Solving Kepler's Equation Over Three Centuries

[Will Flannery]

TL;DR

'Solving Kepler's Equation Over Three Centuries' is the title of a book by Peter Colwell. The question is - are analytical solutions used in physics or engineering today ?

'Solving Kepler's Equation Over Three Centuries' is the title of a book by Peter Colwell. The book covers numerical methods starting with Kepler's method, followed by Newton's method, and many analytic methods starting with Lagrange's method and continuing up till the 1990s.

I spent many years in engineering working on rocket navigation and associated problems, and I didn't even know there were analytic solutions to Kepler's problem. Do analytic solutions play a role in any branches of engineering or physics today?

 

[phyzguy]

I think the Keplerian orbital elements are still an important method to calculate planetary positions, although there are higher accuracy methods available. This page from JPL gives you an idea how accurate the Keplerian calculations are.

 

[Will Flannery]

I think the Keplerian orbital elements are still an important method to calculate planetary positions, although there are higher accuracy methods available. This page from JPL gives you an idea how accurate the Keplerian calculations are.

Following the link in the JPL page leads to this page on the formulas used, https://ssd.jpl.nasa.gov/txt/aprx_pos_planets.pdf
and (apparently) the orbit position as a function of time, i.e. E as a function of M, is calculated using Newton's iterative method, Eq. 8-37 on the page.

I also note that Colwell sometimes compares analytic methods to Newton's method as a measure of their accuracy ... e.g. on pg. 132 Note: the analytic methods are all infinite series

With e chosen as in the table below we have calculated for each
Mj = ...
the quantities
Ek = ... formula for E using 10 terms of the Levi-Civita solution
and En, the solution by Newton's method iterated to the point where the successive iteration differ by less than 10^-10
B(e) = sum of errors for each Mj = sum 100[Ek-En]/En
Table with columns for e and Be

 

[phyzguy]

Kepler's equation, M = E - e \sin(E), cannot be solved analytically. Newton's method is just an iterative method for finding roots of equations, and is often used to find numerical solutions to equations. It usually converges quicly to high accuracy.

 

[Will Flannery]

Kepler's equation, M = E - e \sin(E), cannot be solved analytically.

That's what I thought too, but ... I was wrong. The best reference is Colwell's book, Lagrange was the first to produce an analytic solution. The basic idea is this - if you can expand $\phi(z)$ as a Taylor series, then you can invert the series to obtain $\phi^{-1}(z)$ The heuristic argument is given in ... from A Course in Modern Analysis, Whitaker pdf ...
Given $\phi(a) = b$ and $\phi'(a)<>0$ and a Taylor series expansion
$\phi(z) - b = \phi'(a)(z - a) + \frac{\phi''(a)}{2!}(z-a)^2 + ...$
then
$z - a =\frac{\phi(z) - b}{\phi'(a)} - \frac 1{\phi'(a)}\frac{\phi''(a)}{2!}(z-a)^2 + ...$
with
$\phi'(z) = \frac{\phi(z)-b}{z-a}$
and
$z - a =\frac{\phi(z) - b}{\phi'(a)} - \frac 1{\phi'(a)}\frac 1{2!}\frac{\phi^{(2)}(z)}{\phi'(z)^2}(\phi(z)-b)^2 - ...$

I think that's right (Whitaker is overly concise). In any case, Whitaker follows with a proof giving the actual values of the series coefficients.

 

[gneill]

It might be noted that a series solution is not the same thing as a closed-form solution. It's sort of like an iterative solution written out term by term.

Since the original equation is transcendental in nature, I believe that this precludes a closed-form algebraic solution.

 

[phyzguy]

It might be noted that a series solution is not the same thing as a closed-form solution. It's sort of like an iterative solution written out term by term.

Since the original equation is transcendental in nature, I believe that this precludes a closed-form algebraic solution.

Exactly. A power series solution is not an analytic solution. Solving an equation analytically is usually taken to mean finding a closed form solution in terms of elementary functions. The solution given by Newton's method can also be written as a power series.

