Proving Kepler's 1st Law

Question

I am currently taking a Multivariable Calculus class, and my professor says there are going to be proofs on the next exam. He says that proving one of the three Kepler's law is going to be on it, but I do not know how to prove Kepler's 1st Law.

Kepler's First Law states that When orbiting, the orbited object (i.e. the Sun) is at one of the focus of the elliptical orbit.

This is what I started with.

$\vec{a}\times\vec{h}=\frac{-GM}{r^2}\vec{u}\times\left(r^2\ \vec{u}\times\vec{u'}\right)=-GM\left[\left(\vec{u}\cdot\vec{u'}\right)\vec{u}-(\vec{u}\cdot\vec{u})\vec{u'}\right]=GM\vec{u'}$ because $\vec{u}\cdot\vec{u'}=0$ and $\vec{u}\cdot\vec{u}=1$ because $\vec{u}$ is a unit vector.

I have gotten this far, but I do not know how to continue from here. Can anyone steer me along the right direction and tell me if I have the correct idea?

Answer 1

I assume that:

• $\vec r=r\vec u$ is the position vector pointing from the Sun to a planet.
• $\vec v=(\vec r)'$ is the velocity of the orbiting object.
• $\vec a=(\vec v)'$ is the acceleration of the orbiting object.

Also, a property (Triple product) from vector calculus:

• For vectors $\vec a, \vec b, \vec c\in\Bbb R^3$ we have $$\vec a\cdot(\vec b\times\vec c)=\vec c\cdot(\vec a\times\vec b)=\vec b\cdot(\vec c\times\vec a)$$

---

You have found that $$\vec a\times(\vec r\times\vec v)=GM(\vec u)'\tag{1}$$

The trick is to show that the vector $\vec r\times\vec v$ is constant. Notice that $$(\vec r\times\vec v)'=\vec v\times\vec v+\vec r\times\vec a\tag{2}$$ By Netwons 1. law you already know that $\vec r,\vec a$ are parallel. This, with the fact that $\vec v\times\vec v=0$ yields $(\vec r\times\vec v)'=0$, so the vector is constant, say $\vec r\times\vec v=\vec h$. This will help us to work on eq. (1). Since $\vec h$ is constant, we have $$(\vec v\times\vec h)'=GM(\vec u)'$$ In other words $$(\vec v\times\vec h-GM\vec u)'=0\tag{3}$$

This means that the vector $\vec v\times\vec h-GM\vec u$ is also constant, so lets make $$\vec v\times\vec h-GM\vec u=\vec c\tag{4}$$ Now, let:

• $z=0$
• $\vec c$ be aligned along the $x$-axis, that is, $\vec c=(c,0,0)$
• $\vec r=(r\cos(\theta),r\sin(\theta),0)$

Consider $\vec r\cdot (\vec v\times\vec h)=\vec h\cdot(\vec r\times\vec v)$ (equality follows from the triple product), and compute it in two ways. We have:

• $\vec r\cdot(\vec v\times\vec h)=\vec r\cdot(\vec c+GM\vec u)=rc\cos(\theta)+rGM=r(\cos(\theta)+GM)$

• $\vec h\cdot(\vec r\times\vec v)=\vec h\cdot\vec h=|h|^2$

Combining the two bullets, we arrive at $$r=\frac{|h|^2}{GM+c\cos(\theta)}\tag{5}$$ Eq. (5) describes an ellipse.

---

Source:

• https://mathcs.clarku.edu/~djoyce/ma131/kepler.pdf