The depth at which acceleration due to gravit | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The acceleration due to gravity at a depth $d$ below the surface of the Earth is given by the formula: $g^{\prime} = g \left(1 - \frac{d}{R}\right

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The acceleration due to gravity at a depth $d$ below the surface of the Earth is given by the formula: $g^{\prime} = g \left(1 - \frac{d}{R}\right)$ Given that the acceleration due to gravity at depth $d$ is $\frac{g}{n}$,we substitute this into the equation: $\frac{g}{n} = g \left(1 - \frac{d}{R}\right)$ Dividing both sides by $g$: $\frac{1}{n} = 1 - \frac{d}{R}$ Rearranging the terms to solve for $d$: $\frac{d}{R} = 1 - \frac{1}{n}$ $\frac{d}{R} = \frac{n-1}{n}$ $d = \frac{R(n-1)}{n}$

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

Similar Questions

If acceleration due to gravity at distance $d < R$ from the centre of the Earth is $\beta$,then its value at distance $d$ above the surface of the Earth will be [where $R$ is the radius of the Earth].

$A$ body weighs $49 \, N$ on a spring balance at the North Pole. What will be its weight recorded on the same weighing machine if it is shifted to the equator (in $, N$)? (Use $g = \frac{GM}{R^2} = 9.8 \, m/s^2$ and radius of the Earth,$R = 6400 \, km$.)

At what distance above and below the surface of the earth will a body have the same weight? (Take the radius of the earth as $R$.)

Derive the equation for the variation of acceleration due to gravity $g$ with depth $d$ below the surface of the Earth.

Write the difference between $G$ and $g$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module