Physics - Gravitation Question with Solution | TestHub

At a depth $d$ below the surface of the earth, $$\begin{array}{rl}& m{g}^{\prime }=G\frac{M^{\prime }m}{(R-d{)}^2}\\ & M^{\prime }=\frac{4}{3}\pi (R-d{)}^3\rho \end{array}$$ $$\begin{array}{rl}\Rightarrow g^{\prime }& =\frac{G{M}^{\prime }}{(R-d{)}^2}\\ & =\frac{4}{3}G\pi (R-d)\rho \end{array}$$ If $d=\frac{R}{2}$ then $$g^{\prime }=\frac{4}{3}G\pi \rho =\frac{2\pi GR\rho }{3}$$ On the surface of the earth $$\begin{array}{rl}g& =\frac{Gm}{RV}=\frac{4}{3}\pi GR\rho \\ \therefore g^{\prime }& =\frac{g}{2}\end{array}$$ $\therefore$ The body weighed $250\mathrm{N}$ on the surface of the earth would weigh $\frac{1}{2}\times 250=125\mathrm{N}$, half way down towards the centre of the earth.