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Only Sun and Earth in a system

 
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Jun7-12, 04:35 PM   #1
 

Only Sun and Earth in a system


For a hypothetical system of a Sun and Earth (other planets absent), how can I compute analytically (or where can I find data on) the length of the year on Earth?
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Jun7-12, 06:13 PM   #2
 
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uhh??
it's a year!
Jun7-12, 06:34 PM   #3
 
Do you mean, given its orbital characteristics, could you calculate its revolution about the sun from first principles?
Jun7-12, 06:37 PM   #4
 

Only Sun and Earth in a system


Quote by tiny-tim View Post
uhh??
it's a year!
Well, say, the length of a year (i.e. one revolution) in seconds...
Jun7-12, 06:42 PM   #5
 
Quote by DaveC426913 View Post
Do you mean, given its orbital characteristics, could you calculate its revolution about the sun from first principles?
Let's say so. I know that 3. Kepler law gives the time of revolution, but it true for circular orbit. Well, Earth's orbit is almost circular, so may be the result would be good?

So, yes, for known distance from the Sun, to calculate the time of revolution if there were only Sun and Earth in the system (two body problem). Analytically. I have done it numerically, so I want to compare the results.
Jun7-12, 06:51 PM   #6
D H
 
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Quote by nikolafmf View Post
Let's say so. I know that 3. Kepler law gives the time of revolution, but it true for circular orbit. Well, Earth's orbit is almost circular, so may be the result would be good?
Kepler's 3rd law applies to elliptical orbits, circular orbits being just a special case. This is very close to what you want. A slight refinement due to Newton says you need to account for the mass of the planet as well. With this slight modification,
[tex]P=2\pi\sqrt{\frac {a^3}{G(M_s+M_p)}} = 2\pi\sqrt{\frac {a^3}{GM_s(1+M_p/M_s)}}[/tex]
There's a slight problem with this expression. G and the sun's mass are each known to a measly four decimal places. The product of the two is known to nine places. It's better to use the product, denoted as [itex]\mu_s[/itex] rather than G and Ms. This yields
[tex]P=2\pi\sqrt{\frac{a^3}{\mu_s(1+M_p/M_s)}}[/tex]
Jun7-12, 06:54 PM   #7
 
Quote by D H View Post
Kepler's 3rd law applies to elliptical orbits, circular orbits being just a special case. This is very close to what you want. A slight refinement due to Newton says you need to account for the mass of the orbiting object as well. With this slight modification,
[tex]P=2\pi\sqrt{\frac{a^3}{G(M_s+M_e)}}=2\pi\sqrt{\frac{a^3}{GM_s(1+M_e/M_s)}}[/tex]
Oh, thank you very much :) That is I was looking for :)
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