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 ✕ 1026 kg.

Answer 1

Given

saturn like planet has mass m =4.25*10^26 kg

inner radius r1 = 69000000 m

outer radius is r1 = 155000000 m

for the time period we know the Kepler's third law law of periods

T^2 = (4*pi^2*R^3)/(G*M)

T = sqrt((4*pi^2*R^3)/(G*M))

now the time period for inner radius  

T1 = sqrt((4*pi^2*69000000^3)/(6.6726*10^-11*4.25*10^26)) s

T1 = 21385.110264162264 s

T1 = 21385.110264162264/3600 hr

T1 = 5.940 hr

T1 = 0.2475 days

and time period for outer radius  

T2 = sqrt((4*pi^2*155000000^3)/(6.6726*10^-11*4.25*10^26)) s

T2 = 72000.482205599896 s

T2 = 21385.110264162264/3600 hr

T2 = 20 hr hr

T2 = 0.8333 days