Saturn is located at a distance of 9.54 AU from the Sun what is its orbital period
Accepted Solution
A:
I hope you remember the relation between Time period of a planet and its orbital radius(obviously assuming it to be circular) given by Keplar,
T^2 = (R^3 x 4 x pi^2)/G x Mass of the sun
In general, mass of that body about which satellite revolves.
1 Astronomical Unit = 149597870700 metres Therefore, 9.54 AU = 9.54 x above quantity = 1.427163686 x 10^12 metres
Plugging G = 6.67 x 10-11 m3 kg-1 s-2 R = 1.427163686 x 10^12 metres pi = 3.14 Mass of the sun = 1.9891 × 10^30 kg and taking square root of T,
T comes out to be.... T = 929563576 seconds!
1 minute = 60seconds. so 929563576 seconds =
929563576 seconds x minutes/60seconds seconds will cancel out.... T = 15492726.27 minutes. i.e T = ~29.476 years= ~29.5 years (if you dunno how I did, tell me in comment section :) ) So Saturn takes ~ 29.5 Earth years to complete one revolution around Sun
That means, 1 year on Saturn is equivalent to 29 years and 6 months on Earth!