# Proving Kepler's second law in a different way

1. Is it possible to find the linear equations for all of the 12 lines stretched from the ellipse's focus to the ellipse's perimeter in the Kepler's second law? If yes, how? (assume we have divided the year to 12 months).
2. Can Kepler's second law be generalized to all of ellipses (horizontal or vertical) with any given semi-major and semi-minor axes or it is only applicable to planet orbits?

I've tried to get a parametrical answer to the first question by letting one of the lines cross the ellipse focus (A point): $$A=(\sqrt{a^2-b^2},0)$$ and one point on the ellipse perimeter (M point): $$M=(m,\frac{b}{a}\sqrt{a^2-m^2})$$. my goal was to calculate $$m$$. so I found the area between $$m$$ and $$a$$ (semi-major axis) plus a triangle area with the base of: $$m-\sqrt{a^2-b^2}$$ and height of: $$\frac{b}{a}\sqrt{a^2-m^2}$$ with the help of integral; then I equaled the result to $$\frac{\pi{ab}}{12}$$. since I had $$\arcsin(\frac{m}{a})$$ in the result, I substituted $$m$$ with $$a\sin\theta$$ and the final result was $$\frac{\pi}{3}=\theta+\cos\theta\sqrt{1-(\frac{b}{a})^2}$$. then the $$\theta$$ was $$\frac{\pi}{3}-\sqrt{1-(\frac{b}{a})^2}<\theta<\frac{\pi}{3}+\sqrt{1-(\frac{b}{a})^2}$$ . now remember that $$m=a\sin\theta$$. so we have $$a\sin(\frac{\pi}{3}-\sqrt{1-(\frac{b}{a})^2}).

Is this a valuable way?

• I don't understand what those 12 lines are supposed to be. Could you explain, please? Dec 3, 2019 at 17:14
• Kepler has said that when a planet orbits its sun, the area that is made in each month are equal. each year has 12 months. the sun is located on the ellipse's focus point (the orbit is an ellipse, not a circle). @Aretino Dec 3, 2019 at 18:37
• Kepler's law states that "a line joining a planet and the Sun sweeps out equal areas during equal intervals of time". The intervals of time needn't be months, they can be any length. And months, after all, don't last all the same time. Dec 3, 2019 at 21:13
• In addition, I don't understand your second question: Kepler's second law has to do with motion, it isn't a geometrical property. It holds for any motion derived from a central force, not necessarily gravitation. It is in fact equivalent to the conservation of angular momentum. Dec 3, 2019 at 21:23

Let $$c=\sqrt{a^2-b^2}$$, $$e=\frac{c}{a}$$ to simplify notation. As the center of the ellipse is at the origin and the center of gravity in the focal point $$(c,0)$$, the angle $$E$$ named "eccentric anomaly" parametrizes the ellipse as $$x=a\cos E$$, $$y=b\sin E$$.

The 12 sections of the ellipse are a representation of the 12 months of the year and serve to visualize the different speeds along the orbit, fast at the pericenter and slow at the apocenter. The connection between time and angle is given by the "Kepler equation" $$2\pi\frac{t}{T}=M=E-e\sin(E)$$ What you want is to set $$t_k=k\frac{T}{12}$$, $$k=0,1,..,11$$ and solve this equation to find $$E_k$$ $$\frac{k\pi}{6}=M_k=E_k-e\sin(E_k)$$ There are multiple ways to approximate this equation. For a numerical solver a decent starting point is $$E_k\approx M_k+e\sin(M_k)$$. Doing just this to produce an example plot gives

from scipy.optimize import fsolve
a=5; e=0.6; c=a*e; b=(a**2-c**2)**0.5

plt.figure(figsize=(8,4))
plt.gca().set_aspect("equal")

phi = np.linspace(0,2*m.pi,300)
plt.plot(a*np.cos(phi),b*np.sin(phi),'b', lw=3)
for k in range(-6,6):
Mk = m.pi*k/6
Ek = fsolve(lambda E: Mk-E+e*m.sin(E),Mk+e*m.sin(Mk))
plt.plot([c,a*m.cos(Ek)],[0,b*m.sin(Ek)],'r',lw=2)
plt.plot([c],[0],'oy', ms=8)
plt.show()

• Thanks. are the areas of the 12 slices in the picture equal then? Apr 6, 2021 at 12:01
• Yes, visually equal and numerically equal up to the default accuracy of the fsolve algorithm. Apr 6, 2021 at 12:21