David's Blog

Living a quiet life in Coquitlam, B.C.

Location: Coquitlam, British Columbia, Canada

Sunday, November 04, 2012

Keplers Equation: Eccentric Anomaly Values when sin(E) = sin(M)

In this post I present some information about the solution to Kepler’s Equation of Elliptical Motion (KE) at the point at which sin(E) = sin(M). This point is significant because it is the only point at which the value of sin(E) might be found which is a linear expression in time. Plots of E and M intersect only at t = 0, π, and 2π. However, plots of sin(E) and sin(M) intersect at one more point, and because sin(M) operates on an operand that is linear in time, the value for sin(E) at this time would also be known. I will refer to M and E at this point as MsE=sM and EsE=sM, respectively.

According to KE, E = M + e sin(E)
Since sin(EsE=sM) = sin(MsE=sM), this equation can be re-written as follows:

EsE=sM = MsE=sM + e sin(MsE=sM)

The only time at which this condition holds is when E and M are symmetric around π/2. Also at this time, M + E = π.


EsE=sM - π/2 = π/2 – MsE=sM
= (e/2) sin(MsE=sM) = (e/2) sin(EsE=sM)

From STATEMENT 1, expressions can be made in terms of either EsE=sM or MsE=sM:

EsE=sM - (e/2) sin(EsE=sM) = π/2         EQUATION 1
MsE=sM + (e/2) sin(MsE=sM) = π/2       EQUATION 2

EQUATION 1 is very similar to KE itself, and can be solved numerically. A plot of sin(EsE=sM) is shown in the graph below.

From a diagram of this geometry, note that

cos((e/2) sin(MsE=sM)) = sin(MsE=sM)
sin((e/2) sin(MsE=sM)) = cos(MsE=sM)

The half-angle identities could now be used:

cos(B/2) = √((1 + cos(B))/2)
sin(B/2) = √((1 - cos(B))/2)


cos((e/2) sin(MsE=sM)) = √((1 + cos(e sin(MsE=sM)))/2)
sin((e/2) sin(MsE=sM)) = √((1 - cos(e sin(MsE=sM)))/2)

These forms don’t really help, because they are even more complicated equations to solve.
At this point, the most concise expression for sin(EsE=sM) is as the solution to the equation

cos[(e/2) θ] = θ, where θ = sin(EsE=sM).

This equation is analogous to the expression for sin(ET/4), which is as the solution to the equation

cos[e θ] = θ, where θ = sin(ET/4)

Some numerical data for e = 0.786151377748:
EsE=sM = 1.9377086781046062 rad,
MsE=sM = 1.20388397548 rad,
ηsE=sM = 2.290150905621918 rad,
rη = 0.792451718158,
sin(EsE=sM) = 0.933439441046,
cos(MsE=sM) = 0.358735013494,
√(1 - e²) sin(EsE=sM) = 0.576897301019

Also, on the graph below, note that a plot of cos(0.45e) looks very close to a plot of sin(EsE=sM).

Some notes for the following image:
i) sin(ET/4) is labeled sET4;
ii) sin(EsE=sM) is labeled sqEM;
iii) c1e = cos(0.45e);
iv) as e goes to 1, the equation goes to cos(θ/2) = θ,
    where θ = sin(MsE=sM). This equation is solved numerically:
    sin(MsE=sM) = 0.900367222589,
    cos(MsE=sM) = 0.435130859037,
    MsE=sM = 1.1206127155 rad,
    EsE=sM = 2.02097993809 rad.

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