Lankide:Keta/Kepler

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

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...

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

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.

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:

$v_{az}={\frac {\pi ab}{T}}={\frac {h}{2}}\quad \Longrightarrow \quad T={\frac {2\pi ab}{h}}$

Orain, b a-ren menpe dagoenez...

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

Beraz, ber bi eginez

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