Kep-Ler something something

Begin LogEntry_0017 
StarDate 197749.60


We spent our last log entry looking at our star on it's own, but now it's time to look at it with it's planets again, including this newly industrialised one.


We've spent some time trying to understand the works of the mathematician and astronomer Jo-hannes Kep-Ler, and will document our work in this log entry. 


The first law is quite evident from our earlier log entries, and simulations, it states:

The orbit of every planet is an ellipse, with the sun in one of the foci

This however is not entirely accurate, which was found with more modern equipment and methods than what was available to Kep-Ler, but given the equipment and data he worked with, it was as accurate as one could expect. In our work we figured out that they both orbit around a shared center of mass, but we will go into further detail about this in a later log entry.


Kep-Ler's 2nd law states that:

A line joining a planet and the Sun sweeps out equal areas during equal intervals of time

This sounds very technical, but it can be simplified with an illustration.

Bildet kan inneholde: linje, sirkel, parallell.
Example of areas swept out by the lines between the planet and the sun. Not to scale, the areas are just examples and are not equal.

He we can more easily see what is meant by this law, the change in area divided by the change in time should be constant, or more mathematically:

\(\frac{dA}{dt} = C\)

Where \(C\) is a constant. If we consider the infinitesimal area \(dA\) swept out by the change in position \(d\vec{r}\)over the infinitesimal timestep \(dt\), we can approximate it using a right angled triangle:

Bildet kan inneholde: tekst, linje, håndskrift, mønster, design.


From this we can easily see that the change in angle \(\theta\)\(d\theta\) is dependent upon the timestep \(dt\), and that we can solve this geometrically. 


In our case we can see that the shortest side can be approximated as the length of the longest side \(|\vec{r}|\), or \(r\), times the change in angle \(d\theta\). This can be done because the shortest side, let's call it \(x \), can be found using some Py-thago-rean geometry.


Sinus of the angle \(d\theta\) can be found by taking the opposite side and diving it by the adjacent side: \(sin(d\theta) = \frac{x}{r}\). We also know that the change in angle \(d\theta\) will be very small, and thus we can make the simplification \(sin(d\theta) \approx d\theta\), since \(lim_{\theta \rightarrow 0} sin(\theta) = \theta\). Now we can put this in and then rearrange to find that the shortest side is \(x = rd\theta\).

Now it's easy to find the area of the triangle, since it can be found with \(A = \frac{1}{2}lh\), where \(l\) is the length of the longer side and \(h\) is the length of the shorter side, or the height.

Putting this in we get that \(dA = \frac{1}{2}r^{2}d\theta\)


This however does not prove Kep-Ler's 2nd law, since it's not clear from this that \(\frac{dA}{dt}\) is constant. We have to look at the angular momentum per mass, \(h\), which is defined as \(h = |\vec{h}| = \frac{|\vec{r}\times \vec{p}|}{m_2} \), this can be further simplified as \(h = |\vec{r}\times \vec{v}| = |\vec{r}\times \dot{\vec{r}}|\). Taking this cross product, we get:

\(\begin{bmatrix} \hat{u}_{r} & \hat{u}_{\theta} &\hat{u}_{k} \\ r & 0 & 0 \\ 0 & r\dot{\theta} & 0 \end{bmatrix} = r^{2} \dot{\theta} \hat{u}_{k}\)


And then we get that \(h = r^{2}\dot{\theta}\). In our earlier work on the equations of motion we derived an expression on the form \(\frac{d}{dt} \Big(r^{2}\dot{\theta} \Big) = 0\), which we now can relate back to the angular momentum per mass \(h\), and see that it now is constant over time.

But how do we relate this to \(\frac{dA}{dt}\)?

Bildet kan inneholde: tekst, linje, mønster, design.


We can here see the line \(d\vec{S}\), which is the line between the points \(\vec{r}_{n}\) and \(\vec{r}_{n+1}\), this line can be approximated as \(d\vec{S} = \vec{v}dt\), and then we can find the area between \(\vec{r}_{n}\) and \(\vec{r}_{n+1}\) with the equation \(dA = \frac{1}{2} |\vec{r} \times d\vec{S}| = \frac{1}{2}|\vec{r} \times \vec{v}dt|\), we can simplify to \(\vec{r}\) since \(\vec{r}_{n} \approx \vec{r}_{n+1} \approx \vec{r}\).

Dividing both sides by \(dt\) gives us \(\frac{dA}{dt} = \frac{1}{2} |\vec{r}\times \dot{\vec{r}}| = \frac{1}{2}h \), and thus we have proven that \(\frac{dA}{dt}\) is constant in our system!


So we went back to the piles and piles of paper with orbital data, and checked that our simulations were consistent with Kep-Ler's 2nd law. 


Now we can move on to this Kep-Ler fellow's third law, it states:

The ratio of the square of an object's orbital period with the cube of the semi-major axis of its orbit is the same for all objects orbiting the same primary.

This statement mathematically means that the period \(P\) is \(P^{2} = a^{3}\), where \(a\) is the semimajor axis. This however is not entirely accurate, it is however very close and again amazing work given the tools Kep-Ler had to work with.


We can show this by integrating \(\frac{dA}{dt}\) for a full time period \(P\), this gives us \(\int_{0}^{P}\frac{dA}{dt}dt = \frac{1}{2} \int_{0}^{P} r^2 \dot{\theta} dt\), which with some basic integration gives us \(\frac{1}{2} P r^{2} \dot{\theta}\), this expression is then equal to the entire area of the ellipse, which is \(\pi ab\), where \(a\) again is the semimajor axis and \(b\) is the semiminor axis. 

We now have the expression \(\pi a b = \frac{1}{2}Pr^2 \dot{\theta}\), which can be rearranged to give us \(P = 2\pi \frac{ab}{r^{2}\dot{\theta}} = 2\pi \frac{ab}{h}\), to get further we have to use some expressions we found in some of our other work.

First we have \(M = G(m_1 + m_2)\), which is the gravitational constant \(G\) times the total mass of the system, then we have \(p\)(not momentum in this case), which is defined as the square of the angular momentum per mass, divided by \(M\), giving us \(p = \frac{h^2}{M}\). This variable \(p\), can also be shown to be equal to \(a(1-e^2)\), and lastly the semiminor axis \(b\) can be shown to be equal to \(a \sqrt{1-e^2}\).

If we now square our expression for the period \(P\), we get:

\(P^2 = 4\pi^2 \frac{a^2b^2}{h^2} = 4\pi^2 \frac{a^4 (1-e^2)}{ma(1-e^2)}\)

Which simplifies to:

\(P^2 = a^3\frac{4\pi^2 }{G(m_1 + m_2)} \)

From the definition of \(G\), and the fact that one mass will be significantly bigger than the other in our system, we can see that the \(\frac{4\pi^2 }{G(m_1 + m_2)}\) will be very close to 1, it would be 1 if we ignored the smaller mass. Not accounting correcting factor for this was the error in Kep-Ler's third law, but again, given the data and tools he was working with this error was practically impossible to find. When New1000kg came along some years later, his work on gravitation naturally led to Kep-Ler's laws, it confirmed and corrected them.


End LogEntry_0017
Publisert 30. sep. 2020 20:45