Physics orbiting?

Question

The rings of a Saturn-like planet are composed of chunks of ice that orbit the planet. The inner radius of the rings is 69,000 km, while the outer radius is 155,000 km.The mass of this planet is 4.25 *10^26 kg.
Find the period of an orbiting chunk of ice at the inner radius

Answer 1

F = -GM1M2/R^2 = -M2V^2/R
G = 6.6742*10^-11N-m^2/kg^2
(Rω)^2 = GM1/R
P = 2π/ω, ω = (2π)/P
4π^2/P^2 = GM/R^3
P = 2π/√(GM/R^3)
P = 2π/√((6.6742*10^-11N-m^2/kg^2)(4.25 *10^26 kg)/(69,000 km)^3)
P = 2π/√((6.6742*10^-11 m^3/s^2)(4.25 *10^26)(1km^3/10^9m^3/(69,000 km)^3)
P = 2π/√((6.6742*10^-11/ s^2)(4.25 *10^26)/(69*10^6)^3)
P = 2π/√((6.6742*10^-3 )(4.25)/(69)^3) seconds
P = 21,383 seconds
P = 356.38 minutes
P = 5.9396 hours

Answer 2

v = ?(GM/r) ... the place M is the mass of the Earth and G is the gravitational consistent. Convert your 'r' in to meters = 6800000m Throw the values into the equation and there you have it :) For the term, replace the above equation into that for around action (T = (2?r)/v) this provides: T = ?(4 ?² r ³)/GM ;)

Answer 3

Or you could just say

T = √(4π²R³/GM) = √(4π²*(69E6)³/(6.67E-11*4.25E26)
= 21389 sec

T = 5.941 hr