Yervant Terzian/Terry Herter
This is a simplified derivation of Kepler’s third law and its application to finding planets. It is meant to supplement the Binary Star and Eclipsing Binary Java applets and serve as a bit of a tutorial for those wanting to know more.
Consider two massive objects orbiting one another. This could be two stars or a star and a planet. Suppose for now we consider two stars which may have comparable masses. Let use designate these by masses m1 and m2. Then the center of mass of the system, by definition, will not move as the stars orbit one another (it may drift through space due to the joint motion of the stars), and in fact the stars orbit about the center of mass of the system. Consider two objects balanced on a seesaw, as shown in the diagram below.
The balance point is the center of mass. In this diagram the center of mass is located a distance r1 from star 1 and r2 from star 2. We have then the condition that at all times
Let r be the total distance between the stars, then
We now assume circular orbits so r1 and r2 do not change as the stars rotate about one another. The force of gravity keeps the objects together, so we have Newton’s law of gravity
G is Newton’s gravitational constant. The force on each star is equal and opposite (the center of mass can not move).
We need one more equation to complete the picture. The stars are undergoing a constant acceleration. Since we are in a circular orbit this acceleration manifests itself as a change in direction rather than a change in speed of the object. (Recall that a force results in a change in the velocity of an object. This can be a change in either the speed or direction, or both of the object.). The centripetal acceleration for star one is given by
[Aside: This equation for the centripetal acceleration is most easily derived using some elementary trigonometry and calculus. For an object traveling in a circular orbit of radius R, we can write the x and y coordinates of the object in terms of R and the angle q about the origin (see figure).
where R a constant. We differentiate with respect to time (as indicated by the dot above the appropriate symbol) to get the velocities in x and y
The angular rate () is constant for a circular orbit. Differentiating once more with respect to time and using the fact that the angular speed is constant we get the accelerations in x and y
Using the standard trigometric identity, , the total acceleration is then
where we have used the fact that . End Aside.]
Now from Newton law relating forces and acceleration we have the centripetal force for each of the two stars given by
Each of these must equal the force generated by the Newton’s law of gravity and again are in opposite directions. We’re not worried about the sign since we know that the two stars are always on the opposite sides of the center of mass.
By setting the centripetal force equal to the gravitational force we have for star one
Where we have used the relationship between the orbital period of star 1, its velocity, and the circumference of its orbit, . From our definition of r and the results from the fixed center of mass we also have
Putting this all into our force balancing equation gives
which just gives us Newton’s form of Kepler’s third law.
which tells us that the square of the orbital period is proportional to the cube of the separation between the two stars. This equation is valid for non-circular orbits as well. When applied to the solar system, this equation can be used to find the mass of the sun (from the orbital motion of a planet around the sun) or the masses of any planets with moons.
Choosing the correct units can make things easier. If the masses are in solar masses, the distance in astronomical units (the distance from the earth to the sun), and the period is in years then this equation becomes
The detection of Exosolar planets is a bit trickier. Since, as yet, the planet itself cannot be seen, astronomers must rely on observing the reflex motion of the star to the planet. (Note: there are a few direct detections of exoplanets but this is very rare). From our analogy with the seesaw above we can see that since stars are usually much more massive than planets the motion of the star will be very small. Two possible ways to detect this stellar motion are through the Doppler shift and though stellar wiggle (the star moving back and forth on the sky). Since stars drift in space, we must look for a “periodicity” to the stellar motion associated with the orbiting planet.
Using the equations we have set up above we can derive the magnitude of these two effects. Let mass 1 represent the star and mass 2 represent the (unseen) planet. Starting with our equation equating the centripetal and gravitational forces we have
Here we now want to know the effect in terms of the planets distance, r2. Thus using our equations relating r, r1, r2, m1 and m2 from before but eliminating now r and r2
so that now we have
We can take the square root of this expression to get the peak velocity that will be observed however we need to account for the fact that the motion causing the Doppler shift may not all be along our line-of-sight. Letting i be the inclination angle of the orbit to the line-of-sight and simplifying further by assuming that m2 << m1. We have the peak (or maximum) radial velocity of the star along the line-of-sight (vr) given by
To make this more transparent we use the following notation: m2 = mp (planet mass in Jupiter masses, Msun = 1047 Mjup ), m1 = ms (star mass in solar masses), and r2 = a (the semi-major axis of the planet orbit in astronomical units). Our the radial velocity equation is now
Using Kepler’s third law and again assuming ms >> mp we can write a in terms of the period, P to get this equation in a different form
For Jupiter mass object locate about 5 AU from a solar type star, the period will be 12 years and the peak velocity will be ~13 m/sec.
The equation for the wiggle on the sky is easier. The angular movement on the sky of the star as the planet orbit around it is just given by
where d is the distance to the star. Again with m2 = mp (planet mass in Jupiter masses), m1 = ms (star mass in solar masses), r2 = a (the semi-major axis of the planet orbit in astronomical units), and taking the distance, d, in parsecs, we have the angular displacement
Again we can use Kepler’s third law to write a in terms of the period, P to get
We can now see two reasons why detecting exosolar planets in this fashion is so hard. First the displacements are very small and unlike the Doppler Effect get larger with distance. Second, larger angular movements happen with larger separations. However, this corresponds to longer orbital periods. Thus it may take many years to have enough time to possible detect the reflex motion of the star in this way. As an example, for a Jupiter mass object located 5 AU from a star like the sun which is 10 pc away the wobble will be 0.0005 arcseconds with a period of about 12 years!