Question 366: Gravitation

Question

The depth at which acceleration due to gravity becomes $$\frac{\mathrm{g}}{\mathrm{n}}$$ is [ $$\mathrm{R}$$ = radius of earth, $$\mathrm{g}=$$ acceleration due to gravity, $$\mathrm{n}=$$ integer $$]$$

Accepted Answer

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

Answer

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

Answer

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

Answer

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

Answer

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

Gravitation

The depth at which acceleration due to gravity becomes $\frac{\mathrm{g}}{\mathrm{n}} is [ \mathrm{R} = radius of earth, \mathrm{g}= acceleration due to gravity, \mathrm{n}= integer ]

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