Question

Assuming the earth to be a sphere of uniform mass density, how much would a body weight (in N) half way down to the center of the earth if it weighed $250IN$ on the surface?

Answer 1

The acceleration due to gravity on the surface of earth is g. By going below the surface of earth acceleration due to gravitational force decreases because of reduction in depth. And weight of a body is magnitude of multiplication of mass (m) and acceleration on the body due to gravitational force i.e. mg. Half way down means at $depth(d) = \dfrac{{radius(R)}}{2}$.

Complete step by step solution:
Here it is given that the weight of the body is $250N$.
$W = mg = 250N$
Acceleration (${g'}$) due to gravity at depth d is given by-
${g'} = \left( {1 - \dfrac{d}{R}} \right) \times g$
$\Rightarrow$ ${g'} = (1 - \dfrac{{R/2}}{R}) \times g$
$\Rightarrow$ ${g'} = (1 - \dfrac{1}{2}) \times g$
$\Rightarrow {g'} = \dfrac{g}{2}$
So gravitational acceleration on the body at depth $d = \dfrac{R}{2}$ is $\dfrac{g}{2}$
Weight (${W'}$) of the body at depth d is-
${W'} = mg'$
$\Rightarrow {W'} = m \times \dfrac{g}{2}$
$\Rightarrow {W'} = \dfrac{{250}}{2} = 125N$

Note: The above formula of acceleration due to gravitational force is only valid for a depth below the earth surface not for a height or above the surface. If we want to find the gravitational acceleration at a height above the surface then we can find using the formula –
${g'} = \left( {1 - \dfrac{{2h}}{R}} \right) \times g$ here, $h$ is height above the earth surface.