Multiplying by dr/dt and integrating with respect to t

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SUMMARY

The discussion focuses on the derivation of the equation $$m\ddot{r}=\frac{L^2}{mr^3} -V'(r)$$ and the subsequent integration process. By multiplying this equation by the radial velocity $$\dot{r}$$ and integrating with respect to time $$t$$, the result is $$\frac{1}{2}m\dot{r}^2+\frac{L^2}{2mr^2}+V(r) = C$$. The clarity of this derivation is enhanced by differentiating both sides of the integrated equation with respect to $$t$$, leading to a rearrangement that confirms the original equation. This method provides a definitive understanding of the relationship between the terms involved.

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BearY
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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?
 
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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.
 
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andrewkirk said:
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.
It is very clear when it's done this way. Thanks a lot.
 

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