Assuming the earth to be a sphere of uniform mass density, how much would a body weigh halfway down to the centre of the earth if it weighed $250 N$ on the surface ?

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Solution

Let, $g_{d}$ be the acceleration due to gravity at depth $d$ from the ground.

We know that $\frac{g_{d}}{g} = (1 - \frac{d}{R})$

Here, $d = \frac{R}{2}$

$∴ \frac{g_{d}}{g} = ⎛ ⎜ ⎜ ⎜ ⎝ 1 - \frac{\frac{R}{2}}{R} ⎞ ⎟ ⎟ ⎟ ⎠ = \frac{1}{2}$

If $W$ be the weight of the body on the surface of Earth and $W_{d}$ be the weight of the body at depth $d$

then, $\frac{W_{d}}{W} = \frac{m g_{d}}{m g} = \frac{g_{d}}{g} = \frac{1}{2}$

or $W_{d} = \frac{1}{2} W = \frac{1}{2} \times 250 = 125 N$

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Newton's Law of Gravitation

Standard XII Physics

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Assuming the earth to be a sphere of uniform mass density, how much would a body weigh halfway down to the centre of the earth if it weighed 250 N on the surface ?

Assuming the earth to be a sphere of uniform mass density, how much would a body weigh halfway down to the centre of the earth if it weighed $250 N$ on the surface ?