Finding distance of a planet whose orbital period is 2x Earth

[JohnGalt]

A small planet has an orbital period that is twice that of Earth. What is the planet's orbital distance?

From Kepler's 3rd law of planetary motion, I can assume that:

r_p³ / r_E³ = T_p² / T_E²

where r_p is the orbital radius of the planet and T_p is the orbital period of the planet

Therefore,

r_p³ = (T_p² / T_E²) r_E³

= [ (2 x T_E)² / T_E² ] r_E³

Therefore,

r_p = ³√ ([ (2 x T_E)² / T_E² ] r_E³)

= 1 060 071 250 m

which differs from the given answer. What error have I made?

 

[DaveC426913]

I imagine this is a shortcut but, if the period is 2x, isn't the radius simply √2x?

(No, it's other way round. If the radius is 2x, then the orbital velocity is √2x.)

 

[Janus]

The equation looks okay, did you use the right values?

Also, reducing the equation further might help as it reduces chances for math errors.

(hint: by some slight re-arranging, you can eliminate a variable from the equation.)