Physics Lab #12

 

Kepler’s Third Law

Theory

Johannes Kepler was a pretty poor astronomer and a bit of an oddball mystic, but a very good mathematician.  He was fortunate enough to inherit the observational data of a very good astronomer indeed, Tycho Brahe, in 1601. (The year Giordano Bruno was burned at the stake by the Church for insisting that the Earth revolved around the Sun and women were the equal of men.)  Between that time and 1617 he published his three famous laws:

 

1. Planets, including the Earth, revolve around the Sun in elliptical orbits where the Sun sits at one of the foci.
2. The imaginary line between the planet and the Sun sweeps out equal areas of space in equal units of time for each planet.
3. The semi-major axis of the elliptical orbit cubed divided by the period of the orbit squared is a constant for all the planets.

 

(Go here to find out more about each of the laws.  In particular look at the simulation of the orbits of the planets to see what it is each of the laws is referring to.  Note the elliptical orbits, the slowing down and speeding up as the planets change distance from the Sun, and the fact that as distance increase so does period.)

 

Kepler was coming up with these planetary laws at the same time as Galileo was pointing the first telescope to the skies.  Both men’s work had a great influence on the world as a whole and the young Isaac Newton in particular.  One of Newton’s great successes was to be able to show that each of Kepler’s Laws derive quite neatly from Newton’s laws of motion together with his law of universal gravitation.  

 

Here’s an easy version of how Newton did it for the third law of Kepler:

Imagine a gravitational system with two objects, one of much greater mass than the other (M being the bigger and m the smaller mass), where the smaller mass undergoes circular motion.  For such a system we can equate the gravitational force to the mass of the smaller object times its centripetal acceleration by using N2.  We get:

 

GMm/r² = 4π²mr/P²

 

(Where everything is the usual: r is distance between object centers, P is period of orbit and if your printer printed out a p, it should have been the pi symbol.)  After canceling the m above and rearranging algebraically we get:

 

                                               r³/P² = GM/4π²

 

Which is the result that Kepler got if we take the elliptical orbits to be circles (a special type of ellipse) which is close enough to the case.  Newton’s derivation has the added advantage of giving a value for the constant that the ratio on the left hand side is equal to.  In fact, this method is how astronomers “weigh” stars in binary systems, galaxies and extra-solar planets.  It is also how we happen to know that there is dark matter out there.  (Can you see how?)

 

Procedure

Go here to the virtual planetary system applet and do the following:

1. Making sure the eccentricity is set to 0.0 and the mass to 1.0, take measurements of the period (in years) for orbits of different radius (in astronomical units = A.U.s.   1 A.U. = 1.5 x 10¹¹ m = the average distance between the Sun and Earth).  Find the period for more than one orbit to increase your accuracy just as in the pendulum lab where you took the period of ten orbits of the conical pendulum.
2. Go to the data here and add it to your table.  (This data is the agreed upon data for the planets of the solar system.)
3. Graph the data to prove Kepler’s 3^rd law.  Be careful about what you should actually be graphing!  It is not r vs. P.
4. The slope of the graph should be equal to 1.0 since the units are in Earth orbit radii and years.  (Convince yourself of this.)  If you do a unit conversion to MKS units you can use the equation from Newton above to measure the mass of the Sun.  Do it.