A body weighs 250,N on the surface of the ear | 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 $g' = g(1 - \frac{d}{R})$,where $g$ is the a

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(A) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula $g' = g(1 - \frac{d}{R})$,where $g$ is the acceleration due to gravity at the surface,$R$ is the radius of the earth,and $d$ is the depth.Given that the body is halfway down to the center of the earth,the depth $d = \frac{R}{2}$.Substituting this value into the formula,we get $g' = g(1 - \frac{R/2}{R}) = g(1 - \frac{1}{2}) = \frac{g}{2}$.The weight of the body at depth $d$ is $W' = mg' = m(\frac{g}{2}) = \frac{mg}{2}$.Since the weight at the surface is $W = mg = 250\,N$,the weight at half the depth is $W' = \frac{250}{2} = 125\,N$.

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

Similar Questions

The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh $\frac{1}{6}$ th as much as at present is ($g=$ gravitational acceleration,$R=$ equatorial radius of the earth).

$A$ spherical planet far out in space has a mass $M_0$ and diameter $D_0$. $A$ particle of mass $m$ falling freely near the surface of this planet will experience an acceleration due to gravity which is equal to

If the value of $g$ at the surface of the earth is $9.8 \, m/s^2$,then the value of $g$ at a place $480 \, km$ above the surface of the earth will be .......... $m/s^2$. (Radius of the earth is $6400 \, km$)

How does the rotation of the Earth affect the acceleration due to gravity on its surface?

Vedclass Test Series

Exam Paper Generator

Online Exam Module