This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: r 2 = k 1 R 1 r , (7) where r = 0 at r = R . Now solve this equation for r , and rearrange it into the form s 2 k R 3 dt = dr R 3 / 2 q 1 r 1 R = dx q 1 x 1 , (8) 1 where x = r/R , (9) and we choose the minus sign because t will increase as r decreases. If we make the variable substitution, x = sin 2 , dx = 2 sin cos d , (10) the integration of Eq.(7) becomes straightforward. We integrate the left side from t = 0 to t = t c and the right side from = / 2 ( x = 1 ) to = 0 ( x = 0 ) to obtain s 2 k R 3 t c = 2 Z / 2 sin 2 d = Z / 2 (1 cos 2 ) d = 2 . (11) Thus, the time until the particles collide is t c = 2 r R 3 2 k = 4 2 , (12) where the right hand side is found after we substitute Eq.(5) for R. 2...
View
Full
Document
 Fall '10
 Wilemski
 mechanics, Angular Momentum, Mass, Momentum

Click to edit the document details