Question

How does the earth revolve around the sun?

Answer 1

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 `SE = (a(1-e^2))/(1+e cos(theta))`

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 = 2pi` and the time-period is 365.265363 days, nearly.

At `theta=0`, SE is the minimum ( perihelion ) = `a(1-e)` = 147.1 Million km, nearly. This happens around Jan 3.
At `theta=pi`, SE is the maximum (aphelion) = `a(1+e)` = 152.1 Million km, nearly. This happens around July 2.
`theta > 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.