Difference between d/dt and d(theta)/dt? Why is it dr or ds/dt?

[Anonymous001]

TL;DR Summary

In a physics pdf about kepler laws of motion I found these things. I also solo-learned calculus recently so I am not used to notations.

So, first of all, why and how are we taking the derivative of the vector r or s as d/dt if t is not a parameter of the equations?

Second question is what is the difference between d/dt(r) and d(theta)/dt(r) and also between d/dt(s) and d(theta)/dt(s)? Like, both of these appear at the bottom of the second image (and further in the pdf) and I don't see a difference.

P.S.:I attached the whole file to this post just in case.

Edit: Also, why is the orbit described as R=(r*sin(theta);r*cos(theta))? Isn't that a circle? Shouldn't it be considered an ellipse, parabola or hyperbola or anything else? What I mean is that there is drawn an ellipse but the vector form describes a circle. Why?

 

[fresh_42]

So, first of all, why and how are we taking the derivative of the vector r or s as d/dt if t is not a parameter of the equations?

The planets are moving, i.e. they change position in time $t$. Therefore $r\, , \,s,$ and $\theta$ change in time, i.e. are functions dependent on time. It is implicitly assumed without noting $\vec{r}=\vec{r(t)}\, , \,\vec{s}=\vec{s(t)},$ and $\theta=\theta(t).$

Second question is what is the difference between d/dt(r) and d(theta)/dt(r) and also between d/dt(s) and d(theta)/dt(s)? Like, both of these appear at the bottom of the second image (and further in the pdf) and I don't see a difference.

$\dfrac{d}{dt} f(t) = \dfrac{df(t)}{dt} = f'(t) =\dot{f}(t)$ are all different notations of the derivative of the function $f=f(t)$ with respect to the variable $t$ that is usually used for time.

Here are even more notations:
https://www.physicsforums.com/insights/the-pantheon-of-derivatives-i/

Edit: Also, why is the orbit described as R=(r*sin(theta);r*cos(theta))? Isn't that a circle?

It is only a circle if the radius $r$ does not change in time.

Shouldn't it be considered an ellipse, parabola or hyperbola or anything else? What I mean is that there is drawn an ellipse but the vector form describes a circle. Why?

It describes an ellipse if the radius $r$ changes.

 

[Anonymous001]

The planets are moving, i.e. they change position in time $t$. Therefore $r\, , \,s,$ and $\theta$ change in time, i.e. are functions dependent on time. It is implicitly assumed without noting $\vec{r}=\vec{r(t)}\, , \,\vec{s}=\vec{s(t)},$ and $\theta=\theta(t).$$\dfrac{d}{dt} f(t) = \dfrac{df(t)}{dt} = f'(t) =\dot{f}(t)$ are all different notations of the derivative of the function $f=f(t)$ with respect to the variable $t$ that is usually used for time.

Here are even more notations:
https://www.physicsforums.com/insights/the-pantheon-of-derivatives-i/
It is only a circle if the radius $r$ does not change in time.It describes an ellipse if the radius $r$ changes.

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?

 

[fresh_42]

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.