El problema de Kepler
El problema de Kepler
El problema de Kepler
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• <strong>El</strong>ección <strong>de</strong>l plano θ = π/2<br />
⎧<br />
⎪⎨<br />
⎪⎩<br />
• Potencial efectivo<br />
E = 1<br />
2 ( ˙r2 + r 2 ˙ϕ 2 ) − GM<br />
r<br />
J = r 2 ˙ϕ (⇒ ˙ A = J/2)<br />
E = 1<br />
2 ˙r2 + Φef(r) , Φef(r) ≡<br />
⎧<br />
2<br />
⎪⎨<br />
J 1<br />
Φef(r) = 0 ⇒ r = ≡<br />
2GM 2 p<br />
⎪⎩ Φ ′ ef(r) = 0 ⇒ r =<br />
J 2<br />
GM<br />
= p<br />
4<br />
2 J GM<br />
−<br />
2r2 r<br />
(figura)<br />
(figura)<br />
⇒ Φef(p) = − GM<br />
2p