After the discussion in the previous two installments, we have basically grasped the motion of the two-body problem. We have found all the integrals for the two-body problem when the orbit is an ellipse, provided the proof for common formulas such as the "vis-viva equation," and left some unanswered questions. Specifically, the final integral for when the orbit is a parabola or a hyperbola has not yet been found. Now we will solve these two problems. The key integral remains: \dot{r}^2 = \frac{2\mu}{r} - \frac{\mu a(1-e^2)}{r^2} - \frac{\mu}{a} \tag{12}
I. Parabolic Orbit
For a parabolic orbit, we have e=1 and a \to \infty, but the relationship at the periapsis remains unchanged. That is to say, the periapsis distance is still a(1-e)=p. Since a tends toward infinity, the term \frac{\mu}{a} can be ignored, so (12) can be simplified to: \dot{r}^2 = \frac{2\mu}{r} - \frac{2p\mu}{r^2} It is not difficult to derive: \sqrt{2\mu}dt = \frac{rdr}{\sqrt{r-p}} By letting R^2 = r-p, the above equation becomes: \sqrt{2\mu}dt = \frac{(R^2+p)d(R^2+p)}{R} = (2R^2+2p)dR Integrating both sides: \sqrt{2\mu}t = \frac{2}{3} R^3 + 2pR + K_3 Taking t=0 as the time of periapsis passage, we get K_3=0. Thus, the parabolic orbit problem is solved: \sqrt{\frac{\mu}{2}}t = \frac{1}{3} R^3 + pR, \quad r = R^2 + p \tag{16} (Note: This is different from the solution in tutorials like "Introduction to Celestial Mechanics" and other textbooks. This method directly solves for the length of the radius vector r and then finds the angle, which is the reverse of the original method. I personally find this method easier to understand and more in line with our thinking. Additionally, this is a cubic equation in R, and we do not need to find its root formula; we can just use iterative methods for specific data to find the corresponding R.)
II. Hyperbolic Orbit
For a hyperbola, we have e > 1 and a < 0, so we let A = -a to avoid taking the square root of a negative number. Equation (12) becomes: \dot{r}^2 = \frac{2\mu}{r} - \frac{\mu A(e^2-1)}{r^2} + \frac{\mu}{A} \tag{17} This leads to: \sqrt{\frac{\mu}{A}}dt = \frac{rdr}{\sqrt{(A+r)^2 - A^2 e^2}} This is similar to the equation for an elliptical orbit. The method is also to use substitution, but instead of trigonometric functions, we use "hyperbolic functions." Let A+r = Ae \cosh(E), then it can be transformed into: \sqrt{\frac{\mu}{A^3}}dt = (e \cosh E - 1)dE Integrating both sides: \sqrt{\frac{\mu}{A^3}}t = e \sinh E - E + K_4 Similarly, taking t=0 as the time of periapsis passage, we get K_4=0. Thus, we obtain Kepler’s equation for hyperbolic orbits: \sqrt{\frac{\mu}{A^3}}t = e \sinh E - E \tag{18} This is somewhat similar to Kepler’s equation for elliptical orbits and is also a transcendental equation.
Appendix: A brief introduction to hyperbolic functions
According to Euler’s representation, we can express the cosine as: \cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2} where e = 2.71828\dots and i^2 = -1. If we remove the "i" from the expression, we can define a new type of function called the "hyperbolic cosine function," denoted as \cosh. That is: \cosh \theta = \frac{e^{\theta} + e^{-\theta}}{2} Similarly: \begin{aligned} \sinh \theta &= \frac{e^{\theta} - e^{-\theta}}{2} \\ \tanh \theta &= \frac{\sinh \theta}{\cosh \theta} \end{aligned} Regarding derivatives: \begin{aligned} (\sinh \theta)' &= \cosh \theta \\ (\cosh \theta)' &= \sinh \theta \end{aligned} For more information, please refer to: Wikipedia - Hyperbolic functions
Finally, the two-body problem part ends here. There are other small issues such as how to solve the Kepler equation, but you don’t need to think of them as so complicated. For equations that are difficult to solve directly, there are general numerical methods in mathematics that can be used to find specific solutions of any desired accuracy; for astronomical applications, this is sufficient. Additionally, the theme "Equations and the Universe" may not be updated for a long time because I need to prepare for the Astronomy Olympiad and want to study the perturbation solutions of the three-body problem, so I hope everyone will wait. Of course, the blog will continue to be updated regularly.
Reprinting please include the address of this article: https://kexue.fm/archives/582
For more detailed reprinting matters, please refer to: Scientific Space FAQ