Circular Orbits

In this section we assume that the central body of mass $M_E$ is the earth and is fixed (or moving uniformly). And the second body of mass $m$ is a small artificial satellite. The Earth's mass $M_E= 5.9742\times 10^{24}{\rm kg}$ Sputnik I on the other hand weighed 83.6 kg. So its gravitational pull on the Earth can probably be neglected. (This is certainly not true for its effect on world history or American politics). We will also assume that the satellite moves on a circular orbit. and that earth is at the center of this orbit. Let $r$ be the radius of the circular orbit. The distance traveled after a time $t$ is given by

$$s(t)= r(\phi(t)-\phi(0)).$$

Hence,

$$v=r\phi'=r\omega$$

The area swept in a time $t$ is given by

$$A(t)= r^2\phi(t).$$

Kepler's second law implies that this area is proportional to $t$. It follows that

$$\phi(t)=\omega t$$

for some constant angular velocity $\omega$. Thus we have the rule:

Satellites on circular orbits must move at a constant angular velocity.

In polar coordinates we have then

$${\bf r}(t)=\left(\begin{array}{c} r\cos\omega t \\ r\sin\omega t\end{array}\right),$$

and therefore

$${\bf r}''(t)=-\omega^2\left(\begin{array}{c} r\cos\omega t \\ r\sin\omega t\end{array}\right)=-\omega^2{\bf r}(t).$$

(4) can thus be written as

$$m\omega^2{\bf r}(t)=\frac{GM_E m}{r^3}{\bf r}(t),$$

or simply as

$$\omega^2 r^3 =G M_E.$$ (6)

If $T$ is the period of this motion then

$$\omega T= 2\pi.$$

substituting this into (6) we get

$$\frac{r^3}{T^2}=\frac{G M_E}{4\pi^2},$$ (7)

this is of course Kepler's third law in the special case of circular orbits. The expression $G M_E$ is called the geocentric gravitational constant. Its value is

$$G M_E= 3986004\times 10^8 {\rm m}^3/{\rm s}^2.$$