高级：椭圆轨道

Suppose that an object with mass $m$ orbits an object of mass $M$. Kepler's First Law of Planetary Motion states that the orbit is in the shape of an ellipse. We would like to know more about the nature of this orbit.

The point of closest approach toward object $M$ is called the perigee, and the point on the orbit farthest away from object $M$ is called the apogee.

Let us denote the following:

$a=$ semi-major axis

$b=$ semi-minor axis

$P=$ perigee

$A=$ apogee

$Q=$ location of mass $M$

$r=$ distance from $m$ to $M$

$v=$ speed of object $m$

It turns out that the equations for elliptical orbits are nearly identical to those of circular orbits. However, they are very difficult to derive, so I will not attempt to prove them here.

First, a very simple and obvious fact: $$OP+OA=PA=2a$$ Now, we have the period $T$: $$T^2=\frac{4\pi^2a^3}{MG}$$ Then, we have the total energy: $$E_{total}=K+U$$ $$E_{total}=\frac{1}{2}mv^2-\frac{mMG}{r}$$ It turns out that this total energy can also be written in a single term: $$E_{total}=\frac{1}{2}mv^2-\frac{mMG}{r}=-\frac{mMG}{2a}$$ Now, by setting this total energy to zero, we may figure out the escape velocity at a particular point: $$E_{total}=0$$ $$\frac{1}{2}mv^2-\frac{mMG}{r}=0$$ $$v_{esc}=\sqrt{\frac{2MG}{r}}$$ Using these equations and a set of initial conditions, one can calculate everything for a given elliptical orbit.

This post is about the vis-viva equation, which states that the velocity of an object of mass $m$ orbiting an object of mass $M$ at a distance of $r$ is, $$v = \sqrt{ GM \left( \frac{2}{r} - \frac{1}{a} \right) }$$ Here, $a$ is the semi-major axis of the orbit. Note that $a$ is infinite for parabolas and negative for hyperbolas.

We will now prove this equation. First, we will show that the total energy in an orbit is $-\frac{GMm}{2a}$.

We will prove this for elliptical orbits since the result is the same for others. Let $r_1$ and $r_2$ be the distances of the vertices of the major axis from the object it is orbiting. Let $L$ be the angular momentum. From energy conservation, we have, $$\frac{L^2}{2mr_1^2} - \frac{GMm}{r_1} = \frac{L^2}{2mr_2^2} - \frac{GMm}{r_2}$$ From this, we can simplify that $\frac{L^2}{2mr_1r_2} = \frac{GMm}{a}$. If we add the two energies at this points and let the total energy be $E$, we get, $$2E = \frac{L^2}{2m}\left( \frac{1}{r_1^2} + \frac{1}{r_2^2} \right) - GMm \left( \frac{1}{r_1} + \frac{1}{r_2} \right)$$ Substituting what we know, we get, $2E = - \frac{GMm}{a}$, which means that $E =- \frac{GMm}{2a}$.

Now, by energy conservation, $$\frac{1}{2} m v^2 - \frac{GMm}{r} = - \frac{GMm}{2a}$$ Solving, we get $v^2 = GM \left( \frac{ 2}{r} - \frac{1}{a} \right)$, which reduces to the equation above.