Kepler's Third Law orbiting planet

1. Aug 26, 2010

natski

Dear all,

A very simple (almost embarrassingly so) question here, but I just want to double-double check what is going on.

Kepler's Third Law states for a planet of mass Mp orbitting star of mass M* at a semi-major axis $a$ with a period P, that:

[P/(2pi)]^2 = a^3/G (Mp+M*)

Now what I want to know is what is $a$ referring to? Is it the planet to star distance (assuming a circular orbit) or the planet to centre-of-mass distance? I think it is the former.

What if we have a second planet as well and assume perfect Keplerian orbits for the two planets... now if $a$ really does correspond to the planet-star separation then then this must be a variable since the star-barycentre distance is now changing as a result of the presence of the second planet.

Can anyone clear this up?

Natski

2. Aug 26, 2010

Gear300

3. Aug 30, 2010

natski

I think maybe you misunderstood my question. I am speaking of which frame of reference that definition is valid.

4. Aug 30, 2010

D H

Staff Emeritus
You two are talking past one another. Natski, you are not speak of the frame of reference. Length is an invariant in Newtonian mechanics. It doesn't vary from one frame to another.

Kepler's laws apply only to planets of negligible mass, and only to two body systems. Make the mass of the planet significant (e.g., Jupiter) and Kepler's laws are not correct. Throw in other planets with non-negligible mass and the assumption of elliptical orbits is not correct.

5. Aug 30, 2010

Janus

Staff Emeritus
With the equation:

$$\left(\frac{P}{2 \pi}\right)^2= \frac{a^3}{G(M_p+M_*}$$

"a" is the sum of the semi-major axes of the two bodies' orbits, which would be the distance between them in the case of an circular orbit.

6. Sep 5, 2010

cmos

Another way of thinking about what 'a' is would be to go back to the derivation of the equation. In setting up the appropriate equation of motion (with the center of mass coincident with the origin of coordinates), you use a position vector with a length that is equal to the separation between the two masses. The solution yields that the evolution of this vector sweeps out an ellipse. Thus 'a' is the semi-major axis of this ellipse. Careful substitutions show that the masses also sweep out ellipses about the center of mass, however, with different major axes than the first ellipse.

7. Sep 5, 2010

cmos

I don't see how the first part (...planets of negligible mass) of this statement is correct. Neglecting relativity, Kepler's laws make no restriction on the masses of the bodies. Indeed, mass terms show up in the equations. Remembering that these laws follow from Newton's Law of Universal Gravitation, the only assumption the is made is that the bodies are point particles. Considering the great distances between planets relative to their diameters, this assumption is valid.

I do, however, agree with the statement that Kepler's laws only apply to two-body systems. However, the gigantic mass of our sun makes Kepler's laws approximately valid for most of the planets in our solar system (i.e. the (n>2)-body effects are negligible and you can get away with only consider the sun-planet interaction).

8. Sep 5, 2010

D H

Staff Emeritus
Sure it does. Kepler's third law is that P2a3. Newton's laws on the other hand tell us that P2=a3(2π)^2/(GMp+GM). Kepler's third law would dictate that the right-hand side of this latter equation is constant.
The derivation from Newton's laws also assumes the planets are of negligible mass.

9. Sep 5, 2010

cmos

The second equation is commonly called Kepler's third law (whether or not he actually wrote it done precisely) because it implies the first equation. For elliptical orbits, given three of the variable quantities, you are able to determine the fourth. Even if you did not know the masses of the two bodies, you can still deduce how much the period must change for a given change in the semi-major axis or vice-versa.

Again, this is never invoked. Mass plays a central role in Newton's laws.