 

[Will Flannery]

I gave the wrong link in my previous post, the correct link is A First Course in Modern Analysis - Whitaker - pdf. This book was written in 1902 I think and is a fantastic book and appears to be a standard reference.

And ... the last line in my previous post should read ...
$z - a = \frac{\phi(a)-b}{\phi'(a)} - \frac{1}{\phi'(a)}\frac{1}{2!}\frac{\phi^{(2)}(a)}{\phi'(a)^2}(\phi(z) - b)^2 - ...$

 

[George Jones]

I believe that this precludes a closed-form algebraic solution.

Exactly. A power series solution is not an analytic solution. Solving an equation analytically is usually taken to mean finding a closed form solution in terms of elementary functions. The solution given by Newton's method can also be written as a power series.

What is the definition of $\sin x$?

 

[gneill]

What is the definition of $\sin x$?

$$sin(x) = \sum_{n=0}^\infty \frac{\left( -1 \right)^n}{\left( 2n + 1\right)!} x^{2n + 1}$$

 

[George Jones]

$$sin(x) = \sum_{n=0}^\infty \frac{\left( -1 \right)^n}{\left( 2n + 1\right)!} x^{2n + 1}$$

Yes.

Alternatively, $\sin x$ could be defined as the solution of the differential equation
$$\frac{d^2f}{dx^2} + f = 0$$
that satisfies the initial conditions $f\left(0\right) = 0$ and $f'\left(0\right) = 1$.

 

[vanhees71]

I gave the wrong link in my previous post, the correct link is A First Course in Modern Analysis - Whitaker - pdf. This book was written in 1902 I think and is a fantastic book and appears to be a standard reference.

This book is a gem! Whenever you need to understand some special functions, particularly the hypergeometric and confluent hypergeometric functions (which include most of the other types of special function you encounter in theoretical physics) there you find an answer.

 

[MathematicalPhysicist]

What is the definition of $\sin x$?

I'll need a right triangle for that...

 

[vanhees71]

I'd define it as
$$\sin x=\sum_{j=0}^{\infty} \frac{(-1)^j x^{2j+1}}{(2j+1)!}.$$
At Kepler's time they used tables for the trigonometric functions, and how these were calculated is very fascinating:

https://www.sciencedirect.com/science/article/pii/S0315086016000185

 

[mpresic3]

Summary:: 'Solving Kepler's Equation Over Three Centuries' is the title of a book by Peter Colwell. The question is - are analytical solutions used in physics or engineering today ?

I spent many years in engineering working on rocket navigation and associated problems, and I didn't even know there were analytic solutions to Kepler's problem. Do analytic solutions play a role in any branches of engineering or physics today?

Absolutely, they do. My mentor and I worked in this area and we agreed that when you can do a non-trivial problem, (the more non-trivial the better), you have a good way to check solutions from a computer model against a known solution. Any numerical analysis by computer (for example integration routines) could encounter round-off error, and other associated errors. Aside from this I could describe several cases where knowing analytical solutions pointed out problems in numerical methods (not just software bugs) put forward.

A non-trivial check case is worth it's weight in gold. Any difference in the numerical solution to the check case needs to be explained.

For example, (I won't tell of many cases I met), to give credit to another coworker. He told me that a routine that we were using to propagate Kepler's equation (not using integration), did not give the initial position and reverse velocity when taking the final conditions and running them backward in time. We went to the routine developer who fixed the method used in his routine, (it was not a simple software fix). Apparently, no one for years attempted to run the routine backward in time. (I personally could not believe this was overlooked but it turned out to be true).

IAs far as spending many years in engineering and not knowing the analytic solutions to Kepler's problem, I am not too surprised. Seems like many employers are choosing engineers over physicists. My first day of work over 35 years ago, now, I certainly was aware of that fact. I do not want to downplay the discipline of aerospace engineering, because I work with them, but physicists bring an lot to the table when teamed with them. I advised some managers that you do not need a dozen physicists, but you better have at least one on the project.