Physics Help Forum Help with Kepler's Law of Periods

 Periodic and Circular Motion Periodic and Circular Motion Physics Help Forum

 Jul 24th 2011, 01:16 PM #1 Junior Member   Join Date: Jun 2011 Posts: 13 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?
 Jul 24th 2011, 07:49 PM #2 Senior Member   Join Date: Apr 2008 Location: HK Posts: 886 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 __________________ Good results were achieved and the new task is to become a good doctor.
 Jul 24th 2011, 08:57 PM #3 Junior Member   Join Date: Jun 2011 Posts: 13 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.
 Jul 25th 2011, 06:34 PM #4 Senior Member   Join Date: Apr 2008 Location: HK Posts: 886 Is it possible that you should use the values of the sun instead of the earth in order to get the period? __________________ Good results were achieved and the new task is to become a good doctor.
 Jul 25th 2011, 07:29 PM #5 Junior Member   Join Date: Jun 2011 Posts: 13 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.
Jul 27th 2011, 10:07 PM   #6
Senior Member

Join Date: Apr 2008
Posts: 815
 Originally Posted by werehk Is it possible that you should use the values of the sun instead of the earth in order to get the period?
Bingo
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.
__________________
Isaac
If the problem is too hard just let the Universe solve it.

Nov 11th 2011, 09:49 AM   #7
Physics Team

Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,341
 Originally Posted by werehk 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.

 Tags kepler, law, periods

 Thread Tools Display Modes Linear Mode

 Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post tanaki Kinematics and Dynamics 4 Nov 12th 2009 12:58 AM craigmain Advanced Mechanics 3 Mar 16th 2009 12:36 AM meredith xo cheer Special and General Relativity 0 Jan 25th 2009 05:45 PM meredith xo cheer Kinematics and Dynamics 0 Jan 25th 2009 05:31 PM qweiop90 Advanced Mechanics 1 Dec 8th 2008 12:52 PM