How Long Does It Take an Asteroid to Orbit the Sun Compared to Earth?

AI Thread Summary
The discussion revolves around calculating the orbital period of an asteroid using Kepler's laws. The asteroid orbits the Sun at a distance of 4.2 x 10^11 m, while Earth orbits at 1.5 x 10^11 m with a period of 3.2 x 10^7 s. Participants apply the formula R^3/T^2 to find the asteroid's period, leading to differing initial estimates. One user arrives at a period of approximately 1.5 x 10^8 seconds after clarification. The conversation highlights the application of Kepler's laws in determining orbital periods.
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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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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.
 


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
 
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