Question

A body weighed $250N$ on the surface assuming the earth to be a sphere of uniform mass density, how much would it weigh half way down to the centre of the earth?

• A. 195 N
• B. 240 N
• C. 125 N
• D. 210 N.

Answer 1

Answer: (C)

Given: Weight of the body on the earth’s surface $(W)=50N$ and depth $(d)=\frac{R}{2}$ . We know that weight of the body at a distance (d) from the surface of the earth = $W\left(1-\frac{d}{R}\right)=250\times \left(1-\frac{R/2}{R}\right)=250\times \frac{1}{2}=125N$
Aliter : Given: Weight of the body on the surface of the earth
$mg=250N$

When we move down a distance R/2 towards the earth’s centre, the value of acceleration due to gravity decreases. First let’s calculate the value of acceleration due to gratvity at a depth R/2 below the surface. What we have to remember here is that the whole mass of the earth is not going to be effective at a depth of R/2. Let, $\rho$ be the uniform mass density of the earth. Then the effective mass of earth at a depth R/2 below is
$M^{\prime }=\frac{4}{3}\pi {\left(R/2\right)}^3\times \rho =\frac{4}{3}\pi R^3\rho \times \frac{1}{8}=\frac{M}{8}$
Where M = mass of earth on the surface
Now $g^{\prime }=\frac{G{M}^{\prime }}{{\left(\frac{R}{2}\right)}^2}=4\frac{G{M}^{\prime }}{R^2}=4\times \frac{GM}{8R^2}=\frac{g}{2}$
$\Rightarrow m{g}^{\prime }=\frac{mg}{2}=\frac{250}{2}=125N$