# Suppose the Sun were twice as massive as it actually is. What would be the orbital period of a planet at a distance of 10 AU from the Sun?

*print*Print*list*Cite

### 1 Answer

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.

**Sources:**