Acceleration Due To Gravity Below And Above The Surface Of Earth

Consider a point mass m at a height h above the surface of the earth as shown in Fig.1. The radius of the earth is denoted by $R_E$. 

Fig.1: g at a height h above the surface of the earth.

Since this point is outside the earth, its distance from the centre of the earth is ($R_E + h$). If F(h) denoted the magnitude of the force on the point mass m , we get from Eq. $|\mathbf{F}|=G\frac{m_1m_2}{r^2}$:

$F(h)=G\frac{M_Em}{(R_E+h)^2}$     (1)   

The acceleration experienced by the point mass is $F_h/m$ ≡ g(h) and we get

$g(h)=\frac{F(h)}{m}=\frac{GM_E}{(R_E+h)^2}$   (2)

This is clearly less than the value of g on the surface of earth :

 $g=\frac{F(h)}{m}=\frac{GM_E}{R_E^2}$

For , h R << E we can expand the RHS of Eq. (2) :

$g(h)=\frac{GM_E}{R_E^2(1+h/R_E)^2}=g(1+h/R_E)^2$

For $\frac{h}{R_E}<<1$, using binomial expression,

$g(h)\cong \left(1-\frac{h}{R_E}\right)$    (3)

Equation (3) thus tells us that for small heights h above the value of g decreases by a factor $\left(1-\frac{h}{R_E}\right)$.

Fig. 2: g at a depth d. In this case only the smaller sphere of radius ($R_E-d$) contributes to g.

Now, consider a point mass m at a depth d below the surface of the earth (Fig. 2), so that its distance from the centre of the earth is ($R_E-d$) 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,

$\frac{M_s}{M_E}=\frac{(R_E-d)^3}{R_E^3}$     (4)

Since mass of a sphere is proportional to be cube of its radius.

Thus the force on the point mass is

$F(d)=\frac{GM_sm}{(R_E-d)^2}$      (5)

Substituting for $M_s$ from above, we get

$F(d)=\frac{GM_Em(R_E-d)}{R_E^3}$     (6)

and hence the acceleration due to gravity at a depth d

$g(d)=\frac{F(d)}{m}$ is

$g(d)=\frac{GM_E(R_E-d)}{R_E^3}$

$g(d)=g\left(\frac{R_E-d}{R_E}\right)=g\left(1-\frac{d}{R_E}\right)$      (7)

Thus, as we go down below earth’s surface, the acceleration due gravity decreases by a factor ($1-d/R_E$) 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.

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