What will be the acceleration due to gravity at a depth d , where g...

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