PWiz said:
if someone goes away more confused than they originally were
I certainly did, if not exactly because of your input.
I came to this thread with the intention of helping the OP, only to realize half-way through writing a response that I don't understand the issue myself, and the subsequent posts failed to clear it up for me (and, I dare to presume, perhaps also for the OP).
For one, I don't see the issue
@sophiecentaur has with the direction in the derivation - you get the same result for ##U(b)## regardless of the direction you move the test particle.
For m moving from infinity to b:
$$ \Delta U= - \int_\infty^b \vec F \cdot d \vec r = - \int_\infty^b F dr = - \int_\infty^b \frac{GMm}{r^2} dr = -GMm(-\frac{1}{b}) = \frac{GMm}{b} $$
$$ \Delta U = U(b) - U(\infty) = U(b) $$
$$ U(b) = \frac{GMm}{b} $$
For m moving from b to infinity:
$$ \Delta U = - \int_b^\infty \vec F \cdot d \vec r = - \int_b^\infty - F dr = - \int_b^\infty \frac{GMm}{r^2} dr = -GMm(\frac{1}{b}) = - \frac{GMm}{b} $$
$$ \Delta U = U(\infty) - U(b) = - U(b)$$
$$ -U(b) = - \frac{GMm}{b} $$
$$ U(b) = \frac{GMm}{b} $$
And of course the sign for U at point b in both cases is positive, and I don't see the reason why it should come out negative. If it's a matter of convention, then it feels artificial to insert it at this stage. The only two sign conventions I'm aware of in this case is the negative sign for the force if it's not in the same direction as the displacement when switching from the vector form, and the ##\Delta U=-W##. Shouldn't these two alone net the correct result?
I saw this derivation in the wikibooks section:
http://en.wikibooks.org/wiki/Physic.../Gravitational_Potential_Energy#Another_proof
that nets a negative result, but I don't understand why they switch the integration limits for no apparent (to me at least) reason in the third line. They certainly do not mention any extra conventions being applied at any stage.
I checked in Kleppner-Kolenkow(p.171), and they just leave the derivation at
$$U(r)=\frac{A}{r}$$
again, with the positive sign.
So, to repeat the question the OP asked: where's the mistake?