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

(D) The gravitational acceleration at a depth $d$ below the Earth's surface is given by the formula: $g_d = g \left(1 - \frac{d}{R}\right)$.Given that

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The gravitational acceleration at a depth $d$ below the Earth's surface is given by the formula: $g_d = g \left(1 - \frac{d}{R}\right)$.Given that the acceleration due to gravity at depth $d$ is $g_d = \frac{g}{2n}$.Substituting this into the formula:$\frac{g}{2n} = g \left(1 - \frac{d}{R}\right)$.Dividing both sides by $g$:$\frac{1}{2n} = 1 - \frac{d}{R}$.Rearranging the terms to solve for $d$:$\frac{d}{R} = 1 - \frac{1}{2n}$.$\frac{d}{R} = \frac{2n - 1}{2n}$.Therefore,$d = R \left(\frac{2n - 1}{2n}\right)$.

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).

Similar Questions

The weight of a body on the surface of the earth is $63 \ N$. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth? (in $N$)

The angular speed of the Earth,such that an object on the equator may appear weightless,is ($g = 10\,m/s^2$,radius of the Earth $R = 6400\,km$).

The linear speed of a particle at the equator of the earth due to its spin motion is $V$. The linear speed of the particle at latitude $30^{\circ}$ is:

$A$ spring balance is graduated at sea level. If a body is weighed with this balance at consecutively increasing heights from the Earth's surface,the weight indicated by the balance:

What will be the acceleration due to gravity at a height $h$ above the Earth's surface,where $R$ is the radius of the Earth and $g$ is the acceleration due to gravity on the surface of the Earth?

Vedclass Test Series

Exam Paper Generator

Online Exam Module