2
$\begingroup$

The definition of Gravitational Potential at a point is the work done per unit mass in moving it from infinity to that point. However the work is positive and if you perform the integral you get a positive value as you should. $$W=\int_\infty^r\left(-\frac{GMm}{r^2}\right)dr=\frac{GMm}{r}$$ $$\frac{W}{m}=\frac{GM}{r}=-V$$ $F=-\frac{GMm}{r^2}\hat{r}$ as the unit vector, $r$, is pointing towards the mass $M$. So I know why the Gravitational potential has to be negative so shouldn't the definition be "The magnitude of the gravitational potential at a point is the work done per unit mass in moving it from infinity to that point."

$\endgroup$
  • $\begingroup$ The gravitational potential energy comes out with the proper sign because we employ $$\Delta U=U(r)-U(\infty)=-W= - \int_\infty^r (-\frac{GMm}{r^2}dr)=-\frac{GMm}{r}=U(r)$$ because $U(\infty)=0$. The force must be pointing towards inward to get the proper sign. If we employ the same force (pointing inwards) for the derivation of the Gravitational Potential, then we get a positive gravitational potential according to the definition. $\endgroup$ – Jeff Aug 27 at 14:06
  • $\begingroup$ Just divide everything by $m$ and it is all the same signs for potential energy and potential. In your definition it just depends on if you add the "per unit mass" or not $\endgroup$ – Aaron Stevens Aug 27 at 15:33
0
$\begingroup$

First, the definition should be

The work done per unit mass in moving it slowly from infinity to that point

since we don't want to include kinetic energy here (for example, I could do more work than needed to get the mass to position $r$, so if we don't include slowly here then this definition does not give a unique potential energy).

Second, you are mixing up your signs and/or who/what should be doing the work we consider. You have calculated the work done by gravity. Therefore, you have correctly found that $W_\text{grav}=-\Delta U$, which is why you are getting a positive sign.

To use your definition, you have to consider the work done by you to move the mass from infinity. This force points radially outwards because we want to move the mass in a way that its very slow velocity remains constant, so your integral should be $$\Delta U=W_\text{you}=\int_\infty^r\frac{GMm}{r^2}dr=-\frac{GMm}{r}$$

Since $U(\infty)=0$, we can just say that $\Delta U=U(r)=-\frac{GMm}{r}$, and this is probably what you were expecting.


Honestly, I prefer the definition of potential energy that just relies on the conservative force, and not this idea of the work an external agent does against this conservative force. $$\mathbf F=-\nabla U$$ $$U(\mathbf b)-U(\mathbf a)=-\int_{\mathbf a}^{\mathbf b}\mathbf F\cdot\text d\mathbf r$$

$\endgroup$
  • 1
    $\begingroup$ You are defining potential energy which does have the correct sign as you just showed but I'm referring to gravitational potential. Ok, so the definition should be "The work done per unit mass (not by gravity) but by an external force in moving it without acceleration from infinity to that point?" $\endgroup$ – Jeff Aug 27 at 13:32
  • $\begingroup$ @Jeff Right, sorry, It is a simple fix. $V=U/m$. The answer still applies $\endgroup$ – Aaron Stevens Aug 27 at 13:38
  • $\begingroup$ Yes that would fix the sign but whoever is in charge of these definitions, didn't. they intend it to be gravity doing the work? $\endgroup$ – Jeff Aug 27 at 13:39
  • $\begingroup$ @Jeff When you move the mass slowly from infinity, both you and gravity are doing work at the same time. You can either say that the work done by gravity is the negative change in potential energy, or you can say the work done by you is the change in potential energy. They are equal and opposite, and they both relate to the potential energy (or potential for energy per unit mass). $\endgroup$ – Aaron Stevens Aug 27 at 13:41
  • $\begingroup$ Yes I understand that Work is the negative of the change in PE but that isn't what I'm arguing. Everything in your statements are correct. But if you apply it to just the gravitational potential (not the energy, regardless of what you prefer) the sign isn't correct. Imagine this argument for Coulombs law where the sign of V determines whether it's attractive or repulsive. Since the definition should be consistent with the definition of $W=-\Delta PE$, you get the wrong sign. $\endgroup$ – Jeff Aug 27 at 13:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.