Acceleration Due To Gravity Of The Earth

The earth can be imagined to be a sphere made of a large number of concentric spherical shells with the smallest one at the centre and the largest one at its surface. A point outside the earth is obviously outside all the shells. Thus, all the shells exert a gravitational force at the point outside just as if their masses are concentrated at their common centre according to the result stated in section 8.3. The total mass of all the shells combined is just the mass of the earth. Hence, at a point outside the earth, the gravitational force is just as if its entire mass of the earth is concentrated at its centre.

For a point inside the earth, the situation is different. This is illustrated in Fig. 2.

Fig.1 The mass m is in a mine located at a depth d below the surface of the Earth of mass $M_E$ and radius $R_E$. We treat the Earth to be spherically symmetric.

Again consider the earth to be made up of concentric shells as before and a point mass m situated at a distance r from the centre. The point P lies outside the sphere of radius r. For the shells of radius greater than r, the point P lies inside. Hence according to result stated in the last section, they exert no gravitational force on mass m kept at P. The shells with radius ≤ r make up a sphere of radius r for which the point P lies on the surface. This smaller sphere therefore exerts a force on a mass m at P as if its mass $M_r$ is concentrated at the centre. Thus the force on the mass m at P has a magnitude

$F=\frac{Gm(M_r)}{r^2}$     (1)

We assume that the entire earth is of uniform density and hence its mass is 

$M_E=\frac{4\pi}{3}R_E^3\rho$

where $M_E$ is the mass of the earth $R_E$ is its radius and ρ is the density. On the other hand the mass of the sphere $M_r$ of radius r is

$\frac{4\pi}{3}\rho r^3$

and hence

$F=Gm\left(\frac{4\pi}{3}r\right)\frac{r^3}{r^2}$

$F=Gm\left(\frac{M_E}{R_E^3}\right)\frac{r^3}{r^2}$

   $=\frac{GmM_E}{R_E^3}r$   (2)

If the mass m is situated on the surface of earth, then r = $R_E$ and the gravitational force on it is, from Eq. (2)

$F=\frac{GM_Em}{R_E^2}$    (3)

The acceleration experienced by the mass m, which is usually denoted by the symbol g is related to F by Newton’s 2$^{nd}$ law by relation F = mg. Thus

$g=\frac{F}{m}=\frac{GM_E}{R_E^2}$   (4)

Acceleration g is readily measurable. $R_E$ is a known quantity. The measurement of G by Cavendish’s experiment (or otherwise), combined with knowledge of g and $R_E$ enables one to estimate $M_E$ from Eq. (4). This is the reason why there is a popular statement regarding Cavendish : “Cavendish weighed the earth”.

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