Examples in the Use of Kepler's Equation

We will show how to use Kepler's Equation which is

M = E - e sin E

to get the time, t, or position, n, in orbit. An addition relationships

cos n = [cos E - e]/[1 - e cos E]


cos E = [cos n + e]/[1 + e cos n]

will be needed.
EXAMPLES: An orbit with e = 0.20
  1. Determine the fraction of a period required to travel 1/6 of an orbit past perihelion (60o).
  2. Determine the angular position past perihelion after 1/6 of the orbital period.

Example 1.
n = 60o
cos E = (cos 60o + 0.2)/(1 + e cos 60o) = 0.46667
E = 1.08528 radians = 62o.18
M = 1.08528 - 0.20 sin(1.08528 radians) = 0.90839
t/P = M/(2p) = 0.14458

Example 2.
t/P = 1/6 = 0.16667
M = 2p/6 = p/3
Iterative Solution: Kepler's equation must be solved for the eccentric anomaly, E, but the equation is transandental and has no algebraic solution. The solution must be done numerically. As long as the eccentricity is not near 1, an interative solution as follows works well.

This is repeated until two succeeding values of E are the same within the precision desired.

Using the values of M and e for this example we get

Calculate the true anomaly:
cos(n) = [cos(1.2361) - 0.20]/[1 - 0.20 cos(1.2361)] = 1.3759
n = 1.4328 radians = 82o.09

Compare these results with a circular orbit, ie e = 0.
In that case the angle would be 60o and the time would be 0.16667 period.

The geometry of the example is shown in the diagram to the right. Note how little an eccentricity of 0.2 changes the shape of the ellipse from a circular shape. The biggest difference between the reference circle and the elliptical orbit is the center of reference. The focus of the ellipse is 0.2 of the radius from the center of the ellipse and circle. The angles and times calculated in these examples are shown here. In only 0.1446 of a period the orbiting object will be 60o from perihelion while if one waits 1/6 of a period it will be 82o from the perihelion.