Sorry if this comes as a nuisance at all, but I'm desperate, so here's a pop quiz. Let's pretend there's a new planet whose orbit is closer to the sun than Mercury's. We'll call it planet Tron. Tron is located 0.1 AU from the sun. How would you determine Tron's distance from EARTH (whose disatance is 1 AU from the sun) at superior conjunction, after determining Tron's orbital period around the sun (using P^2=a^3, AKA Kepler's third law)? Is there some forumla for this that I missed out on?

All I can think of to do is figure Earth's distance from the sun MINUS Tron's distance from the sun, but that calculation doesn't take superior conjunction into account. It has to be a bit more complicated than such simple subtraction....

Thanks in advance for any help!

It would be earth's distance plus Tron's distance as they are on opposite sides of the sun. Around 1.1 AU if you're not particularly concerned with accuracy.

Seeing as how we can't see it at superior conjuction then we must infer the orbit from observed orbital elements. Make them what you wish.

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When I am done here I think I will go create something from metal.

I would think that you can assume an inclination of 0 and just deal with semimajor axes. Though, I always get skeptical on problems where I think it seems too simple too.

Thanks all! That was all too simple. I feel silly for having asked! Anyway, thanks to you, I have a good chance of passing my exam!

Welcome, and Good luck! :wink:

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