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KLASIˇCNA MEHANIKA - Studentske web stranice

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206 POGLAVLJE 7. GRAVITACIJA I CENTRALNE SILE<br />

(2): Drugi način je ovaj: uvedimo novu varijablu<br />

koju shvaćamo kao funkciju kuta ϕ<br />

˙ρ = d ρ<br />

d t = d ρ<br />

d ϕ<br />

¨ρ = d ˙ρ<br />

d t = d d t<br />

u(ϕ) = 1/ρ<br />

L 0<br />

m = ρ2 ˙ϕ = 1 u 2 ˙ϕ ⇒ ˙ϕ = L 0<br />

m u2<br />

d ϕ<br />

d t = d u−1 −1 d u ˙ϕ =<br />

d ϕ u 2 d ϕ<br />

(<br />

− L )<br />

0 d u<br />

= − L 0<br />

m d ϕ m<br />

= − L 0<br />

m ˙ϕ d2 u<br />

d ϕ 2 = − L2 0<br />

m 2 u2 d2 u<br />

d ϕ 2 .<br />

Sada jednadžba gibanja (7.50), poprima oblik<br />

L 0<br />

m u2 = − L 0 d u<br />

m d ϕ ,<br />

d d u<br />

d ϕ d t = −L 0<br />

m<br />

d<br />

d ϕ<br />

( d u<br />

d ϕ<br />

)<br />

d ϕ<br />

d t<br />

(<br />

m − L2 0 d2 u<br />

m 2 u2 d ϕ − 1 2 u<br />

m(¨ρ − ρ ˙ϕ 2 ) = f(ρ)<br />

)<br />

L 2 0<br />

m 2 u4<br />

= f(1/u)<br />

/<br />

·<br />

−m<br />

L 2 0u 2 ,<br />

d 2 u<br />

d ϕ + u = − m f(1/u)<br />

, (7.52)<br />

2 L 2 0 u 2<br />

gdje je u = u(ϕ), tj. eliminirano je vrijeme. Gornja se jednadžba zove Binetova 20 formula.<br />

Ako se sada vratimo u varijablu ρ = 1/u,<br />

d u<br />

d ϕ = d<br />

d ϕ<br />

d 2 u<br />

= d<br />

d ϕ 2 d ϕ<br />

1<br />

ρ = −1 d ρ<br />

ρ 2 d ϕ ,<br />

( ) −1 d ρ<br />

ρ 2 d ϕ<br />

= 2 ( ) 2 d ρ<br />

− 1 d 2 ρ<br />

ρ 3 d ϕ ρ 2 d ϕ , 2<br />

Binetovu formulu (7.52) možemo napisati i u varijabli ρ = ρ(ϕ)<br />

( ) 2<br />

2 d ρ<br />

− 1 d 2 ρ<br />

ρ 3 d ϕ ρ 2 d ϕ + 1 2 ρ = m ρ 2 f(ρ)<br />

L 2 0<br />

/<br />

· (−ρ 2 ),<br />

d 2 ρ<br />

d ϕ − 2 ( 2 d ρ<br />

− ρ = 2 ρ d ϕ) m ρ 4 f(ρ) (7.53)<br />

L 2 0<br />

20 Jacques Philippe Marie Binet, 1786. - 1856., francuski matematičar.

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