(B) The acceleration due to gravity at a depth $d$ is given by the formula: $g_d = g\left(1 - \frac{d}{R}\right)$.Given that $g_d = \frac{g}{n-1}$,we

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

(B) The acceleration due to gravity at a depth $d$ is given by the formula: $g_d = g\left(1 - \frac{d}{R}\right)$.Given that $g_d = \frac{g}{n-1}$,we substitute this into the equation:$\frac{g}{n-1} = g\left(1 - \frac{d}{R}\right)$.Dividing both sides by $g$:$\frac{1}{n-1} = 1 - \frac{d}{R}$.Rearranging the terms to solve for $d$:$\frac{d}{R} = 1 - \frac{1}{n-1}$.$\frac{d}{R} = \frac{n-1-1}{n-1} = \frac{n-2}{n-1}$.Therefore,$d = R\left(\frac{n-2}{n-1}\right)$.

Class 11 Physics Gravitation Acceleration Due to Gravity and its Variation

Medium

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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A
$R\left(\frac{n}{n-1}\right)$
B
$R\left(\frac{n-2}{n-1}\right)$
C
$R\left(\frac{2n-1}{n}\right)$
D
$R\left(\frac{n-1}{2n-1}\right)$

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Acceleration Due to Gravity and its Variation

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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).

Similar Questions

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

$A$ body weighs $250\,N$ on the surface of the earth. How much will it weigh halfway down to the center of the earth (in $,N$)?

The mass of the moon is $\frac{1}{81}$ of the earth,but the gravitational pull (acceleration due to gravity) is $\frac{1}{6}$ of the earth. This is due to the fact that:

The angular speed of the Earth in $rad/s$,so that bodies on the equator may appear weightless is: [Use $g = 10\, m/s^2$ and the radius of the Earth $R = 6.4 \times 10^3\, km$]

What should be the angular speed with which the Earth must rotate on its axis so that a person on the equator would weigh $\frac{3}{5}$ th as much as their present weight?

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