JEE Main & Advanced Physics Gravitation / गुरुत्वाकर्षण Variation in g With Depth

Acceleration due to gravity at the surface of the earth

\[g=\frac{GM}{{{R}^{2}}}=\frac{4}{3}\pi \rho GR\]                      ...(i)

Acceleration due to gravity at depth d from the surface of the earth

\[{g}'=\frac{4}{3}\pi \rho G(R-d)\]                         ...(ii)

From (i) and (ii)  \[{g}'=g\left[ 1-\frac{d}{R} \right]\]

(i) The value of g decreases on going below the surface of the earth. From equation (ii) we get \[{g}'\propto (R-d)\].

So it is clear that if d increase, the value of g decreases.

(ii) At the centre of earth \[\frac{g\,{{R}^{2}}{{T}^{2}}}{4{{\pi }^{2}}}={{\left( R+h \right)}^{3}}\] \[\therefore \ \ {g}'=0\], i.e., the acceleration due to gravity at the centre of earth becomes zero.

(iii) Decrease in the value of g with depth

Absolute decrease \[\Delta g=g-{g}'=\frac{dg}{R}\]

Fractional decrease \[\frac{\Delta g}{g}=\frac{g-{g}'}{g}=\frac{d}{R}\]

Percentage decrease \[\frac{\Delta g}{g}\times 100%=\frac{d}{R}\times 100%\]

(iv) The rate of decrease of gravity outside the earth (\[\text{if}\,\ h<<R\]) is double to that of inside the earth.