## Hunus 3 years ago Help

1. Hunus

In 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 $\vec G \cdot \vec r = -d \phi$

2. Hunus

Can someone explain this to me?

3. hamza_b23

alright do you get the first step atleast where he brought the m over?

4. hamza_b23

i 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 side-dU/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