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

1. Jul 28, 2008

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:

rp3 / rE3 = Tp2 / TE2

where rp is the orbital radius of the planet and Tp is the orbital period of the planet

Therefore,

rp3 = (Tp2 / TE2) rE3

= [ (2 x TE)2 / TE2 ] rE3

Therefore,

rp = 3√ ([ (2 x TE)2 / TE2 ] rE3)

= 1 060 071 250 m

2. Jul 28, 2008

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.)

3. Jul 28, 2008

Janus

Staff Emeritus
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.)