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Old Jul 24th 2011, 01:16 PM   #1
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Help with Kepler's Law of Periods

I can use some help regarding Kepler's Law of Periods... I'm trying to apply it to the Earth, and I'm getting a seemingly incorrect answer:


T^2 = 4pi^2*r^3/(GM)

G = 6.67E-11
M = 5.98E24
r = 15E10


When I plug everything in, I get:

T = 1.82769299 E10

If the units of T is seconds, then this comes out to approximately 580 years.


Where did I make a mistake?
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Old Jul 24th 2011, 07:49 PM   #2
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If you want to find out the period of earth( i assume that you mean the period going around the sun), you should better use

instead of T^2 = 4pi^2*r^3/(GM)

The reason is that for T^2 = 4pi^2*r^3/(GM), it is only suitable for very large planet( primary body). In the other words, the earth should be much larger than the sun in order that the equation will be working. This is what i thought and please correct me if i've made mistakes
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Old Jul 24th 2011, 08:57 PM   #3
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Well, I need to learn how to apply that particular formula. I would think it should work in this scenario, but it's not even a matter of being slightly off - it's obscenely off.

I can't help but think that I must have applied it incorrectly, but I don't know how.
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Old Jul 25th 2011, 06:34 PM   #4
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Is it possible that you should use the values of the sun instead of the earth in order to get the period?
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Old Jul 25th 2011, 07:29 PM   #5
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To the best of my knowledge, it simply applies to any body in orbit, without needing to take into account another body. That's why the formula only references one mass, not two. It requires the semi-major axis radius, which would not apply if we were talking about the "center" of the orbit.
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Old Jul 27th 2011, 10:07 PM   #6
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Originally Posted by werehk View Post
Is it possible that you should use the values of the sun instead of the earth in order to get the period?
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Check out wikipedia's article. Keplers laws are an approximations that would work perfectly if:
Originally Posted by Wikipedia
Kepler's laws are strictly only valid for a lone (not affected by the gravity of other planets) zero-mass object orbiting the Sun; a physical impossibility. Nevertheless, Kepler's laws form a useful starting point to calculating the orbits of planets that do not deviate too much from these restrictions.
See Kepler's laws of planetary motion - Wikipedia, the free encyclopedia.
As for your formula in your first post, I think it should work. If you use the mass of the Sun rather than the one of the Earth.
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Old Nov 11th 2011, 09:49 AM   #7
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Originally Posted by werehk View Post
Is it possible that you should use the values of the sun instead of the earth in order to get the period?
Correct! The equation T^2 = 4 pi^2 R^3/(GM) refers to the mass of the body about which the satellite (or moon, or planet) is in orbit. By using M = 5.98E24 Kg you have found the orbital period of a satellite orbiting earth at a distance of 15E10 meters (93 million miles). Change M to the mass of the sun (2E30 Kg) and see what you get.
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