El problema de Kepler
El problema de Kepler
El problema de Kepler
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• Estados ligados: E < 0 ⇒ 0 ≤ e < 1<br />
(x + ae) 2<br />
a 2<br />
+ y2<br />
= 1<br />
b2 ⎧<br />
⎪⎨ a ≡<br />
⎪⎩<br />
p GM<br />
=<br />
1 − e2 −2E<br />
p<br />
b ≡ √<br />
1 − e2 =<br />
J<br />
√<br />
−2E<br />
(ϕ0 <strong>de</strong>termina el periastro: mínimo <strong>de</strong> r)<br />
r =<br />
(figura , ϕ0 = 0)<br />
p<br />
e cos(ϕ − ϕ0) + 1 ⇒<br />
⎧<br />
r0 ⎪⎨<br />
=<br />
⎪⎩<br />
p<br />
e + 1 = a(1 − e2 )<br />
= a(1 − e) = a − c<br />
1 + e<br />
rπ/2 = p<br />
rπ = p<br />
= a(1 + e) = a + c<br />
1 − e<br />
c = a2 − b2 = p<br />
1 − e2e <br />
⇒ e = c<br />
<br />
a<br />
8
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