1.8 Gravitation

Aside from inventing classical mechanics and calculus, Isaac Newton’s most well-known contribution to science is his discovery of the universal law of gravitation. If we consider two objects, one of which is located at the origin, then the gravitational force between the two bodies has magnitude

F=\frac{Gm_{1}m_{2}}{r^{2}}.

(1.80)

In vectorial form, these forces form a Newton III pair.

	$\displaystyle\boldsymbol{F}_{1\to 2}$	$\displaystyle=-\frac{Gm_{1}m_{2}}{\lvert\boldsymbol{r}_{1\to 2}\rvert^{3}}% \boldsymbol{r}_{1\to 2}$		(1.81)
	$\displaystyle\boldsymbol{F}_{2\to 1}$	$\displaystyle=-\frac{Gm_{1}m_{2}}{\lvert\boldsymbol{r}_{2\to 1}\rvert^{3}}% \boldsymbol{r}_{2\to 1}=\frac{Gm_{1}m_{2}}{\lvert\boldsymbol{r}_{1\to 2}\rvert% ^{3}}\boldsymbol{r}_{1\to 2}=-\boldsymbol{F}_{2\to 1},$		(1.82)

where $\boldsymbol{r}_{1\to 2}=\boldsymbol{r}_{2\to 1}=\boldsymbol{r}_{2}-\boldsymbol% {r}_{1}$ .

Consider an object of mass $m$ near the surface on the Earth. The radius of the Earth is very large so we can approximate the distance between the Earth and the object as simply the radius of the Earth. The gravitational force which the Earth exerts on the object is given by

F_{G}\approx\frac{GM_{\text{Earth}}}{R_{\text{Earth}}^{2}}m=mg,

(1.83)

where $g=\frac{GM_{\text{Earth}}}{R_{\text{Earth}}^{2}}$ . Thus we have recovered the weight force that we have been using for the force due to gravity. If the height of the object is so large that the approximation no longer holds, then $g$ depends on the height $R_{\text{Earth}}+h=r$ . This is the same as the general case, the force only depends on the distance between the centre of the Earth and the object. Now consider the case where the object is below the surface of the Earth. In this case, the mass of the Earth depends on the distance between the centre and the object. The mass enclosed within a radius $r$ is given by the density multiplied by the enclosed volume:

M(r)=\frac{4}{3}\rho\pi r^{3}.

(1.84)

If we assume that the density of the Earth is constant, then it is given by the total mass divided by total volume:

\rho=\frac{M_{\text{Earth}}}{\frac{4}{3}\pi R_{\text{Earth}}^{3}}.

(1.85)

Thus, $g(r)$ is given by

	$\displaystyle g(r)$	$\displaystyle=\frac{GM(r)}{r^{2}}=\frac{G}{r^{2}}\frac{M_{\text{Earth}}}{R_{% \text{Earth}}^{3}}r^{3}$		(1.86)
		$\displaystyle=\frac{GM_{\text{Earth}}}{R_{\text{Earth}}^{3}}r.$		(1.87)

Example 1.8.

Calculate $g$ at the surface of Earth.