- #1
JohnGalt
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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?
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?