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Q.
A body weighed $250\, N$ 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?
Given: Weight of the body on the earth’s surface $(W) = 50 \,N $ 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)=\quad250\times\left(1-\frac{R / 2}{R}\right)=250\times\frac{1}{2}=125\,N$
Aliter : Given: Weight of the body on the surface of the earth
$mg = 250 \,N$
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' = \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 $g' = \frac{GM'}{\left(\frac{R}{2}\right)^{2}}=4 \frac{GM'}{R^{2}}=4\times\frac{GM}{8R^{2}}=\frac{g}{2}$
$\Rightarrow \, mg'=\frac{mg}{2}=\frac{250}{2}=125 \,N$
