Mismas, one of Saturn's moons, has an orbital radius of 1.87 * 10^8 meters and an orbital period of about...

Question

Mismas, one of Saturn's moons, has an orbital radius of 1.87 * 10^8 meters and an orbital period of about 23.0 hours. Use Newton's version of Kepler's third law to find Saturn's mass.

Please help me with this problem.
I'm having trouble understanding it.
I asked this once before, but I was confused with the answers and the formula's used.

Answer 1

As you know, Newton's variation of Kepler's law is something like this:

P^2*G*(M1+M2)/4(pi)^2=a^3

Kepler's law in its elementary form is P^2=a^3 meaning that the square of the orbital period is equal to the cube of the semimajor axis (the average distance between he sun and the planet)
Kepler's law works for the sun and any object orbiting it, but does not work for planets and moons. That is what Newton's law corrects.

You are given the radius of the orbit, and that is the semimajor axis. You also have the period. But you must remember that the semimajor axis was originally measured in AU and the period was measured in years, so you will have to make some conversions into meters and hours.
4 x Pi^2 is a constant and should be no trouble. The real trick here is the mass of Mimas and Saturn. To be completely accurate, they should have given you the mass of Mimas so that the M1+M2 term would be correct. M1 would be the mass of Saturn and M2 would be the mass of Mimas.
However, in this case, the mass of Mimas, when compared to the mass of Saturn is insignificant, so you can pretty much just assume that M1+M2 is almost like saying M1+0. So in the end all you really have to do is just solve the equation for M1 and assume the value of M2 is 0.
Have fun!

Answer 2

Use 4(?²)r³ = G(M)(p²) the place r is the Moon's orbital radius 3.80 4 x 10^8 meters, p it era in seconds (2,358,720 seconds) and G = 6.sixty seven x 10^-11 Nm^2/kg^2 M = 4(?²)r³/(G(p²)) M = a million.86 x 10^24 kilograms