Gravitation

VARIATION OF g WITH DEPTH

Let us assume that the earth to be a homogeneous uniform sphere of radius ‘R’,density \(\rho\) and mass ‘M’Now
                        
 

Let g  be the  acceleration due to gravity at a place on the surface of the earth.When a body is on the surface of the earth,its entire mass M attracts that body towards its centre.We know that
     \( g = \frac{{Gm}} {{R^2 }} \)---------(1)
Now consider a point mass m at a depth d from the surface of the earth.The outer shell of thickness d which is exterior to the body exerts no resultant gravitational force on the body.The force on the body will only be due to the mass of the earth confined in the inner solid sphere of radius (R-d).The mass of inner solid sphere is m_s 
then   \( \frac{{M_s }} {M} = \frac{{\left( {R - d} \right)^3 }} {{(R)^3 }}(or)M_s = \frac{{M\left( {R - d} \right)^3 }} {{(R)^3 }} \) 
The gravitational force on the point mass is
        \( F = \frac{{GM_s m}} {{(R - d)^2 }} = \frac{{GM}} {{R^3 }}\frac{{\left( {R - d} \right)^3 m}} {{(R - d)^2 }} = \frac{{GMm(R - d)}} {{R^3 }} \)  The acceleration due to gravity is the force for the unit mass and let it be g_d at the depth d.
       \( \therefore g_d = \frac{F} {m} = \frac{{GM}} {{R^3 }}(R - d) \) ----------(2)
From Eqs.1 and 2 we get
         \( \frac{{g_d }} {g} = \frac{{R - d}} {R} \Rightarrow g_d = g\left[ {1 - \frac{d} {R}} \right] \) ---(3)

from the above equation it is clear that the value of ‘g’ decreases with the depth .
Eq.3 gives the variation of ‘g’ with depth.

i) ‘g’ decreases as depth d increases
ii)’g’ becomes zero at the centre of the earth
iii)’g’ is maximum on the surface of the earth and it decreases with height and depth
iv)Decrease in ’g’ at small heights is more than the decrease in ‘g’ at small objects
v)at large heights decreases in ‘g’ is less than the decrease in ‘g’ at large depths.