. (b)), so that its distance from the centre of the earth is ( as shown in the figure. The earth can be thought of as being composed of a smaller sphere of radius ( R E – d ) and a spherical shell of thickness d . The force on m due to the outer shell of thickness d is zero because the result quoted in the previous section.
As far as the smaller sphere of radius ( R E – d ) is concerned, the point mass is outside it and hence according to the result quoted earlier, the force due to this smaller sphere is just as if the entire mass of the smaller sphere is concentrated at the centre. If M s is the mass of the smaller sphere, then, M s /M E = ( R E – d) / R E ( . ) Since mass of a sphere is proportional to be cube of its radius. Fig.
. (b) g at a depth d. In this case only the smaller sphere of radius (R E –d) contributes to g. Thus the force on the point mass is F ( d ) = G M s m / ( R E – d ) ( .
) Substituting for M s from above , we get F ( d ) = G M E m ( R E – d ) / R E ( . ) and hence the acceleration due to gravity at a depth d , g ( d ) = F d is F d g d ( ( . ) Thus, as we go down below earth’s surface, the acceleration due gravity decreases by a factor ( ). The remarkable thing about acceleration due to earth’s gravity is that it is maximum on its surface decreasing whether you go up or down.
. GRAVITATIONAL POTENTIAL ENERGY We had discussed earlier the notion of potential energy as being the energy stored in the body at its given position. If the position of the particle changes on account of forces acting on it, then the change