How to Compute the Length of a Year on Earth in a Two-Body System?

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To compute the length of a year on Earth in a two-body system with the Sun, Kepler's 3rd law can be applied, which is valid for both circular and elliptical orbits. The formula for the orbital period (P) incorporates the semi-major axis (a) and the masses of the Sun (M_s) and Earth (M_e). A more precise calculation uses the gravitational parameter (μ_s) instead of the individual constants for G and M_s. The discussion emphasizes that while Earth's orbit is nearly circular, the elliptical nature can still be accounted for in calculations. This analytical approach allows for a comparison with numerical results previously obtained.
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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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uhh?? :confused:

it's a year! o:)
 
Do you mean, given its orbital characteristics, could you calculate its revolution about the sun from first principles?
 
tiny-tim said:
uhh?? :confused:

it's a year! o:)

Well, say, the length of a year (i.e. one revolution) in seconds...
 
DaveC426913 said:
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.
 
nikolafmf said:
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,
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)}}
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 \mu_s rather than G and Ms. This yields
P=2\pi\sqrt{\frac{a^3}{\mu_s(1+M_p/M_s)}}
 
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D H said:
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,
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)}}

Oh, thank you very much :) That is I was looking for :)
 
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