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.
2006-12-10 05:24:36 · 2 answers · asked by swimmertommy 1 in Science & Mathematics ➔ Physics
First of all, the correct spelling is Mimas.
The equation you're looking for is:
P^2 = 4 pi^2 a^3 / (GM)
Where...
P = the period of Mimas' orbit in years
P = 23 hours = 82800 seconds
pi = 3.14159265358989...
a = the semimajor axis of Mimas' orbit
a = 1.87E+8 meters
G = the gravitational constant
G = 6.6725E-11 m^3 kg^(-1) sec^(-2)
M = Saturn's mass in kilograms
Solving for M,
M = 4 pi^2 a^3 / (G P^2)
M = 5.64E+26 kilograms
M = 94.4 Earth masses
2006-12-10 05:36:06 · answer #1 · answered by Anonymous · 0⤊ 0⤋
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
2016-12-30 05:33:25 · answer #2 · answered by ? 3 · 0⤊ 0⤋