Question 322: Gravitation

Question

A mine is located at depth $$R / 3$$ below earth's surface. The acceleration due to gravity at that depth in mine is ($$R=$$ radius of earth, $$g=$$ acceleration due to gravity)

Accepted Answer

Here, $$g^{\prime}$$ be the acceleration due to gravity in mine.
Then, $$g^{\prime}=g\left(1-\frac{d}{R}ight)$$ ...... (i)
Here,
$$\begin{aligned}
d(	ext { depth }) & =\frac{R}{3} \
R & =	ext { Radius of earth }
\end{aligned}$$
Now, from Eq. (i), we get
$$\begin{aligned}
\Rightarrow \quad g^{\prime} & =g\left(1-\frac{R}{3 	imes R}ight) \
g^{\prime} & =\frac{2 g}{3}
\end{aligned}$$

Answer

$$g$$

Answer

$$3 g$$

Answer

$$\frac{2 g}{3}$$

Answer

$$\frac{g}{3}$$

Gravitation

A mine is located at depth $R / 3 below earth's surface. The acceleration due to gravity at that depth in mine is (R= radius of earth, g=$ acceleration due to gravity)

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