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

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Discussion Overview

The discussion revolves around calculating the length of a year on Earth within a simplified two-body system consisting of the Sun and Earth. Participants explore both analytical methods and the implications of orbital characteristics, particularly focusing on Kepler's laws and their applicability to Earth's orbit.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant inquires about the analytical computation of the length of a year, seeking data or methods related to a two-body system.
  • Another participant humorously suggests that a year is simply a year, indicating a possible misunderstanding of the question's intent.
  • There is a clarification that the inquiry pertains to calculating the time of revolution based on orbital characteristics.
  • Participants discuss the applicability of Kepler's 3rd law, noting that it is valid for elliptical orbits and that Earth's orbit is nearly circular, which may yield satisfactory results.
  • A mathematical expression is presented, incorporating the mass of the planet and the Sun, with a suggestion to use the product of gravitational constant and solar mass for improved accuracy.
  • A later reply expresses gratitude for the mathematical formulation provided, indicating it aligns with the original inquiry.

Areas of Agreement / Disagreement

Participants generally agree on the relevance of Kepler's laws to the problem, but there is no consensus on the best method for calculation or the implications of the assumptions involved.

Contextual Notes

The discussion highlights the limitations of using Kepler's laws in the context of Earth's slightly elliptical orbit and the precision of constants involved in the calculations.

nikolafmf
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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,
[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]
 
Last edited:
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,
[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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