Kepler's Equation: A Listing of Some Derivatives

This post serves as a compilation of some derivatives for Kepler's Equation of Elliptical Motion. Nothing fancy here. Just posting some basic characteristics of the motion that might be useful later.

For a start, here are the M and E curves plotted on the same graph:

From Kepler's Equation of Elliptical Motion, the derivative of E with respect to (w.r.t.) t is found:

For the model I've been using throughout this work, ω = 1, so this derivative can be simplified further. In fact, I'm going to examine derivatives w.r.t. M instead of time. RETURNING TO A GENERIC FORM IN WHICH ω IS NOT 1, WE HAVE TO REMEMBER TO PUT ω BACK IN.

Here is the derivative of E w.r.t. M:

Here is a graph of dE/dM:

Here is the second derivative of E w.r.t. M:

Here is the third derivative of E w.r.t. M:

Each term in the numerator of this expression has a factor of e; when e is 0, the third derivative is 0.

This derivative also has a zero at a non-intuitive point. To find this root, rearrange this equation so that it is solely in terms of cos(E) or sin(E). The simpler approach is to express it in terms of cos(E):

In this form, it is a quadratic expression in terms of cos(E); solving it yields two roots. One root is complex, which is meaningless in the present context. The other root follows:

Numerically, this corresponds to:
cos(E) = 0.947351608247
E = 0.325935469297
sin(E) = 0.320195144173
M = 0.074213615557

I haven't seen these numbers up to now. (I wonder what their significance is in terms of the motion?)

Carrying on, here is the fourth derivative of E w.r.t. M:

The numerator of this expression is already in the form of a quadratic equation in terms of cos(E). Again, one root is complex, which is meaningless in the present context. The other root follows:

Numerically, this corresponds to:
cos(E) = 0.85192985408
E = 0.551136659451
sin(E) = 0.523655921123
M = 0.139463835594

These derivatives may be useful in creating a Taylor polynomial and/or creating interpolating functions over certain intervals.

One comment I would like to make immediately is that, in using the quotient rule to perform the differentiation, the term (1 - e cos(E)) always appears in the numerator before simplification--it's not present in just the denominator. Simplifying the numerator may reveal some insight into the motion; however, if the derivatives are going to be used in a Taylor polynomial, and computed numerically, it may be a better idea to leave the (1 - e cos(E)) term as is. After all, this term has to be computed for the denominator anyhow. The better course of action may be to compute the term (1 - e cos(E)) once, save its value in a variable, and re-use it as needed. Just a thought. It appears in all the derivatives that follow too.

Some features to note:
(i) When the motion crosses the y-axis (i.e., E = π/2 rad),
dE/dt = ω (i.e., dE/dt is parallel to M; i.e., dE/dM = 1);
(ii) dE/dM is a maximum at t = 0, at which time it takes the value 1/(1 - e);
(iii) dE/dM is a minimum when M = E = π, at which time it takes the value 1/(1 + e).
(iv) from the second derivative of E w.r.t. M, the point of inflection for the E curve occurs when sin(E) is 0.

Carrying on, here are the cos(M) and cos(E) curves plotted on the same graph:

The red line is cos(E) and the blue line is cos(M).

The derivative of cos(E) w.r.t. M is found as

Here are plots of the first derivatives of the cosine curves on the same graph:

The second derivative of cos(E) w.r.t. M is

Here are plots of the second derivatives of the cosine curves on the same graph:

Note that the second derivative of cos(E) takes a zero value
when cos(E) = e; in other words, the point of inflection occurs when the motion is at the semi-latus rectum (M = 0.180371160753 rad).
Also note that when elliptical motion reduces to circular motion,
(i.e., e = 0), the point of inflection occurs at the usual times:
every time cos(M) = 0 (i.e., M = π /2, 3 π /2, etc.).

Now repeat the process for the sine curves. Here are the sin(M) and sin(E) curves plotted on the same graph:

The red line is sin(E) and the blue line is sin(M).

The derivative of sin(E) w.r.t. M is

Here are plots of the first derivatives of the sine curves on the same graph:

The second derivative of sin(E) w.r.t. M is

Note that this derivative is similar to the second derivative of E w.r.t. M; it's just missing the scaling factor e.

Here are plots of the second derivatives of the sine curves on the same graph:

Note that the second derivative of sin(E) takes a zero value
when sin(E) = 0; in other words, points of inflection occur each time M = 0, π, 2π, etc.
Again note that when elliptical motion reduces to circular motion,
(i.e., e = 0), the point of inflection occurs at the usual times:
every time sin(M) = 0 (i.e., M = 0, π , 2 π , etc.).

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.)

t (algebraic)		t (numeric)		E (algebraic)		sin(E) (algebraic)		sin(E) (numeric)
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

*1
Consider the vector, r_η, 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, r_η is at (x, y) = (0, √(1 - e²)).
As the motion proceeds, r_η 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(E_t=T/4), and the cos(E) expression is always greater than or equal to cos(E_t=T/4). They are equal when e = 0. This expression can serve as a bracketing function for sin(E_t=T/4).