MATH SOLVE

4 months ago

Q:
# 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 Γ 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!

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 Γ 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!