
#1
Mar2210, 07:05 PM

P: 81

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 ? 



#2
Mar2210, 07:22 PM

Sci Advisor
HW Helper
P: 8,961

Do you have a reason for thinking that?
Is it the same reason that Kepler had? 



#3
Mar2210, 07:26 PM

P: 27

use kepler's laws. R^3/T^2 is always constant when orbiting the same parent body.




#4
Mar2210, 07:32 PM

P: 81

Asteroid
Hello,
So it would be (1.5x10^11) ^3 / (3.2 x10^7) ? 



#5
Mar2210, 07:32 PM

P: 81

Hello,
So it would be (1.5x10^11) ^3 / (3.2 x10^7) ? 



#6
Mar2210, 07:47 PM

P: 27

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. 



#7
Mar2210, 07:50 PM

P: 81

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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