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Hunus
 4 years ago
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Hunus
 4 years ago
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Hunus
 4 years ago
Best ResponseYou've already chosen the best response.0In a book I'm reading, the author keeps moving quantities to the other side of the differential element. \[ m\vec G \cdot d \vec r =  dU\] He then says to divide by m \[\vec G \cdot d \vec r = \frac{dU}{m}\]he then says \[\phi = \frac{U}{m}\] so that \[\vec G \cdot \vec r = d \phi\]

Hunus
 4 years ago
Best ResponseYou've already chosen the best response.0Can someone explain this to me?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0alright do you get the first step atleast where he brought the m over?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i am going to assume you know that because it is a constant and you can divide it if you like the author chooses too thats not important , ok so now to the second line we got G.dr = dU/m right after what the author did was he took the integral of both sides lets do the left hand side first G.dr the integral of that in respect to r is G.r because dr says we have a function of r. Now lets do the right sidedU/m the integral of that is U/m, we dont do anything to 1/m because that is just a constant next then he says phi = U/m where they let G.r=dphi
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