The period of a planet orbiting a round a star, for example, the Sun, relates to th=e radius of the orbit according to the third Kepler's Law:

`T^2 = (4pi^2)/(GM) a^3`

Here, T is the period of the planet ( 1year for the Earth orbiting the Sun), M is the...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

The period of a planet orbiting a round a star, for example, the Sun, relates to th=e radius of the orbit according to the third Kepler's Law:

`T^2 = (4pi^2)/(GM) a^3`

Here, T is the period of the planet ( 1year for the Earth orbiting the Sun), M is the mass of the Sun, G is the gravitational constant and a is the distance between the planet and the star. This distance between the Earth and the Sun is 1 AU (this is an approximation - see the attached reference link.)

From this formula you can see that the square of the period varies inversely with the mass of the star and directly with the cube of the radius of the orbit. So the square of the new period will be

`T_n ^ 2= (4pi^2)/(G(2M)) (10 AU)^3 = (1000)/2 (4pi^2)/(GM) (1 AU)^3=500T^2`

Since T = 1 year, the new period will be

`T_n = sqrt(500)T = 22.36` years.

**Further Reading**