## Asteroid

An asteroid is orbiting around the sun at a distance of 4.2x 1011 m. If the Earth orbits around the Sun at a distance of 1.5 x 1011 m with a period of 3.2 x107 s, what is the period of the asteroid?

Im thinking it was 9.0 x 10^7 s ?
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 Recognitions: Homework Help Science Advisor Do you have a reason for thinking that? Is it the same reason that Kepler had?
 use kepler's laws. R^3/T^2 is always constant when orbiting the same parent body.

## Asteroid

Hello,

So it would be (1.5x10^11) ^3 / (3.2 x10^7) ?
 Hello, So it would be (1.5x10^11) ^3 / (3.2 x10^7) ?
 An asteroid is orbiting around the sun at a distance of 4.2x 1011 m. If the Earth orbits around the Sun at a distance of 1.5 x 1011 m with a period of 3.2 x107 s, what is the period of the asteroid? Im thinking it was 9.0 x 10^7 s ? (4.2x10^11)^3/(t^2) = (1.5x10^11)^3/(3.2x10^7)^2 I'm looking at 1.5x10^8 sec.
 Ooo I get it now! wow, thank you soo much :):):) I don't know how I didn't see that from the beginning