Keplerrek ez zekien zergatik bere legeak ondo zeuden, Newtonek...
Indarra zentrala denez...
Koordinatu polarretan...
Berdinduz
![{\displaystyle {\ddot {r}}-r{\dot {\theta }}^{2}=f(r)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f39f8e8a7a92e9a8b64d46084a2655a1c00da21b)
![{\displaystyle r{\ddot {\theta }}+2{\dot {r}}{\dot {\theta }}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8bc0c4f8ba4831ce15c0b313a26ef8146f3d391c)
Bigarrenetik...
![{\displaystyle 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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dfded04659f2ab52c58f3350ca2c80433c38f8b3)
Ebazten...
![{\displaystyle \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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9e97797c22947c4567f9bfbbed6ff239058764d3)
konstantea da, momentu angeluar espezifikoa... Aldagaia aldatuz...
![{\displaystyle r={\frac {1}{u}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d9563737c81b9b5605cbdc8a08ad997d098aac7f)
![{\displaystyle {\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 }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/58104e95f48bcb29ce2b80e2e48c76712696c619)
![{\displaystyle {\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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ece899c7631a7e415f64463b898ea95809adfa66)
Bestalde, Newtonen grabitazio...
![{\displaystyle f\left({\frac {1}{u}}\right)=f(r)=-{\frac {GM}{r^{2}}}=-GMu^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d741d070ce63bd6711b2183d3b24a572c2ba3e7)
Ondorioz, r direkzioan...
![{\displaystyle -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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c68633a1c87d8719865e203167ddd80d89e7d4d)
Ekuazio honen soluzio orokorra...
![{\displaystyle u={\frac {GM}{h^{2}}}{\bigg [}1+e\cos(\theta -\theta _{0}){\bigg ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/59fd469e69b2e9aec3d1b882701b64578d78b88f)
Azkenik, θ0=0...
![{\displaystyle r={\frac {1}{u}}={\frac {h^{2}/GM}{1+e\cos \theta }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c0bf71cbfdd2d939cd6c49383f343cf9e99760c)
Momentu angeluarra, L, definizioz:
Deribatzen...
F ii r delako... Beraz,
Azalera paralelogramoaren erdia da:
Guzti hori konstantea denez, azalera ere konstantea da.
Erabili beharrekoa, momentu angeluar espezifikoa, h:
Periodoa, T, ateratzeko, abiadura areolarra kte denez:
Orain, b a-ren menpe dagoenez...
Beraz, ber bi eginez