### Keplers Equation: Eccentric Anomaly Values at the Quarter-period

In this post I present some information about the solution to Kepler’s Equation of Elliptical Motion (KE) at the point t = T/4 (where T is the period of the motion). I will refer to E at t = T/4 as E

_{T/4}.

If

*e*is 0, meaning a circle, the solution reduces to the usual value for sinusoidal motion at the quarter period: π/2. However, for elliptical motion, E is greater than π/2 at this time.

According to KE,

E = M +

*e*sin(E)Since the magnitude of sin(E) is between 0 and 1, that means E

_{T/4}is restricted to the range M < E < (M +

*e*):

π/2 < E

_{T/4}< π/2 + 1In fact, over the range

*e*= 0 to

*e*= 1, sin(E

_{T/4}) ranges from 1 to cos(D),

or D, where D is the "Dottie Number", the solution to the equation

cos(D) = D

(D is an irrational number; it does not have an exact value, but must be calculated numerically to as many digits as required: 0.73908513321516064...)

In other words, the range of E

_{T/4}is further restricted:

π/2 < E

_{T/4}< π/2 + DNow recall the sine addition formula:

sin(θ

_{1}+ θ_{2}) = sin(θ_{1}) cos(θ_{2}) + cos(θ_{1}) sin(θ_{2})Therefore, sin(π/2 + D) = cos(D).

So, sin(E

_{T/4}) is restricted to the range

cos(D) < sin(E

_{T/4}) < 1This suggests a function that could bracket sin(E

_{T/4}) over the range

*e*= 0 to

*e*= 1. In fact, for the first graph below, the following functions were included to bracket the curve of sin(E

_{T/4}):

i) a simple straight-line function through the endpoints: 1 - m*

*e*, where m = 1 - D = 0.260914866785, and

ii) cos(D*

*e*).

Both these functions are equal to sin(E

_{T/4}) at the endpoints.

In other words,

(1 - m*

*e*) <= sin(E_{T/4}) <= cos(D**e*)Some notes for the following image:

i) sin(E

_{T/4}) is labeled sQ (the red line);

ii) y1 = 1 - m*

*e*(the green line);

iii) cDe = cos(D*

*e*) (the blue line).

Alternately, consider cos(E

_{T/4}).

cos(E

_{T/4}) goes from 0 for

*e*= 0 to -sin(D) at

*e*= 1. Note that the following graph depicts NEGATIVE cos(E

_{T/4}) (it is my personal preference to work with positive numbers.)

Notes for this image:

i) -cos(E

_{T/4}) is labeled ncQ (the red line).

ii) y1 = m*

*e*(the green line). This time m = sin(D) = 0.673612029183.

iii) sDe = sin(D*

*e*) (the blue line).

Some numerical data for

*e*= 0.786151377748:

E

_{T/4}= 2.20436934578188 rad,

cos(E

_{T/4}) = -0.592028088783,

sin(E

_{T/4}) = 0.805917329564,

tan(E

_{T/4}) = -1.36128225136,

√(1 -

*e*²) sin(E

_{T/4}) = 0.498084301804

Labels: elliptical motion, Keplers Equation