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

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


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

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.


1 Answer 1


Yes, potential energy decreases in the direction of the force. So potential energy decreases as you move closer to the Earth. Any texts that say potential energy increases closer to the Earth shouldn't be taken seriously. The texts could be talking about the magnitude of the potential energy, but that really isn't a useful concept at all.

  • $\begingroup$ Isn't the fundamental idea that $F = dU$? I can't see how this minus sign is not pure convention. $\endgroup$
    – EEEB
    Commented Oct 31, 2020 at 18:48
  • $\begingroup$ @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. $\endgroup$ Commented Oct 31, 2020 at 18:51
  • $\begingroup$ $F=-dU \Rightarrow F=d(-U)$, set $\tilde{U}=-U$ and $F=d\tilde{U}$. $\endgroup$
    – EEEB
    Commented Oct 31, 2020 at 18:54
  • 1
    $\begingroup$ @EEEB That's fine, you can do that. But it's not potential energy anymore. $\endgroup$ Commented Oct 31, 2020 at 18:54

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