}$ times the value at the earth's surface is ($R=$ radius of the earth)

Accepted Answer

$$\begin{aligned}
& g_d=g\left(1-\frac{d}{R}ight) \
& g\left(\frac{1}{n-1}ight)=g\left(1-\frac{d}{R}ight) \quad \ldots\left(	ext { Given: } g_d=g\left(\frac{1}{n-1}ight)ight. \
	herefore \quad & \frac{d}{R}=1-\left(\frac{1}{n-1}ight) \
	herefore \quad & d=R\left(\frac{n-2}{n-1}ight)
\end{aligned}$$

Answer

$R\left(\frac{n}{n-1}\right)$

Answer

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

Answer

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

Answer

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

Gravitation

The depth 'd' at which the value of acceleration due to gravity becomes \frac{1}{n-1} times the value at the earth's surface is (R= radius of the earth)

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