0
$\begingroup$

I have a big project due tomorrow morning regarding Kepler's laws, and I'm stuck on one. I have to prove Kepler's second law via observations I've collected from Stellarium. Problem is, the areas either end up negative or they're not equal.

The project is focused on Jupiter and its moons. I have graphed the oscilating distance between Jupiter and each of its Galilean moons. Through that I have found the a and c of every ellipse, and through those - the b. With those, I have the ellipse function for every orbital path, and the focal points of each one (I decided that Jupiter is on the positive focal point).

Now I have to prove the second law. I have radii for each moon's path measured in equal time intervals. I will use three to build 2 "triangles." If the function of the ellipse is: f(x)=((b^2)-((b^2*(x^2)/(a^2))^0.5 then:

F(X)=b*((a*((sin(x/a))^-1))+((x*((-(x-a)*(x+a))^(1/2)))/abs(a)))/2

In an attempt to simplify, I chose three distances of which the x that is on the ellipse perimeter is always bigger than c. I can find the x of each radii using: r1=a-(cx/a). Then, I find their y's through f(x). THEN I move onto the areas. I have gone over it a few times, and what I have, when all the x's are bigger than c, is:

S1: F(x1)-F(x2)-((y1*(x1-c))/2)+((y2*(x2-c))/2)
S2: F(x2)-F(x3)+((y3*(x3-c))/2)-((y2*(x2-c))/2)

Problem is, the areas don't turn out the same! This is one of the few instances where they aren't in the negatives as well.

For example, with Europa:

x=(((678661260.6^2)-(678661260.6*d))/7006365.5)

f(x)=((678625093.5^2)-((((678625093.5^2)*(x^2))/(678661260.6^2))))^0.5

F(x)=678625093.5*((678661260.6*((SIN(x/678661260.6))^-1))+((x*((-(x-678661260.6)*(x+678661260.6))^(1/2)))/ABS(678661260.6)))/2

x1:173340290.7

y1:656116120.5

x2:407290065

y2:542830084.5

x3: 596849455.5

y3: 323018129.8

In the end, I get:

S1 = 5.04111e17

S2=1.09607e17

The ratio isn't even close to 1! I don't know what went wrong.

Please help. I can send excel sheets, photos, graphs, whatever is needed.

I need this for tomorrow. Please.

Image link for example:

http://imgur.com/a/VzHHp

  • 2
    You only has one day to do this assignment? that was harsh on the teacher.. – 2017-02-27
  • 4
    Everything about this is a surefire way to not get an answer. You didn't format your question properly. You are making it clear that you are simply interested in getting an answer for a project. Your username is "KeplerHelp" strongly indicating you have no interest in further contributions to the site in the future. I won't downvote because I don't believe you've broken any rules, but I would strongly suggest you edit your question if you want a better chance at actually getting an answer. – 2017-02-27
  • 0
    @Chinny84 Well, there’s one more day before it’s due, but it looks like the OP has spent some time before that gathering data. – 2017-02-27
  • 0
    @david Sorry, I'm a bit desperate. What should I change about it? I really need help. – 2017-02-27
  • 1
    You could start by formatting your question to make the math more readable. This site uses MathJax, for which you can find a quick tutorial [here](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference). – 2017-02-27
  • 0
    As a suggestion for how to attack the calculations, you might have an easier time if you use the polar equation of an ellipse with the origin (Jupiter) at a focus. The area of the triangle swept out from $\theta_1$ to $\theta_2$ is $\frac12 r(\theta_1)r_2(\theta_2)\sin(\theta_2-\theta_1)$. You have the two radii. You can recover the cosines of the angles from the ellipse equation and then either use trig identities or $\arccos$. – 2017-02-27
  • 0
    @amd I don't understand polar coordinates, we haven't studied them. They always pop up for the proofs. I'll try to wrap my head around it, thanks. Maybe that's all I need. – 2017-02-28
  • 0
    @amd wait, are you rounding up for a triangle? That's sort of my last resort, I'm trying to avoid doing that. Is it a good idea to learn polar coordinates at 2 am? I feel like if I can explain why this is equal to the area, I can implement my information, get two equal areas and be done with it: http://imgur.com/a/Oxr7u – 2017-02-28
  • 0
    The expression I gave is for the area of the inscribed triangle, which for a small enough time interval will be very close to the actual area. Newton proved the 2nd Law using triangles, FWIW. I’m not sure that there is a nice expression for the actual area of the sector swept out by a vector from one of the foci. The area of the triangle is half the area of the parallelogram defined by the two vectors, so it’s half the length of their cross product. You could compute this in Cartesian coordinates as $\frac12(x_1y_2-x_2y_1)$, but why compute $x$ and $y$ if you don’t really need to? – 2017-02-28

0 Answers 0