Lankide:Keta/Kepler

Wikipedia(e)tik

Keplerrek ez zekien zergatik bere legeak ondo zeuden, Newtonek...

[aldatu] Lehenengo legea

Indarra zentrala denez...

 \vec{a} = \frac{d^2\vec{r}}{dt^2} = f(r)\vec{u_r}.

Koordinatu polarretan...

\frac{d\vec{r}}{dt} = \dot{r} \vec{u_r} + r \dot{\theta} \vec{u_\theta}

\frac{d^2\vec{r}}{dt^2} = (\ddot{r} - r {\dot{\theta}}^2) \vec{u_r} + (r\ddot\theta + 2\dot r \dot\theta) \vec{u_\theta}

Berdinduz

\ddot{r} - r \dot{\theta}^2 = f(r)
r \ddot{\theta} + 2\dot{r}\dot{\theta} = 0

Bigarrenetik...

r \frac{d\dot{\theta}}{dt} + 2\frac{dr}{dt} \dot{\theta} = 0 \quad \Rightarrow \quad \frac{d\dot{\theta}}{\dot\theta} = -2\frac{dr}{r}

Ebazten...

\log \dot\theta = -2 \log r + \log h \quad\Rightarrow\quad \log h = \log r^2 + \log\dot\theta \quad\Rightarrow\quad h = r^2 \dot\theta = kte.

h\, konstantea da, momentu angeluar espezifikoa... Aldagaia aldatuz...

r = \frac {1}{u}
\dot r = \frac{dr}{du}\frac{du}{d\theta}\frac{d\theta}{dt} = -\frac{1}{u^2}\dot\theta\frac{du}{d\theta} = -h\frac{du}{d\theta}
\ddot r = -h \frac{d}{dt} \left ( \frac{du}{d\theta} \right ) = -h\dot\theta\frac{d^2 u}{d\theta^2} = -h^2 u^2 \frac{d^2 u}{d\theta^2}

Bestalde, Newtonen grabitazio...

 f \left ( \frac{1}{u} \right ) = f(r) = - \frac{GM}{r^2} = - GMu^2

Ondorioz, r direkzioan...

 -h^2 u^2 \frac{d^2 u}{d\theta^2}\ -\ h^2 u^3 = f(r) \quad\Rightarrow\quad \frac{d^2 u}{d\theta^2}\ +\ u = -\frac{1}{h^2 u^2} f \left ( \frac{1}{u} \right ) \quad\Rightarrow\quad \frac{d^2 u}{d\theta^2}\ +\ u = \frac{GM}{h^2}

Ekuazio honen soluzio orokorra...

 u = \frac{GM}{h^2} \bigg[ 1 + e\cos(\theta - \theta_0) \bigg]

Azkenik, θ0=0...

 r = \frac{1}{u} = \frac{h^2 / GM}{1 + e\cos\theta}

[aldatu] Bigarren legea

Momentu angeluarra, L, definizioz:

\vec{L} \equiv \vec{r} \wedge \vec{p} = \vec{r} \wedge (m\vec{v}) = \vec{r} \wedge m\frac{d\vec{r}}{dt}

Deribatzen...

\frac{d\vec{L}}{dt} = (\vec{r} \wedge m\frac{d^2 \vec{r}}{dt^2}) + \left ( \frac{d\vec{r}}{dt} \wedge m\frac{d\vec{r}}{dt} \right ) = (\vec{r} \wedge \vec{F}) + (\vec{v} \wedge \vec{p}) = 0

F ii r delako... Beraz,

|\vec{L}| = kte.

Azalera paralelogramoaren erdia da:

dA = \frac{1}{2} |\vec{r} \wedge d\vec{r}| = \frac{1}{2} \left| \vec{r} \wedge \frac{d\vec{r}}{dt} dt \right| = \frac{|\vec{L}|}{2m} dt \Longrightarrow \vec{v_{az}} = \frac{dA}{dt} = \frac{|\vec{L}|}{2m}

Guzti hori konstantea denez, azalera ere konstantea da.

[aldatu] Hirugarren legea

Erabili beharrekoa, momentu angeluar espezifikoa, h:

 \vec{h} = \vec{r} \wedge \vec{v} = \frac{\vec{L}}{m} = kte.

 a = \frac{h^2}{GM(1-e^2)} \quad;\quad b = \frac{h^2}{GM\sqrt{1-e^2}} \quad\Longrightarrow\quad b = \frac{h}{\sqrt{GM}}\sqrt{a}

Periodoa, T, ateratzeko, abiadura areolarra kte denez:

 v_{az} = \frac{\pi a b}{T} = \frac{h}{2} \quad\Longrightarrow\quad T = \frac{2\pi a b}{h}

Orain, b a-ren menpe dagoenez...

 T = \frac{2\pi a b}{h} = \frac{2\pi}{h}a\frac{h}{\sqrt{GM}}\sqrt{a} = \frac{2\pi}{\sqrt{GM}}a^{2 / 3}

Beraz, ber bi eginez

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