Asteroid


by Brittykitty
Tags: asteroid
Brittykitty
Brittykitty is offline
#1
Mar22-10, 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 ?
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mgb_phys
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#2
Mar22-10, 07:22 PM
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Do you have a reason for thinking that?
Is it the same reason that Kepler had?
alexk307
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#3
Mar22-10, 07:26 PM
P: 27
use kepler's laws. R^3/T^2 is always constant when orbiting the same parent body.

Brittykitty
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#4
Mar22-10, 07:32 PM
P: 81

Asteroid


Hello,

So it would be (1.5x10^11) ^3 / (3.2 x10^7) ?
Brittykitty
Brittykitty is offline
#5
Mar22-10, 07:32 PM
P: 81
Hello,

So it would be (1.5x10^11) ^3 / (3.2 x10^7) ?
alexk307
alexk307 is offline
#6
Mar22-10, 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.
Brittykitty
Brittykitty is offline
#7
Mar22-10, 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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