I recently learned how to use differential equations in physics(just the basics), so I tried to prove First Kepler's Law as a challenge. This is my proof, I would like to know if it is correct:
Proof
The first thing I did was decomposing the problem on $x$ and $y$. From the similidute of triangles we have:
$$ a_x:x=a:\sqrt{x^2+y^2}\Rightarrow a_x=a\frac{x}{\sqrt{x^2+y^2}} $$ $$ a_y:y=a:\sqrt{x^2+y^2}\Rightarrow a_y=a\frac{y}{\sqrt{x^2+y^2}} $$
Since $a_x=x''$ and $a_y=y''$ , we have:
$$x''=a\frac{x}{\sqrt{x^2+y^2}}$$ $$y''=a\frac{y}{\sqrt{x^2+y^2}}$$
Since:
$$a=\frac{GM}{r^2}=\frac{GM}{x^2+y^2}$$
We'll have: $$x''=\frac{GM}{x^2+y^2}\frac{x}{\sqrt{x^2+y^2}}$$ $$y''=\frac{GM}{x^2+y^2}\frac{y}{\sqrt{x^2+y^2}}$$
Here I got stumped for some days. Then I noticed that multiplying both sides:
$$x''y''=\frac{G^2M^2xy}{(x^2+y^2)^3}$$
I know it isn't formal, but I thought that the difficult part was the denominator. So I made this substitution: $$x=c_1 \cos(c_2 t+c_3) $$ $$y=c_1 \sin(c_2 t+c_3) $$ This makes the denominator disappear and we have:
$$k_1\cos(c_2 t+c_3)\sin(c_2 t+c_3)=k_2\cos(c_2 t+c_3)\sin(c_2 t+c_3)$$
This proves that my solution was correct and that solution is the parametric equation of an ellipse(however this doesn't prove that it's unique, but I don't know how yet how to solve a differential equations system, so I had to use intuition and a bit of luck).
Thank you for your time
:)
To prove Kepler's first law, take the total energy in polar coordinates $(r,\varphi)$, $$E_\text{total}=\frac12m\dot r+\frac{|\vec L|^2}{2mr^2}-G\frac{mM}{r},$$ where $L=|\vec L|$ is the angular momentum and solve this for $\dot r$:
$$\dot r=\left(\frac2m\left(E_\text{total}-\frac{L^2}{2mr^2}+G\frac{mM}{r}\right)\right)^{1/2}$$ and use $\dot\varphi=\frac{L}{mr^2}$ to get $$\frac{\mathrm d\varphi}{\mathrm dr}=\frac{\mathrm d\varphi}{\mathrm dt}\frac{\mathrm dt}{\mathrm dr}=\frac{L}{mr^2}\left(\frac2m\left(E_\text{total}-\frac{L^2}{2mr^2}+G\frac{mM}{r}\right)\right)^{-1/2}.$$ Now integrate this to get $\varphi(r)$ and solve this for $r$ to finally obtain $$r(\varphi)=\frac{a(1-b^2)}{1+b\cos\varphi}\qquad\text{for}\qquad a=-G\frac{mM}{2E_\text{total}}\quad\text{and}\quad b=\left(1+\frac{2E_\text{total}L^2}{G^2m^3M^2}\right)^{1/2}.$$
This is the equation of the conic section in polar coordinates. For $E_\text{total}<0$, this becomes the equation of an ellipse.