Question 367: Gravitation

Question

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

Accepted Answer

$$\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}ight)$$
If $$g^{\prime}=\frac{g}{n}$$ then $$\frac{g}{n}=g\left(1-\frac{d}{R}ight)$$
$$\begin{aligned}
& 	herefore \frac{1}{\mathrm{n}}=1-\frac{\mathrm{d}}{\mathrm{R}} \
& 	herefore \frac{\mathrm{d}}{\mathrm{R}}=1-\frac{1}{\mathrm{n}}=\frac{\mathrm{n}-1}{\mathrm{n}} \quad 	herefore \quad \mathrm{d}=\mathrm{R}\left(\frac{\mathrm{n}-1}{\mathrm{n}}ight)
\end{aligned}$$

Answer

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

Answer

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

Answer

$$\frac{R}{n}$$

Answer

$$\frac{\mathrm{R}}{\mathrm{n}^2}$$

Gravitation

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

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