Question
Obtain the formula for acceleration due to gravity at the depth ‘d’ below the Earth’s surface.
✓
Answer
- The Earth can be considered to be a sphere made of large number of concentric uniform spherical shells.
- When an object is on the surface of the Earth it experiences the gravitational force as if the entire mass of the Earth is concentrated at its centre.
- The acceleration due to gravity on the surface of the Earth is, $g=\frac{G M}{R^2}$
- Assuming that the density of the Earth is uniform, mass of the Earth is given by $M=$ volume $x$ density $=\frac{4}{3} \pi R^3 \rho$ $\therefore g =\frac{ G \times \frac{4}{3} \pi R ^3 \rho}{ R ^2}=\frac{4}{3} \pi R \rho G$
- Consider a body at a point P at the depth d below the surface of the Earth as shown in figure.
Here the force on a body at P due to outer spherical shell shown by shaded region, cancel out due to symmetry.
The net force on P is only due to the inner sphere of radius OP = R – d. - Acceleration due to gravity because of this sphere is,
$g_d=\frac{ GM ^{\prime}}{( R - d )^2}$
where,
$ M ^{\prime}=\text { volume of the inner sphere } \times \text { density }$
$\therefore \quad M^{\prime}=\frac{4}{3} \pi(R-d)^3 \times \rho$
$\therefore \quad g_d=\frac{G \times \frac{4}{3} \pi(R-d)^3 \rho}{(R-d)^2}$
$\therefore \quad g_d=G \times \frac{4}{3} \pi(R-d) \rho$ - Dividing equation (2) by equation (1) we get,
$ \frac{g_{ d }}{ g } =\frac{ R - d }{ R }$
$\therefore \frac{ g _{ d }}{ g } =1-\frac{ d }{ R }$
$\therefore g_d = g \left(1-\frac{ d }{ R }\right) $
This equation gives acceleration due to gravity at depth d below the Earth's surface.
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