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I'm learning about angular moment and central forces, and in my syllabus I came across this formula:

enter image description here

How did they get 1 over r? I am familiar with the gradient, but I've never seen the gradient calculated for a central potential, so I think I'm a little bit confused here. I would have just thought that the magnitude of $F$ equals $-d(Uc)/dr$, but apparently a factor $1/r$ comes into play, and I don't know why.

Could someone help me?

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  • $\begingroup$ It's a typo. The author probably meant to write $\frac{\vec{\bf r}}{r}=\hat{\bf r}$. $\endgroup$ – Qmechanic Jan 2 '17 at 21:33
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In spherical coordinates, Gradient is calculated like that. It's derived from definition. And it's different from gradient in Cartesian coordinates. That's all.

See here

If you want the reason, David J.Griffith says:

why

Also,

table

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