Anonymous001 said:
Thanks a lot, it really helped. I would have one more question as I went further into the proof. If ##r## changes in time, why does it say that we use a notation ##L=r^2*\frac{d\theta}{dt}## and ##L## is constant? Why is it constant if ##r## changes in time?
##L## does not only depend on ##r.## It also depends on ##\theta .## The combination is constant. It goes like this:
The velocity of the planet is a vector. In our case, a vector in the plane where the planet orbits the sun. It means: two-dimensional. The vectors ##\vec{r}## and ## \vec{s}## build a coordinate system of this plane. We can therefore express the vector ##\vec{v}## by its components, its parts of the vectors ##\vec{r}\, , \,\vec{s}.## The equation is
\begin{align*}
\vec{v}=\dfrac{dr}{dt} \cdot \vec{r} +r\cdot \dfrac{d\theta}{dt} \cdot \vec{s}\tag{2}
\end{align*}
Now, the change of velocity per time is acceleration. This yields (see the calculation in your .pdf)
\begin{align*}
\vec{a}=\underbrace{\left(\dfrac{d^2r}{dt^2}-r\cdot \left(\dfrac{d\theta}{dt}\right)^2\right)}_{=a_1(t)}\cdot \vec{r} +\underbrace{\left(2\cdot \dfrac{d\theta}{dt}\cdot\dfrac{dr}{dt}+r\cdot\dfrac{d^2\theta}{dt^2}\right)}_{=a_2(t)}\cdot \vec{s}\tag{*}
\end{align*}
where ##\dfrac{d^2r}{d^t}=\ddot{r}(t)## and ##\dfrac{d^2\theta}{dt^2}=\ddot{\theta}(t)## are the second derivatives with respect to time.
Next, we look at Newton's law of gravity, where ##G## is the gravitational constant and ##M## the mass of the sun, ##m## the mass of the Earth, and ##\vec{r}## the coordinate direction of the force.
$$
F=-\dfrac{G\cdot M\cdot m}{r^2} \cdot \vec{r}
$$
and his second law of motion, which states that a force is determined by
$$
F=m\cdot \vec{a}
$$
These two equations give us (##m## cancels)
\begin{align*}
-\dfrac{G\cdot M}{r^2}\cdot\vec{r}=\vec{a}
\end{align*}
This means we have two different ways to describe the acceleration
$$
\vec{a}=\left(-\dfrac{G\cdot M}{r^2}\right)\cdot \vec{r}+ 0\cdot \vec{s} =a_1(t)\cdot \vec{r}+a_2(t)\cdot \vec{s}
$$
However, the two directions ##\vec{r}## and ##\vec{s}## form a coordinate system of the plane of motion, the orbital plane. This means we can directly compare their components:
$$
\vec{a}=\underbrace{\left(-\dfrac{G\cdot M}{r^2}\right)}_{=a_1(t)}\cdot \vec{r}+ \underbrace{0}_{=a_2(t)}\cdot \vec{s}
$$
We thus have the two equations
\begin{align*}
-\dfrac{G\cdot M}{r^2}&=a_1(t)=\dfrac{d^2r}{dt^2}-r\cdot \left(\dfrac{d\theta}{dt}\right)^2\tag{3}\\
0&=a_2(t)=2\cdot \dfrac{d\theta}{dt}\cdot\dfrac{dr}{dt}+r\cdot\dfrac{d^2\theta}{dt^2}\tag{4}
\end{align*}
Finally, just for fun, let us differentiate ##r^2\cdot\dfrac{d\theta}{dt}.## This yields by the product (Leibniz) and chain rule
\begin{align*}
\dfrac{d}{dt}\left(r^2\cdot\dfrac{d\theta}{dt}\right)&=\dfrac{d}{dt}(r^2(t))\cdot \dfrac{d\theta}{dt} +r^2(t)\cdot\dfrac{d}{dt}\left(\dfrac{d\theta}{dt}\right)\\
&=2\cdot r(t)\cdot \dfrac{d}{dt}r(t)\cdot\dfrac{d\theta}{dt}+r^2(t)\cdot\dfrac{d^2\theta}{dt^2}\\
&=r(t)\cdot \left(2\cdot \dfrac{d}{dt}r(t)\cdot\dfrac{d\theta}{dt}+r(t)\cdot\dfrac{d^2\theta}{dt^2}\right)\\
&=r(t)\cdot a_2(t) = r(t)\cdot 0 = 0
\end{align*}
This means that ##L:=r^2(t)\cdot\dfrac{d\theta}{dt}## has a zero derivative, i.e. is constant in time. Neither the radius nor the angle is, but ##L## is. Only constant functions have zero derivatives.
