# How does the earth revolve around the sun?

Mar 11, 2016

The path of revolution of the center E of the Earth, about the center S of the Sun, is an ellipse. S is a focus of this ellipse. Distance SE is governed by $S E = \frac{a \left(1 - {e}^{2}\right)}{1 + e \cos \left(\theta\right)}$

#### Explanation:

$\theta$ is the inclination of SE to the initial line $\theta = 0$.
Semi-major of the ellipse a = 149598262 km, nearly. The eccentricity of the ellipse e + 0.01671, nearly.

For one revolution, the period for $\theta = 2 \pi$ and the time-period is 365.265363 days, nearly.

At $\theta = 0$, SE is the minimum ( perihelion ) = $a \left(1 - e\right)$ = 147.1 Million km, nearly. This happens around Jan 3.
At $\theta = \pi$, SE is the maximum (aphelion) = $a \left(1 + e\right)$ = 152.1 Million km, nearly. This happens around July 2.
$\theta > \frac{\pi}{2}$, when SE = a. This is an end of the minor axis of the ellipse.

Importantly, S is away from the center C of the ellipse by CS = ae = 2.5 Million km, nearly.