Does mass affect speed of orbit at a certain distance?

Does the mass of both the parent object, and the child object affect the speed at which the child object orbits the parent object?

I though it didn't (something like T^2 = R^3) until I saw an planet on the iphone exoplanet app, that is closer to it's star than a planet in another system, yet takes longer to complete one orbit. Both of the planets were a similar mass, as were the stars.

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4 Answers

In the limit where $m_2 \ll m_1$, only the mass of the heavy body matters (along with the semi-major axis of the orbit, of course).

Where that limit does not apply, varying the mass of either body changes the reduced mass:

$$\mu = \frac{m_1 m_2}{m_1 + m_2} .$$

Since the system acts as if a negligibly massive object was moving in the field of one having the reduced mass, this does alter the period.

Notice that in the limit above $\mu \to m_2$ and we recover the expected behavior.

Marion and Thorton give the full expression for the period $\tau$ in the form

$$\tau^2 = \frac{4 \pi}{G} \frac{a^3}{m_1 + m_2}$$

where $a$ is the length of the semi-major axis of the orbit and $G$ is the gravitational constant. It should be obvious that in the limit of a heavy primary this reduces to $\tau^2 = \frac{4 \pi}{G} \frac{a^3}{m_1}$.

Side comment: The rule you recall is the one Kepler found for planets in our Solar System. In this case the mass of the sun dominates in every case. Jupiter is about 0.001 solar masses, so the largest correction in at the tenth of a percent level. Observable, but not at all large.

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I think you've got some things mixed up here. It is not true that "the system acts as if a negligibly massive object was moving in the field of one having the reduced mass." In the case $m_2\ll m_1$, the reduced mass $\mu\approx m_2$, so this would say that the system acts like one in which the gravitational field was that of the planet, not the star. Reduced mass isn't really important here. As the Marion-Thornton formula makes clear, all that matters is the total mass $m_1+m_2$. The small-planet approximation arises because $m_1+m_2\approx m_1$, not because $\mu\approx m_2$. – Ted Bunn Apr 15 '11 at 21:39

As @dmckee's answer says, in the limit where the mass of the planet is much less than the mass of the star, the mass of the planet does not have a significant effect on the period. I just want to add a more explicit comment on this part of your question:

I saw an planet on the iphone exoplanet app, that is closer to it's star than a planet in another system, yet takes longer to complete one orbit.

The reason for this is almost certainly not the masses of the planets but rather the masses of the two stars. The systems you're looking at in the app almost certainly satisfy the rule $m_{\rm planet}\ll m_{\rm star}$, so that the planet's mass is unimportant. You say that the masses of the stars are "similar," but I bet they're different enough that that's the explanation for what you're seeing.

One way of writing Kepler's third law, as it applies to planets orbiting other stars, is $$T^2={R^3\over M},$$ which is valid only in a certain choice of units: periods in years, radii in astronomical units, masses in solar masses. dmckee gives the more general formula. This version corresponds to a choice of units that makes the combination of constants $4\pi/G$ come out to 1.

Does the app give you specific information about the numerical values of the various quantities? If so, you could check this. If not, are you sure about your statement that the masses are "similar"?

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You can always think of it like this: Start with one planet of mass m orbiting the star at a certain speed. Now add a 2nd planet of mass m in the same orbit. Same speed, right? Now let them be touching each other in the orbit. Same speed, right? Now spot weld them together. You've got a single planet of mass 2m. Same speed.

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Crudely, if you make the force of gravity equal to the centripetal force (which it is for stationary, well behaving orbits), $\frac{MmG}{r^2}=\frac{mv^2}{r}$, and so $\frac{MG}{r}=v^2=\frac{4 \pi^2r^2}{T^2}$, which contains Kepler's 3rd law.

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