(adsbygoogle = window.adsbygoogle || []).push({}); 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_{p}^{3 }/ r_{E}^{3}= T_{p}^{2}/ T_{E}^{2}

where r_{p}is the orbital radius of the planet and T_{p}is the orbital period of the planet

Therefore,

r_{p}^{3 }= (T_{p}^{2}/ T_{E}^{2}) r_{E}^{3}

= [ (2 x T_{E})^{2}/ T_{E}^{2}] r_{E}^{3}

Therefore,

r_{p}=^{3}√ ([ (2 x T_{E})^{2}/ T_{E}^{2}] r_{E}^{3})

= 1 060 071 250 m

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

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# Homework Help: Finding distance of a planet whose orbital period is 2x Earth

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