Assuming the earth to be a sphere of uniform mass density, a body weighed \(300 \ N\) on the surface of earth. How much …

For a uniform-density Earth, acceleration due to gravity at a depth d is

g_d=g\left(1-\frac{d}{R}\right)

Given:

\(d=\frac{R}{4}\)

Therefore,

\({g}_{d}=g(1-\frac{1}{4})\) \({g}_{d}=\frac{3}{4}g\)

Weight is proportional to g.

\({W}_{d}=300\times \frac{3}{4}\) \({W}_{d}=225\ \text{N}\)