What Is the Orbital Period of an Asteroid with Twice Earth's Mean Radius?

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SUMMARY

The orbital period of an asteroid with a mean orbital radius twice that of Earth's can be calculated using Kepler's Third Law, which states that the square of the orbital period (t) is proportional to the cube of the mean orbital radius (r). Specifically, the formula t² = r³ applies, where r for the asteroid is 2 AU (Astronomical Units). Therefore, the period of the asteroid is 2.83 Earth years, derived from the calculation t = √(r³) = √(2³) = √8.

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Homework Statement


An asteroid revolves around the sun with a mean orbital radius twice that of Earth's. Predict the period of the asteroid in Earth years


Homework Equations



I guess t^2/r^3 where t = period and r = mean orbital radius


The Attempt at a Solution


I missed the lecture in class, and I'm ultimately confused on the subject of Orbital Dynamics and Kepler's Law, any help/explanation would be great, or a link to a useful website, honestly I'm clueless. I think the fact that the r for the asteroid is twice as long as Earth's creates a ratio?
 
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