User:Borbrav/Temp

Proof of Kepler's third law
Define point A to be the periapsis, and point B as the apoapsis of the planet when orbiting the sun.

Kepler's second law states that the orbiting body will sweep out equal areas in equal quantities of time. If we now look at a very small periods of time at the moments when the planet is at points A and B, then we can approximate the area swept out as a triangle with an altitude equal to the distance between the planet and the sun, and the base equal to the time times the speed of the planet.

$$\frac{1}{2}\cdot(1-\epsilon)a\cdot V_Adt=\frac{1}{2}\cdot(1+\epsilon)a\cdot V_Bdt$$

$$(1-\epsilon)\cdot V_A=(1+\epsilon)\cdot V_B$$

$$V_A=V_B\cdot\frac{1+\epsilon}{1-\epsilon}$$

Using the law of conservation of energy for the total energy of the planet at points A and B,

$$\frac{mV_A^2}{2}-\frac{GmM}{(1-\epsilon)a}=\frac{mV_B^2}{2}-\frac{GmM}{(1+\epsilon)a}$$

$$\frac{V_A^2}{2}-\frac{V_B^2}{2}=\frac{GM}{(1-\epsilon)a}-\frac{GM}{(1+\epsilon)a}$$

$$\frac{V_A^2-V_B^2}{2}=\frac{GM}{a}\cdot \left ( \frac{1}{(1-\epsilon)}-\frac{1}{(1+\epsilon)} \right ) $$

$$\frac{\left ( V_B\cdot\frac{1+\epsilon}{1-\epsilon}\right ) ^2-V_B^2}{2}=\frac{GM}{a}\cdot \left ( \frac{1+\epsilon-1+\epsilon}{(1-\epsilon)(1+\epsilon)} \right ) $$

$$V_B^2 \cdot \left ( \frac{1+\epsilon}{1-\epsilon}\right ) ^2-V_B^2=\frac{2GM}{a}\cdot \left ( \frac{2\epsilon}{(1-\epsilon)(1+\epsilon)} \right ) $$

$$V_B^2 \cdot \left ( \frac{(1+\epsilon)^2-(1-\epsilon)^2}{(1-\epsilon)^2}\right )=\frac{4GM\epsilon}{a\cdot(1-\epsilon)(1+\epsilon)} $$

$$V_B^2 \cdot \left ( \frac{1+2\epsilon+\epsilon^2-1+2\epsilon-\epsilon^2}{(1-\epsilon)^2}\right )=\frac{4GM\epsilon}{a\cdot(1-\epsilon)(1+\epsilon)} $$

$$V_B^2 \cdot 4\epsilon=\frac{4GM\epsilon\cdot(1-\epsilon)^2}{a\cdot(1-\epsilon)(1+\epsilon)} $$

$$V_B=\sqrt{\frac{GM\cdot(1-\epsilon)}{a\cdot(1+\epsilon)}}$$

Now that we have $$V_B$$, we can find the rate at which the planet is sweeping out area in the ellipse. This rate remains constant, so we can derive it from any point we want, specifically from point B.

$$\frac{dA}{dt}=\frac{\frac{1}{2}\cdot(1+\epsilon)a\cdot V_B dt}{dt}=\frac{1}{2}\cdot(1+\epsilon)a\cdot V_B=\frac{1}{2}\cdot(1+\epsilon)a\cdot\sqrt{\frac{GM\cdot(1-\epsilon)}{a\cdot(1+\epsilon)}}= \frac{1}{2}\cdot\sqrt{GMa\cdot(1-\epsilon)(1+\epsilon)}$$

However, the total area of the ellipse is equal to $$\pi a \sqrt{(1-\epsilon^2)}a$$. (That's the same as $$\pi a b$$, because $$b=\sqrt{(1-\epsilon^2)}a$$). The time the planet take out to sweep out the entire area of the ellipse equals the ellipse's area, so,

$$T\cdot \frac{dA}{dt}=\pi a \sqrt{(1-\epsilon^2)}a$$

$$T\cdot \frac{1}{2}\cdot\sqrt{GMa\cdot(1-\epsilon)(1+\epsilon)}=\pi \sqrt{(1-\epsilon^2)}a^2$$

$$T=\frac{2\pi \sqrt{(1-\epsilon^2)}a^2}{\sqrt{GMa\cdot(1-\epsilon)(1+\epsilon)}}=\frac{2\pi a^2}{\sqrt{GMa}}= \frac{2\pi}{\sqrt{GM}}\sqrt{a^3}$$

$$T^2=\frac{4\pi^2}{GM}a^3$$

However, if the mass m is not negligible in relation to M, then the planet will orbit the sun with the exact same velocity and position as a very small body orbiting an object of mass $$M+m$$ (see reduced_mass). To integrate that in the above formula, M must be replaced with $$M+m$$, to give $$T^2=\frac{4\pi^2}{G(M+m)}a^3$$. Q.E.D.