Kepler's Third Law

Factual Questions
1

I am puzzled why Kepler’s Third Law works so (seemingly) well for all ellipses - regardless of how eccentric the orbit is. Kepler’s Third relates the square of a planet’s period (revolution) to the cube of the planet’s semimajor axis. Couldn’t two ellipses have the same semimajor axes and different minor axes - hence giving these two distinct ellipses two different eccentricities? Why doesn’t this factor into the equation?

Or, are we all just familiar with just a general equation of which there is a more specific form? - Jinx :confused:

2

The thing about Kepler’s third law that you’ve stated there, is that it only works for planets orbiting the sun. It actually follows rather logically from Newton’s law of gravity - F=GMm/r[sup]2[/sup] (equation 1).

Then, using Newton’s second law of motion, we know that:

F=ma (eqn 2), where a is the acceleration of the body on which the force is being applied, in this case, a planet orbiting the sun.

Now, a=rw[sup]2[/sup] = where w is the angular velocity of the object around the sun. (eqn 3)

So, F=mrw[sup]2[/sup]. (eqn 4)

w=1/P, where P is the rotational period of the planet. (eqn 5)

So, F=mr/P[sup]2[/sup] (eqn 6).

Equating equations 1 and 6 (which we can since the accelerating force is the gravitational force), we obtain:

r/P[sup]2[/sup]=GM/r[sup]2[/sup].

Doing some rearranging, we get:

r[sup]3[/sup]=GMP[sup]2[/sup], Kepler’s Third Law.

Newton also proved that Kepler’s Third Law will work for any two bodies orbiting a common centre of mass.

Also, bear in mind that the total energy in an orbit must remain constant. So, if you have a very eccentric orbit, when the planet is at its furthest away it will be moving slower than when its at its closest point, hence, two orbits with the same semi-major axes, but with different semi-minor axes can have the same orbital period.

3

As an aside-when my high school physics class was introduced to this problem, the teacher related a (probably aprocyphal) story that Newton was confronted by someone asking this questions not long after the publication of Principia, and was apparently genuinely puzzled that someone needed to ask a question with such a simple and obvious answer. This was very encouraging, until we actually sat down and tried to do that homework assignment. :slight_smile:

4

Looks as if I stumbled upon a good thread.
For a calculator for Kepler’s Third Law go to:
http://www.1728.com/kepler3.htm
And for a more advanced calculator go to:
http://www.1728.com/kepler3a.htm

Actually Angua, you can use the T[sup]2[/sup] = r[sup]3[/sup] relationship for any orbital body (not just those orbiting the Sun). (For an example of this go to that first link
http://www.1728.com/kepler3.htm
and scroll to the bottom.

5

Only if the mass of one body is significantly greater than the mass of the other body. Also, the constant of proportionality will be different than to the one used for the solar system, since Kepler’s third law in its form applicable to any orbital body is in fact a statement of proportionality, not equality, ie:

P[sup]2[/sup]=ka[sup]3[/sup]

where k is an arbitrary constant dependant on the system.

For two bodies of roughly similar mass, eg a binary neutron star, then one must use Newton’s form.

6

That only shows it’s true for circular orbits. Jinx’s question is more involved. For an interesting treatment of it, see Feynman’s Lost Lecture

7

You can use energy arguments to show that its true for a closed elliptical orbit. But its slightly more complicated.

8

Angua
Did you go to the website
http://www.1728.com/kepler3.htm
and look at example 2?

It gives data on Jupiter’s satellite Io - orbital period 1.75 days and orbital radius 421,800 kilolmeters.
With the above data, and being told that Europa orbits Jupiter in 3.5 days, you can determine Europa’s orbital radius by T[sup]2[/sup] = r[sup]3[/sup].

Basically, it utilizes Io’s orbit as the standard (just as Earth is the standard when bodies orbitting the Sun are concerned).

Also, since Io’s mass is much smaller than Jupiter’s, the mass of Io does not signifcantly affect the calculations.

As for extreme elliptical orbits (Pluto, comets, etc), T[sup]2[/sup] = r[sup]3[/sup] still holds precisely. Velocities vary greatly though. For example, when Halley’s comet is exactly 1 astronomical unit from the Sun, its velocity is greater than the Earth’s velocity.

There is a velocity formula for ellipses if people think it’s necessary to do the calculations, I’ll see if I can find it and post that too. (That might take a while though).

9

Perhaps the OP is puzzled by the asymmetry in using the semimajor axis but not the semiminor axis? You can think of the semimajor axis as the average distance of the planet from the Sun. Since the Sun is at one focus, and the foci lie on the major axis, the distance from apohelion to Sun plus the distance from Sun to perihelion is equal to the major axis. So the average between the apohelion and the perihelion distances is the semimajor axis.

10

Chronos, that’s it precisely.

If you go to this webpage
http://www.1728.com/ellipse.htm
and scroll down a few screens, you will see Kepler’s first law explained as you described it.

11

Yes, I did. I am well aware of it. But note how you have to use a different standard. Exactly what I said above, with