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I was studying solar geometry and I read the next equation

$$r=r_{0}\left(1+0,017\sin\left[\frac{2\pi(d_{n}-93)}{365}\right]\right) $$

where $r_{0}$ is the average distance between the Earth and the Sun and 0.017 is the eccentricity of the earth orbit, $d_{n}$ is the days.

My question is how to proof the equation and why the argument of the sine. I suppose that is for properties of the ellipse (the orbit of the Earth is a ellipse) but doesn't work.

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Your accept rate or 23% is very eccentric. You should try to round it much closer to 100% so that people will feel like trying to help you. –  DonAntonio Jan 4 '13 at 23:43
    
What does the equation describe? The distance between the Sun and the Earth? –  Neal Jan 4 '13 at 23:43
    
Neal: Yes, DonAntonio: excuse me but you could explain more about that please –  kEoz Jan 4 '13 at 23:51
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@DonAntonio indeed, the accept rate is far more eccentric than the ellipse described –  orlandpm Jan 5 '13 at 0:19
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@kEoz meta.stackexchange.com/questions/5234/… –  c.p. Jan 5 '13 at 0:21

1 Answer 1

The distance as a function of the angle from perihelion is $$ r=\frac{a\left(1-e^2\right)}{1+e\cos(\theta)} $$ where $a$ is the semi-major axis, $e$ is the eccentricity of the orbit, and $\theta$ is the angle from perihelion.

The relation of $\theta$ to $t$, the time since perihelion passage, is $$ \frac{2\pi t}{P}=\phi-e\sin(\phi) $$ and $$ \tan\left(\frac\theta2\right)=\sqrt{\frac{1+e}{1-e}}\tan\left(\frac\phi2\right) $$ where $P$ is the period of revolution ($\phi$ is called the eccentric anomaly).

These formulas are derived in planets.pdf and kepler.pdf.


The formula cited in the question is a first-order deferent approximation assuming that the anomalistic year is 365 days and the earth passes perihelion when $d_n=1.75$.

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