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Asteroid |
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| Mar22-10, 07:05 PM | #1 |
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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 ? |
| Mar22-10, 07:22 PM | #2 |
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Recognitions:
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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 |
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use kepler's laws. R^3/T^2 is always constant when orbiting the same parent body.
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| Mar22-10, 07:32 PM | #4 |
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Asteroid
Hello,
So it would be (1.5x10^11) ^3 / (3.2 x10^7) ? |
| Mar22-10, 07:32 PM | #5 |
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Hello,
So it would be (1.5x10^11) ^3 / (3.2 x10^7) ? |
| Mar22-10, 07:47 PM | #6 |
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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 ? (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 |
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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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