*The painter ChatGPT “A detailed depiction of a stellar system with a bright star at the center, surrounded by several planets.”
Introduction
In the article on Kepler’s equations:
it was explained that the solution to
\[ \small \begin{align*} &M(t) = E(t)-\epsilon\sin E(t) \\ &M(t) = 2\pi\frac{t}{T} = \sqrt{GM}a^{-3/2}t \end{align*} \]
is
\[ \small E(t) = M(t)+2\sum_{n=1}^\infty \frac{J_n(n\epsilon)}{n} \sin nM(t). \]
This was derived from the fact that
\[ \small \epsilon \sin E(t) =2\sum_{n=1}^\infty \frac{J_n(n\epsilon)}{n} \sin nM(t) \]
can be expanded in Fourier form. The subject of this article is to consider how to directly calculate \(\small \cos E(t)\), \(\small r(t),x(t),y(t)\) in equation of motion:
\[ \small \begin{align*} &r(t) = \frac{a(1-\epsilon^2)}{1+\epsilon \cos \omega(t)} \\ &x(t) = r(t)\cos \omega(t) \\ &y(t) = r(t)\sin \omega(t) \\ &z(t) = 0 \\ &\frac{d\omega}{dt} = \frac{|L|}{mr^2} \end{align*} \]
and their time derivatives using series in a similar manner.
Orbital Radius Series
In the section on Kepler’s equations, we stated that since
\[ \small \begin{align*} &r(t) = a(1-\epsilon\cos E(t)) \\ &E(t) = M(t)+2\sum_{n=1}^\infty \frac{J_n(n\epsilon)}{n} \sin nM(t), \end{align*} \]
we can calculate the path of \(\small r(t)\) by substituting the second line into \(\small E(t)\). However, this is difficult to handle because the series is included in the argument of the \(\small \cos\) function. Therefore, we consider expanding this equation as a series. Similarly to \(\small \sin E(t)\), it can be expanded into a Fourier cosine expansion:
\[ \small \frac{r}{a} = 1- \epsilon \cos E(t) = B_0+\sum_{n=1}^\infty B_n \cos nM(t). \]
By performing an inverse Fourier transform to find the coefficients,
\[ \small \begin{align*} B_0 &= \frac{1}{\pi}\int_0^\pi(1-\epsilon \cos E(t))dM(t) = \frac{1}{\pi}\int_0^\pi(1-\epsilon \cos E(t))\frac{dM(t)}{dE(t)}dE(t) \\ &= \frac{1}{\pi}\int_0^\pi(1-\epsilon \cos E(t))^2dE(t)=1+\frac{1}{2}\epsilon^2 \end{align*} \]
and
\[ \small \begin{align*} B_n &= \frac{2}{\pi}\int_0^\pi (1-\epsilon \cos E(t))\cos nM(t)dM(t) \\ &= \left[\frac{2(1-\epsilon \cos E(t))\sin nM(t)}{n\pi} \right]_0^\pi+\frac{2}{n\pi}\int_0^\pi \sin nM(t) \frac{d(\epsilon \cos E(t))}{dM}dM \\ &= -\frac{2\epsilon}{n\pi}\int_0^\pi \sin (nE(t)-n\epsilon\sin E(t)) \sin E(t) dE \\ &= -\frac{2\epsilon}{n}J’_n(n\epsilon) \end{align*} \]
can be obtained. Hence,
\[ \small r(t) = a +\frac{1}{2}a\epsilon^2-2a\epsilon \sum_{n=1}^\infty \frac{J’_n(n\epsilon)}{n} \cos nM(t) \]
holds true. The derivative of the Bessel function can be calculated as:
\[ \small J’_n(x) = \frac{d}{dx}J_n(x) = \frac{J_{n-1}(x)-J_{n+1}(x)}{2}. \]
Since
\[ \small r(t) = a(1-\epsilon\cos E(t)), \]
by inverting the equation, we can obtain
\[ \small \cos E(t) = \frac{a-r}{a\epsilon} = -\frac{1}{2}\epsilon+2 \sum_{n=1}^\infty \frac{J’_n(n\epsilon)}{n} \cos nM(t). \]
Coordinate Series
As explained in the article on Kepler’s equations, \(\small x(t),y(t)\) can be expressed as:
\[ \small \begin{align*} &x(t) = r\cos \omega(t) = a\cos E(t)-a\epsilon \\ &y(t) = r\sin \omega(t) = a\sqrt{1-\epsilon^2}\sin E(t), \end{align*} \]
so by substituting the equation, it can be expressed as
\[ \small \begin{align*} &x(t) = -\frac{3}{2}a\epsilon+2a \sum_{n=1}^\infty \frac{J’_n(n\epsilon)}{n} \cos nM(t) \\ &y(t) = \frac{2a\sqrt{1-\epsilon^2}}{\epsilon}\sum_{n=1}^\infty \frac{J_n(n\epsilon)}{n} \sin nM(t).
