David's Blog

Living a quiet life in Coquitlam, B.C.

Location: Coquitlam, British Columbia, Canada

Sunday, October 21, 2012

Kepler’s Equation: A Summary of Exact Points

I enjoy investigating Kepler’s Equation of Elliptical Motion (KE) and plan to present some information over the course of several posts. I would like to start by posting a tabulation of some exact points for motion on an ellipse. A diagram of the relevant characteristics of an ellipse are depicted at the following link: Kepler's Equation of Elliptical Motion

In the table below, data points are the result of the following procedure:

i) A value for E or sin(E) was specified.

ii) Then KE was used to work backwards to find the time at which this value is reached: ωt = M = E - e sin(E).
In fact, the first column in this table, t (algebraic), simply repeats this expression instead of writing out each point explicitly.

Note that several of the points included in the table are significant in the literature about KE, for example, the point at which E = π/2 radians. Knowing this point, it is easy to work backwards to find the time at which this value was reached:

E - e sin(E) = M
π/2 - e * sin(π/2) = ω*t
π/2 - e = t

In other words, t = π/2 - e

Unfortunately, this does not do us much good in terms of finding a general solution of KE in terms of time. This result isn't very useful by itself--unless we happen to want E at exactly this time. In fact, we could compute as many exact solutions to KE as we wanted in the same way (by working backwards). Again, these results aren't very useful by themselves unless we wanted the values of E at exactly those times. In any case, I have compiled several data sets in the table below. Perhaps this information will be useful for future work.

To better visualize the progress of the motion, some numerical results are also included in the table. These numbers were computed using the following parameters:
eccentricity: e = 0.786151377748
radial velocity of the mean anomaly, M: ω = 1.
(In other words ω = 2π/T = 1, where T is the period of the motion.)


E - esin(E)     0.0916     E     (1 - e²)1     0.3820
E - esin(E)     0.1254     E     (1 - e²)¾     0.4859
E - esin(E)     0.1305     π/6     1/2     0.5
E - esin(E)     0.1804     E     (1 - e²)½     0.6180
E - esin(E)     0.2292     E     cos(e)     0.7066
E - esin(E)     0.2295     π/4     √2/2     0.7071
E - esin(E)     0.2865     E     (1 - e²)¼     0.7862
E - esin(E)     0.3664     π/3     √3/2     0.8660
π/2 – e     0.7846     π/2     (1 - e²)0     1
E - esin(E)     1.2039     E     sin(EsE=sM)     0.9334
E - esin(E)     1.2250     E     sech(e/2)     0.9274
E - esin(E)     1.2377     E     cos(e/2)     0.9237
E - esin(E)     1.2519     E     √[1 - (e/2)2]     0.9195
E - esin(E)     1.4772     E     *1     0.8426
T/4     1.5708     E     sin(Et=T/4)     0.8059

Consider the vector, rη, with origin at the focus (x, y) = (e, 0), that sweeps out equal-area-in-equal-time.
When E = π/2, M = π/2 - e. For e > 0, this happens before t = T/4.
At this time, rη is at (x, y) = (0, √(1 - e2)).
As the motion proceeds, rη crosses the y-axis and at some point intersects the y-axis at a point, h, above the x-axis where h = (1/2) * √(1 - e2).
Solving for the sin(E) and cos(E) values at this time yields:

This expression for sin(E) is always greater than or equal to sin(Et=T/4), and the cos(E) expression is always greater than or equal to cos(Et=T/4). They are equal when e = 0. This expression can serve as a bracketing function for sin(Et=T/4).

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