I wonder if readers still remember that in the chapter "Equations and the Universe", BoJone wrote three full articles to study the two-body problem in celestial mechanics? Although a basic description of the two-body problem was provided, it remains just the tip of the iceberg. In several recent articles, BoJone has emphasized the significant role of "vectors". So, what major events occur when celestial mechanics meets vectors? Did Mars collide with Earth?
In the "Equations and the Universe" series, we derived the fundamental equation of the two-body problem at the beginning:
In the two-body problem, choosing one of the particles as the reference point, the equation of motion for the other particle can be expressed as: \ddot{\vec{r}}=-\frac{\mu \vec{r}}{r^3}
where \mu=G(M+m). Starting from this equation, we will use vectors and a bit of calculus to complete the solution of the two-body problem in celestial mechanics.
First, let us consider \vec{r}\times\dot{\vec{r}}=?. First, find the derivative of this expression: \frac{d}{dt}(\vec{r}\times\dot{\vec{r}})=\dot{\vec{r}}\times\dot{\vec{r}}+\vec{r}\times\ddot{\vec{r}}=0+\vec{r}\times\left(-\frac{\mu \vec{r}}{r^3}\right)=0
Only the derivative of a constant is zero. For vectors, only the derivative of a constant vector (constant in both direction and magnitude) is zero, so: \vec{r}\times\dot{\vec{r}}=\vec{k}
\vec{k} is a constant vector. This is actually a manifestation of the conservation of angular momentum. m\vec{r}\times\dot{\vec{r}} is the angular momentum of the particle relative to the reference particle. Since the torque is zero, angular momentum is conserved. This law, "translated" into the language of celestial mechanics, is "Kepler’s Second Law".
To find other constant vectors, let us consider \vec{k}\times\ddot{\vec{r}}. According to vector knowledge, we have: \vec{k}\times\ddot{\vec{r}}=(\vec{r}\times\dot{\vec{r}})\times\left(-\frac{\mu \vec{r}}{r^3}\right)=-\frac{\mu}{r^3}[(\vec{r}\cdot \vec{r})\dot{\vec{r}}-(\vec{r}\cdot \dot{\vec{r}})\vec{r}]
And it is not difficult to prove: \vec{r}\cdot\dot{\vec{r}}=r\dot{r} (refer to the figure below):
Thus \vec{k}\times\ddot{\vec{r}}=-\mu\left(\frac{\dot{\vec{r}}}{r}-\frac{\dot{r}\vec{r}}{r^2}\right)=\frac{d}{dt}\left(-\frac{\mu\vec{r}}{r}\right).
Integrating both sides, we get: \vec{k}\times\dot{\vec{r}}=-\frac{\mu\vec{r}}{r}-\mu\vec{e}. Here -\mu\vec{e} represents a constant vector. We rewrite this equation as: \vec{k}\times\dot{\vec{r}}+\frac{\mu\vec{r}}{r}=-\mu\vec{e}
Next, we find the integral regarding velocity (vis-viva integral): \frac{d}{dt}\left(\frac{1}{2} \dot{\vec{r}}^2\right)=\dot{\vec{r}}\cdot \ddot{\vec{r}}=\dot{\vec{r}}\cdot \left(-\frac{\mu \vec{r}}{r^3}\right)=\frac{-\mu\dot{r}}{r^2}=\frac{d}{dt}\left(\frac{\mu}{r}\right)
Integrating both sides: \frac{1}{2} \dot{\vec{r}}^2=\frac{\mu}{r}+h, or written as: \frac{1}{2} v^2-\frac{\mu}{r}=h
Next, we have \vec{k}\cdot\vec{e}=\vec{k}\cdot\left(-\frac{\vec{k}\times\dot{\vec{r}}}{\mu}-\frac{\vec{r}}{r}\right)=0, which gives: \mu^2(e^2-1)=2hk^2
Where k=|\vec{k}| and e=|\vec{e}|. So far, we have used vector methods to derive the formulas related to energy and angular momentum in the two-body problem, laying the foundation for the next steps of calculation. The next step is to find the orbit equation, Kepler’s equation, and other solutions. This can be achieved in one go using vector knowledge. (To be continued....)
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