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.

Answer 1

You start with Newton's adaption of Kepler's third law:

(m[1]+m[2])*P^2 = (d[1]+d[2])^3 = R^3

Have you ever gotten to play with this one before? If you have, you know that two axilliary equations help you use Newton's adaption. You won't get far without them:

m[1]*d[1] = m[2]*d[2]
d[1]+d[2] = R

Now, here's what those two equations represent, a two-body orbital system. R is the total distance between the center of the two bodies' masses. m[1] is Saturn's mass. m[2] is Mimas' mass. d[1] is the distance between the center of Saturn's mass and the two-body system's mass. d[2] is the distance between the center of Mimas' mass and the two-body system's mass.

Now, we know that bodies orbit around the center of mass in their system, so the radius of Mimas' orbit is equal to d[2], or the distance between Mimas' center of mass and the system's center of mass. We know the orbital period, or P in the main equation, to be equal to 23 hours. So, lets try pluging these in and see what we have...

(m[1]+m[2])*(596(hrs^2)) = (d[1]+(1.87*10^8(m)))^3 = R^3
m[1]*d[1] = m[2]*(1.87*10^8(m))
d[1]+(1.87*10^8(m)) = R

Will that be enough to solve for m[1], the mass of Saturn? No! There are too many variables left open. No mater how you slice it, you can't get it down to a single variable equation, and if you can't get it down to a single variable equation, you can't solve. What you need to do is find out either the distance between the center of Saturn's mass and the system's mass (d[1]), the distance between the center of mass of the two bodies (R), or the mass of Mimas (m[2]) to solve. Any two these will work, but you'll need to find them out to finish setting up your equation. I'm sorry I can't finish the problem for you, but look on the bright side. I did all of the physics. It's just algebra from here on out.

Answer 2

Use 4(?²)r³ = G(M)(p²) the situation 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