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

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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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