# Homework Help: 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

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

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.)