Question

Weight of a body on earth’s surface is W. At a depth halfway to the centre of the earth, it will be (assuming uniform density in earth)
A. $W$
B. $\dfrac{W}{2}$
C. $\dfrac{W}{4}$
D. $\dfrac{W}{8}$

Answer 1

Hint: We could recall the expression that represents the variation of acceleration due to gravity with depth d. As we are discussing about the point which is halfway to the centre of earth, you could substitute $\dfrac{R}{2}$ in place of d. Multiplying both side with mass m of the body you could compare the body’s weight at surface and depth d and thus get the answer.
Formula used:
Variation of g with depth,
$g'=g\left( 1-\dfrac{d}{R} \right)$

Complete step by step solution:
We are given the weight of the body on the surface of the Earth as W. Assuming the density of Earth to be uniform, we are asked to find the weight of the same body at a depth of half way down the centre of the Earth.
Firstly, let us recall that the acceleration of gravity is position dependent so it varies with depth as well as well as height from the surface of earth. Its variation depth from the surface of earth is given by,
$g'=g\left( 1-\dfrac{d}{R} \right)$ ……………………………… (1)
Where, d is the depth at which g’ is measured and R is the radius of Earth.
At a depth half way down the centre of Earth, the depth can be given by,
$d=\dfrac{R}{2}$
Substituting in (1), we get,
$g'=g\left( 1-\dfrac{\left( \dfrac{R}{2} \right)}{R} \right)$
$\Rightarrow g'=g\left( 1-\dfrac{1}{2} \right)$
$\therefore g'=\dfrac{g}{2}$
Multiplying both sides with mass m of the body,
$mg'=\dfrac{mg}{2}$
But we know that mg is the weight of the body on the surface of earth which is given by W and mg’ is the weight of the body at depth of half of the radius of earth. So,
$W'=\dfrac{W}{2}$
Therefore, at a depth halfway to the centre of the earth, the weight will be $\dfrac{W}{2}$ if W was the weight of the same body on the earth’s surface.

Hence option B is the right answer.

Note:
The variation of acceleration due to gravity with height is given by,
$g'=g\left( 1-\dfrac{2h}{R} \right)$
Where, h is the height from the earth’s surface at which the acceleration due to gravity is to be found. Also, on solving this question we found that though the weight of the body varies with depth the mass remains constant as mass is the measure of an object’s inertial property.