Can you explain the derivation of mr^2 [n X (dn/dt)] in orbital theory?

AI Thread Summary
The discussion revolves around deriving the expression for angular momentum (L) in terms of unit vectors and their derivatives. The user seeks clarification on how the expression L = mr^2[n X (dn/dt)] is obtained from the initial equations involving position (r) and velocity (v). Participants confirm that by substituting the values of r and v in terms of the unit vector n, the expression simplifies correctly. The conversation highlights the importance of recognizing that r^2 can be factored out as a scalar in the calculations. The user expresses relief at finally understanding the derivation after a long period of confusion.
knockout_artist
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Hi,

I need help with the following equations.

n = r/r (unit vector in r direction)


v =d r/dt = dr/dt n + r dn/dt



L = r X (mv) = mr^2 [n X (dn/dt)]

how did this
mr^2 [n X (dn/dt)]
came about?

I would expect some thing from
r X (mv)
to this
m (v X r)
 
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Start from ##\mathbf{L} = \mathbf{r} \times m \mathbf{v}## and substitute for ##\mathbf{r}## and ##\mathbf{v}## their values expressed in terms of ##\mathbf{\hat{n}}##.
 
n = r/r
r=rn

v = r

L = r X mv

v =d r/dt = dr/dt n + r dn/dt
L = m . rn X[ dr/dt n + r dn/dt ]
= m [rn X dr/dt n + rn X r dn/dt]
(red disappear since in same direction)

= m [rn X rdn/dt]

is this right?
 
Correct. And you can take out the ##r^2## from the bracket as it is a scalar.
 
ohh

took me 1 year to figure that our. :)
 
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