\end{align*} \]
Differentiating with respect to \(\small t\) gives us
\[ \small \begin{align*} &\frac{dx(t)}{dt} =-\frac{4a\pi}{T} \sum_{n=1}^\infty J’_n(n\epsilon) \sin nM(t) \\ &\frac{dy(t)}{dt} = \frac{2a\sqrt{1-\epsilon^2}}{\epsilon}\frac{2\pi}{T}\sum_{n=1}^\infty J_n(n\epsilon) \cos nM(t).
\end{align*} \]
Differentiating again gives us
\[ \small \begin{align*} &\frac{d^2x(t)}{dt^2} =-a\frac{4\pi^2}{T^2} \sum_{n=1}^\infty 2nJ’_n(n\epsilon) \cos nM(t) \\ &\frac{d^2y(t)}{dt^2} = -\frac{a\sqrt{1-\epsilon^2}}{\epsilon}\frac{4\pi^2}{T^2}\sum_{n=1}^\infty 2nJ_n(n\epsilon) \sin nM(t). \end{align*} \]
In addition,
\[ \small \frac{dr(t)}{dt} = \frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{r}=\frac{4a\pi\epsilon}{T} \sum_{n=1}^\infty J’_n(n\epsilon) \sin nM(t) \]
must holds true.
We can also apply the series calculation from the previous section to
\[ \small \frac{a}{r} = \frac{1}{1- \epsilon \cos E(t)} = B_0+\sum_{n=1}^\infty B_n \cos nM(t), \]
and in this case it seems possible to calculate
\[ \small \frac{a}{r} = 1+2\sum_{n=1}^\infty J_n(n\epsilon) \cos nM(t). \]
Since
\[ \small \cos\omega(t) = \frac{a}{r}\frac{1-\epsilon^2}{\epsilon}-\frac{1}{\epsilon}, \]
we get
\[ \small \cos\omega(t) = -\epsilon+\frac{1-\epsilon^2}{\epsilon}\sum_{n=1}^\infty 2J_n(n\epsilon) \cos nM(t) \]
by substituting. It seems that \(\small \sin \omega(t)\) can also be calculated using a series, so
\[ \small \sin\omega(t) = \sqrt{1-\epsilon^2}\sum_{n=1}^\infty 2J’_n(n\epsilon) \sin nM(t) \]
holds true (I don’t understand how to derive it though.).
Since the series parts of \(\small \cos \omega(t)\) and \(\small \sin \omega(t)\) are the same as \(\small dx(t)/dt\) and \(\small dy(t)/dt\), substituting gives us
\[ \small \begin{align*} &\frac{dx(t)}{dt} =-\frac{2a\pi}{T} \frac{\sin \omega(t)}{\sqrt{1-\epsilon^2}} \\ &\frac{dy(t)}{dt} = \frac{2a\pi}{T}\frac{\cos\omega(t)+\epsilon}{\sqrt{1-\epsilon^2}}. \end{align*} \]
By calculating the square of the velocity from this equation, we get
\[ \small v^2(t) =\left(\frac{dx(t)}{dt}\right)^2+\left(\frac{dy(t)}{dt}\right)^2 = \frac{4a^2\pi^2}{T^2}\frac{1}{1-\epsilon^2}(2(1+\epsilon\cos \omega(t))+(\epsilon^2-1)). \]
Substituting \(\small T = 2\pi a^{3/2}/\sqrt{GM}\) and rearranging, we get
\[ \small v^2(t) = GM\left(\frac{2}{r(t)}-\frac{1}{a} \right). \]
Therefore, The speed of the elliptical motion can be obtained as:
\[ \small v(t) = \sqrt{GM} \sqrt{\frac{2}{r(t)}-\frac{1}{a}}. \]
Conclusion
The equation of motion of a planet derived from gravitational potential can be obtained from a system of differential equations:
\[ \small \begin{align*} &m\frac{d^2x}{dt^2} = -\frac{GmM}{r^3}x = -\frac{GmM}{(x^2+y^2)^{3/2}}x \\ &m\frac{d^2y}{dt^2} = -\frac{GmM}{r^3}y = -\frac{GmM}{(x^2+y^2)^{3/2}}y. \end{align*} \]
In fact, the solution can be found to be:
\[ \small \begin{align*} &x(t) = -\frac{3}{2}a\epsilon+2a \sum_{n=1}^\infty \frac{J’_n(n\epsilon)}{n} \cos nM(t) \\ &y(t) = \frac{2a\sqrt{1-\epsilon^2}}{\epsilon}\sum_{n=1}^\infty \frac{J_n(n\epsilon)}{n} \sin nM(t).
\end{align*} \]
You may wonder why an asymmetric solution results from a symmetric fundamental equation. This is because the boundary conditions:
\[ \small x(0) = a(1-\epsilon), \quad y(0) =0 \]
are asymmetric. It should be noted that the calculations in this article were performed based on these boundary conditions, and that different boundary conditions would produce different results.
Reference
[1] Bowman, Frank, Introduction to Bessel Functions, Dover Publication Inc., 1958.
[2] Watson, George N., A Treatise on the Theory of Bessel Functions Second Edition, Cambridge University Press, 1944.
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