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I Multiplying by dr/dt and integrating with respect to t

  1. Jun 20, 2017 #1
    I came across this step in a derivation:$$m\ddot{r}=\frac{L^2}{mr^3} -V'(r)$$
    Multiplying by ##\dot{r}## and integrating with respect to t to get $$\frac{1}{2}m\dot{r}^2+\frac{L^2}{2mr^2}+V(r) = C$$
    I am not very clear about how the 1st term came to this. Can some one gives a pointer?
     
  2. jcsd
  3. Jun 20, 2017 #2

    andrewkirk

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    Differentiate both sides of the second equation wrt ##t## and we get
    $$\frac{1}{2}m\left(2\dot{r}\ddot r\right)+\frac{L^2}{2m}\left(-2r^{-3}\dot r\right)+V'(r)\dot r = 0$$
    Re-arranging terms gives us the first equation.
     
  4. Jun 21, 2017 #3
    It is very clear when it's done this way. Thanks a lot.
     
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