- 13 Apr '09 07:45

If we don't know anything more than this, and the Newton's version of Kepler's third law, then we cannot find a reliable answer.*Originally posted by wormer***Minas, a moon of Saturn, has an orbital radius of 1.87x 10^8m and an orbital period of about 23h. Use Newton's version of Kepler's third law and these data to find the mass of Saturn.**

Using the same formulae to the Moon of Earth, or the moon of Pluto, I don't think we will get a very good result.

...unless we assume that the mass of Mimas is near to nothing compared with the mass of Saturn. And that information is not given. - 13 Apr '09 20:08

It is certainly true that the mass of Mimas is much less than Saturn. All of Saturn's moons together are much less.*Originally posted by FabianFnas***If we don't know anything more than this, and the Newton's version of Kepler's third law, then we cannot find a reliable answer.**

Using the same formulae to the Moon of Earth, or the moon of Pluto, I don't think we will get a very good result.

...unless we assume that the mass of Mimas is near to nothing compared with the mass of Saturn. And that information is not given. - 14 Apr '09 05:42

True. I know it, you know it, but I'm sure not all of us knows it without looking it up. Mimas is certainly a small moon.*Originally posted by sven1000***It is certainly true that the mass of Mimas is much less than Saturn. All of Saturn's moons together are much less.**

I ask you: Does it matter if the orbit of Mimas is circular, or does it gives the same result if the orbit is highly elliptical? Do we have to know its eccentricity before we can rely on the answer? - 14 Apr '09 21:26 / 1 edit

We should technically use the semi-major axis of the elliptical orbit, in place of the radius. But the eccentricity itself doesn't matter (so long as the moon doesn't collide with the planet ).*Originally posted by FabianFnas***True. I know it, you know it, but I'm sure not all of us knows it without looking it up. Mimas is certainly a small moon.**

I ask you: Does it matter if the orbit of Mimas is circular, or does it gives the same result if the orbit is highly elliptical? Do we have to know its eccentricity before we can rely on the answer?

Edit: It occurs to me to wonder if the combination of a known period and known semi-major axis around a fixed mass would determine a unique eccentricity. I don't recall off-hand the derivation for eccentricity in terms of these variables. - 26 Apr '09 03:36

You do this...and you'll clean it up!!!*Originally posted by sven1000***And the answer...**

P^2 = C*a^3 is Kepler's third law, with P = period, a = radius of orbit

C = 4*pi^2 / G*M, with G as Newton's constant and M as Saturn's mass, in the Newtonian version.

Using the given numbers for Mimas, I get a mass for Saturn of 5.72x10^26 kg.