Circular Orbits

The object of this section is to correct some of the aspects of the computations in the preceding section. We first drop the assumption that the central body of mass $M$ is at rest or moves uniformly. We replace with the following more sensible assumption.

Assumption: The center of mass of the two body system consisting is at rest or moves uniformly.

One can always arrive at this assumption by decomposing the motion of the two body system into two pieces, a motion of the center of mass and a motion of the two components with respect to the center of mass. If we assume that there is no external forces (except the gravitational forces between the two bodies), Newton's first law tells us that the center of mass is either at rest or moves uniformly. In the case of a satellite moving around Earth, this assumption is not quite correct, since the center of mass moves in the gravitational field of the sun, hence it is not a uniform motion with constant speed. However, on the surface of the Earth the acceleration due to the gravitational pull of the sun is approximately $.0059 m/s^2$ compared to Earth gravity of $9.81m/s^2$, and we will ignore it for the time being.

We center our coordinate system at the center of gravity, which sits on the line connecting the two bodies. In circular motion the distance between the two bodies is constant $r$. We have $r=r_1+r_2$, where $r_1$ is the distance of the body of mass $M$ from the center of mass, and $r_2$ is the distance of the body of mass $m$ from the center of mass. We have from the definition of the center of mass:

$$Mr_1 =mr_2.$$

Furthermore, we have the following balance of centrifugal forces:

$$\begin{eqnarray*} mr_2\omega^2 &=& G\frac{Mm}{r^2}\\ Mr_1\omega^2 &=& G\frac{Mm}{r^2} \end{eqnarray*}$$

Adding these equations and using the property of the center of mass we get:

$$r^3\omega^2=(r_1+r_2)^3\omega^2=G(M+m),$$

or

$$r^3\omega^2=GM\left(1+\frac{m}{M}\right)$$ (8)

This is the corrected form of Kepler's third law for circular orbits. One sees that this is now dependent on the mass of the satellite. In the case Sputnik I, the correction factor

$$\frac{m}{M}=\frac{83.6}{5.9742\times 10^{24}}=1.44\times 10^{-23}$$

which can certainly be ignored.

Examples: The ratio of the mass of the moon to the mass of the earth is

$$\frac{M_L}{M_E}=0.0123.$$

The radius of the moons nearly circular orbit is about 384400 km. This implies that $r_1=0.0123\times 384400 = 472$ i.e the center of the rotation is 472 km above the center of the earth. The orbital period of the moon is

$$27.3\times 86400 =2358720 s$$

i. e. $\omega=2.664\times 10^{-6} s^{-1}$. This produces an acceleration pointed away from the center of the earth and opposite the center of the earth of

$$\omega^2(R_E+r_1)=4.86\times 10^{-5} m/s^2,$$

which is responsible for the second tide.

Another more exciting example is from the search for extra solar planets. Planets do not radiate any energy, and it is therefore impossible to directly observe planets circling around stars. However, if a large planet of mass $m_P$ circles a star of mass $M_S$ with period $T$ we have the following:

$$\begin{eqnarray*} r &=&r_S+r_P\\ M_Sr_S&=&m_Pr_P\\ \frac{r^3}{T^2}&=& \frac{G}{4\pi^2}\left(M_S+m_P\right) \end{eqnarray*}$$

The star will radiate on a different frequency, depending on its velocity relative to the observer. If the star moves on a circular path (in the same plane with the observer) with speed $v=r_S\omega=2\pi r_S/T$, the maximal difference between velocities is $2v$. This can be computed from the shift in frequencies. One can also directly observe the period $T$. From this one can compute the ratio of the two masses and $r_S$. Often one can also estimate $M_S$ from the total radiation of the star. This then completely determines the orbit and size of the planet.