Physics - Gravitation Question with Solution | TestHub

Acceleration due to gravity at the surface of earth, $$g=\frac{GM}{R^2}=\frac{4}{3}\pi \rho GR$$ $$\left[\because M=\frac{4}{3}\pi R^3\cdot \rho \right]$$ Where, $\rho =$ density of earth. Acceleration due to gravity at depth $d$ from the surface of earth, $$g^{\prime }=\frac{4}{3}\pi \rho G(R-d)$$ From Eqs. (i) and (ii), we get $$g^{\prime }=g\left[1-\frac{d}{R}\right]$$