PHYS101: Introduction to Mechanics

We shall derive Kepler's third law, starting with Newton's laws of motion and his universal law of gravitation. The point is to demonstrate that the force of gravity is the cause for Kepler's laws (although we will only derive the third one).

Let us consider a circular orbit of a small mass $m$ around a large mass $M$, satisfying the two conditions stated at the beginning of this section. Gravity supplies the centripetal force to mass $m$. Starting with Newton's second law applied to circular motion,

$F_{\mathrm{net}}=m a_{\mathrm{c}}=m \dfrac{v^{2}}{r}.$

The net external force on mass $m$ is gravity, and so we substitute the force of gravity for $F_{\text {net }}$ :

$G \dfrac{m M}{r^{2}}=m \dfrac{v^{2}}{r}.$

The mass $m$ cancels, yielding

$G \dfrac{M}{r}=v^{2}.$

The fact that $m$ cancels out is another aspect of the oft-noted fact that at a given location all masses fall with the same acceleration. Here we see that at a given orbital radius $r$, all masses orbit at the same speed. (This was implied by the result of the preceding worked example.) Now, to get at Kepler's third law, we must get the period $T$ into the equation. By definition, period $T$ is the time for one complete orbit. Now the average speed $v$ is the circumference divided by the period-that is,

$v=\dfrac{2 \pi r}{T}.$

Substituting this into the previous equation gives

$G \dfrac{M}{r}=\dfrac{4 \pi^{2} r^{2}}{T^{2}}.$

Solving for $T^{2}$ yields

$T^{2}=\dfrac{4 \pi^{2}}{G M} r^{3}.$

Using subscripts 1 and 2 to denote two different satellites, and taking the ratio of the last equation for satellite 1 to satellite 2 yields

$\dfrac{T_{1}^{2}}{T_{2}^{2}}=\dfrac{r_{1} \, ^{3}} {r_{ 2} \,^{3}}.$

This is Kepler's third law. Note that Kepler's third law is valid only for comparing satellites of the same parent body, because only then does the mass of the parent body $M$ cancel.

Now consider what we get if we solve $T^{2}=\dfrac{4 \pi^{2}}{G M} r^{3}$ for the ratio $r^{3} / T^{2}$. We obtain a relationship that can be used to determine the mass $M$ of a parent body from the orbits of its satellites:

$\dfrac{r^{3}}{T^{2}}=\dfrac{G}{4 \pi^{2}} M.$

If $r$ and $T$ are known for a satellite, then the mass $M$ of the parent can be calculated. This principle has been used extensively to find the masses of heavenly bodies that have satellites. Furthermore, the ratio $r^{3} / T^{2}$ should be a constant for all satellites of the same parent body (because $\left.r^{3} / T^{2}=G M / 4 \pi^{2}\right).$ (See Table 6.2).

It is clear from Table 6.2 that the ratio of $r^{3} / T^{2}$ is constant, at least to the third digit, for all listed satellites of the Sun, and for those of Jupiter. Small variations in that ratio have two causes -uncertainties in the $r$ and $T$ data, and perturbations of the orbits due to other bodies. Interestingly, those perturbations can be-and have been-used to predict the location of new planets and moons. This is another verification of Newton's universal law of gravitation.