# Gravitational potential energy sign

Following is a small derivation just so I can explain my question. The gravitational potential energy is:

$$(*)U_g = -\frac{GMm}{r}$$

And:

$$\Delta U =-GMm(\frac{1}{r_{final}} - \frac{1}{r_{initial}})$$

If some mass $$m$$ is taken a height $$h$$ above the ground, we get:

$$\Delta U =-GMm(\frac{1}{R+h} - \frac{1}{R}) = \frac{GMmh}{R(R+h)}$$ approximating $$h\ll R$$ :

$$\Delta U = \frac{GMmh}{R^2}$$ and if we denote $$g=\frac{GM}{R^2}$$ we get the familiar $$\Delta U = mgh$$

That indeed goes hand-in-hand with (*), since the object went further from the center of the earth and therefore gained PE.

Now to the question: Does that mean we should always express the PE to be "more negative" the closer we are to Earth? I see some texts that present PE that gets bigger when you get closer to the Earth and that quite confuses me.

• Isn't the fundamental idea that $F = dU$? I can't see how this minus sign is not pure convention.
• @EEEB Potential energy $U$ is defined so that $\mathbf F=-\nabla U$, where $\mathbf F$ is the conservative force. So if you take out the negative sign then you aren't talking about potential energy anymore. Oct 31, 2020 at 18:51
• $F=-dU \Rightarrow F=d(-U)$, set $\tilde{U}=-U$ and $F=d\tilde{U}$.