How does the earth revolve around the sun?

1 Answer
Mar 11, 2016

Answer:

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.