In our previous discussion, we solved most of the problems and expressed the hope of finding r or \theta as a function of time t. In the final part of that discussion, we performed the following work:
From (7), we obtained \dot{\theta}=h/r^2. Substituting this into (6) yields: \ddot{r} -h^2/r^3=-\frac{\mu}{r^2} \tag{10} This is a second-order differential equation. Its solution is easy to find, but the integration is quite complex: \dot{r}\frac{d\dot{r}}{dr}=h^2/r^3-\frac{\mu}{r^2} \dot{r}d\dot{r}=(h^2/r^3-\frac{\mu}{r^2})dr. Integrating both sides: \dot{r}^2={2\mu}/r-h^2/r^2+K_1 \tag{11} \Rightarrow {dt}/{dr}=\frac{r}{\sqrt{K_1 r^2+2\mu r-h^2}} t=\int \frac{r}{\sqrt{K_1 r^2+2\mu r-h^2}}dr
The final integral is not difficult and can be calculated by looking up integral tables, but its form is extremely complex. Furthermore, it gives t as a function of r; substituting t to find r is inconvenient and not suitable for our purposes. However, Equation (11) remains an important integral. Let us first attempt to find K_1. By substituting a set of actual data, we can determine K_1. Let r=a(1-e) (i.e., at perihelion), then (\theta-\omega)=0, and: \dot{r}=\frac{dr}{d\theta}\cdot \frac{d\theta}{dt} According to Equation (9) from the previous session, we have: \frac{dr}{d\theta}=\frac{h^2/\mu}{[1+e\cos(\theta-\omega)]^2}\cdot e \sin(\theta-\omega) Since we take (\theta-\omega)=0, it follows that \frac{dr}{d\theta}=0 \Rightarrow \dot{r}=0. Substituting this result into (11) and replacing h^2 with \mu a(1-e^2), after simplification, we obtain: K_1=-\frac{\mu}{a} Thus, Equation (11) becomes: \dot{r}^2={2\mu}/r-{\mu a(1-e^2)}/r^2-\frac{\mu}{a} \tag{12}
Before deriving Kepler’s equation, let us perform some preparatory work. This includes deriving Kepler’s Third Law and the formula for linear velocity.
If the orbit of a planet is an ellipse, based on our previous derivation of Kepler’s First and Second Laws, we found that h is essentially twice the areal velocity. Within one period T, the area swept out is the area of the ellipse S=\pi ab=\pi a^2\sqrt{1-e^2}. Thus, h can be expressed as: h=\frac{2\pi a^2\sqrt{1-e^2}}{T} Combining this with h^2=\mu a(1-e^2), we get: \frac{a^3}{T^2}=\frac{\mu}{4\pi^2}=\frac{G(M+m)}{4\pi^2} \tag{13} This is Kepler’s Third Law. Here, the mass of the Sun M is constant, but the masses m of the various planets in the solar system are different. Therefore, strictly speaking, Kepler’s Third Law does not hold perfectly. However, since m is very small compared to M, this tiny difference can largely be ignored, which is why Kepler’s Third Law is still widely used!
Now let us derive the formula for "linear velocity." According to the Pythagorean theorem, we have: x^2+y^2=r^2 Differentiating both sides twice with respect to time t, we get: \dot{x}^2+\dot{y}^2+x\ddot{x}+y\ddot{y}=\dot{r}^2+r\ddot{r} Where \dot{x}^2+\dot{y}^2 is the square of the velocity v^2. According to the differential equations of the two-body problem, we have: x\ddot{x}+y\ddot{y}=-(\frac{\mu x}{r^3}x+\frac{\mu y}{r^3}y)=-{\mu}/r. Substituting this and the result of Equation (12) into the above equation, we obtain: v^2=\mu(2/r-1/a) \tag{14} This is the formula for linear velocity, known as the "Vis-viva Equation" or the "Energy Integral." It is an expression of the Law of Conservation of Energy (Mechanical Energy), indicating that the sum of a celestial body’s kinetic and potential energy is constant. In astronomical competitions, being able to skillfully apply this formula can be very convenient.
Now we come to our main focus: deriving what is known as "Kepler’s Equation." It is not difficult. From (12), we can obtain: \sqrt{\frac{\mu}{a^3}}dt=\frac{rdr}{a\sqrt{a^2 e^2-(a-r)^2}} Friends familiar with calculus will immediately think that for integrals containing a "square root of a sum or difference of squares," trigonometric substitution is generally used. Let a-r=ae \cos E, then the expression simplifies to: \sqrt{\frac{\mu}{a^3}}dt=(1-e \cos E)dE Integrating both sides yields: \sqrt{\frac{\mu}{a^3}}t=E-e \sin E+K_2.
Do you feel a sense of excitement? Yes, the "prototype" of Kepler’s equation has appeared. Following the same logic as before, substitute the data for perihelion to find K_2. When r=a(1-e), t=0 \Rightarrow E=0, thus we find K_2=0.
Furthermore, according to (13), we can change \sqrt{\frac{\mu}{a^3}} into {2\pi}/T, and let M={2\pi}/T \cdot t be called the "Mean Anomaly." Thus, Kepler’s Equation emerges: E=M+e \sin E \tag{15} In celestial mechanics, E is called the "Eccentric Anomaly," and f=(\theta-\omega) is called the "True Anomaly."
In traditional celestial mechanics textbooks, another formula is derived: \tan \frac{f}{2}=\sqrt{\frac{1+e}{1-e}}\tan \frac{E}{2}. However, in my view, this formula is not strictly necessary. We can simply solve for E, then find r, and subsequently find f according to Equation (9).
Is the work finished? Not yet. This formula can only be used for ellipses. Do not forget that in a parabola e=1 and a \to \infty, so Kepler’s equation cannot be used. In a hyperbola, the value of a is negative, so Kepler’s Third Law is also not applicable, requiring other forms. Additionally, regarding the solution of Kepler’s equation and other related issues, BoJone has some new results of his own. However, those will have to wait for the next installment.
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