# Potential energy: where is the error in my logic?

1. Dec 14, 2013

### Nikitin

1) For gravitational potential energy: $F(r)=-\frac{\gamma M m}{r^2}$.

2) For potential energy of a conservative force-field: $-\nabla E_p = F$. So the gradient of the potential energy is always equal to the force in that direction, just with the opposite sign. So when you move in the direction of the force, the potential energy decreases, but if you move against it the potential energy will increase.

3) Thus to find the gravitational potential energy as a function of r, I just need to find how much work I need to do to carry a particle from a distance r to infinitely away from the source of the gravitational force. $dE_p = -\vec{F(r)} \cdot \vec{dr} \Rightarrow E_p(r) = -\int_r^{\infty} -\frac{\gamma M m}{r^2} dr = \frac{\gamma M m}{r}$

Uhh and so I get an expression with the opposite sign of what I'm supposed to get. Help? Can somebody point out the flaw of my logic?

EDIT: Is my flaw that gravitational potential energy is NOT defined as how much work is needed to make something escape the gravitational field? If so, can somebody correct my line of thinking?

EDIT2: Oops I forgot that $-\int_r^{\infty} -\frac{\gamma M m}{r^2} dr = \frac{\gamma M m}{r} = E_p(\infty)-E_p(r)$, and not just $E_p(r)$. Okay, so since the pot. energy is equal to zero at $E_p(\infty)$, I'm all set?

Last edited: Dec 14, 2013
2. Dec 14, 2013

### MalachiK

If you take the integral ∫F.dr you get -GMm $\int_R^{\infty} \frac{1}{r^2}$ dr

1/r vanishes in the limit where r→∞ so we just have -$\frac{GMm}{R}$ which is the work done by gravity moving from R to ∞.

If we swap the limits around to get the work done by gravity moving from ∞ to R then we get a positive value. Gravity does work moving towards the source of the field and we have to do work against gravity in the other direction.