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Asteroid

 
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Mar22-10, 07:05 PM   #1
 

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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Mar22-10, 07:22 PM   #2
 
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Do you have a reason for thinking that?
Is it the same reason that Kepler had?
Mar22-10, 07:26 PM   #3
 
use kepler's laws. R^3/T^2 is always constant when orbiting the same parent body.
Mar22-10, 07:32 PM   #4
 

Asteroid


Hello,

So it would be (1.5x10^11) ^3 / (3.2 x10^7) ?
Mar22-10, 07:32 PM   #5
 
Hello,

So it would be (1.5x10^11) ^3 / (3.2 x10^7) ?
Mar22-10, 07:47 PM   #6
 
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.
Mar22-10, 07:50 PM   #7
 
Ooo I get it now! wow, thank you soo much :):):) I don't know how I didn't see that from the beginning
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