I have seen this graph many times.

enter image description here

From the graph, it would seem that if you calculate the gravitational potential at any given point you obtain a negative, and if you calculate the field strength at the same point you obtain a positive, suggesting the two are in opposite directions.

But, if $g=-\frac{dV}{dr}$, then by using the expression that $V=-G\frac{M}{r}$ and differentiating with respect to r, you should obtain:


Therefore you would expect g to be negative, and so the curve for gravitational field strength would be in the lower half of the graph.

Is this correct?

  • 1
    $\begingroup$ What do you mean by saying they have the same sign? The potential is a scalar, while the field strength is a vector. $\endgroup$
    – Qmechanic
    Oct 11, 2017 at 22:56
  • $\begingroup$ @Qmechanic So if you have a mass M, the potential is negative and approaches zero at infinity. On this scale, we are saying that moving towards infinity is getting bigger (becoming more positive). Therefore the gravitational field strength, being a vector, is acting towards M, so is pointing in the negative direction. Therefore the above graph is wrong? $\endgroup$ Oct 11, 2017 at 22:58
  • $\begingroup$ You calculate gravitational potential with a convention that $V(\infty)=0$. $\endgroup$ Oct 11, 2017 at 23:02
  • $\begingroup$ @SayanMandal yes? $\endgroup$ Oct 11, 2017 at 23:02
  • $\begingroup$ What I meant was that this is the general convention. That's why the potential always has a negative sign. Sorry for the confusion. $\endgroup$ Oct 11, 2017 at 23:06

1 Answer 1


The issue here is gravitational force is truly a vector, $\vec{F}$, so it has a direction. That means if we try to plot it on the same graph as the potential, which is a scalar, we run into issues. One way around it is to talk about the magnitude of the force. This is a scalar, and is found simply by taking the magnitude, that is $\lvert \vec{F} \rvert$. In your $1$d example, this is $$\lvert\vec{F}\rvert=\sqrt{(-g)^2} = g.$$

This is always a positive quantity, and by just looking at the magnitude, we get around the issue of directions. Then it's up to you as the reader to decide which direction that force is pointing in, as you correctly identified.

  • $\begingroup$ Where does the definition that field strength equals the negative gradient of the potential come into it then? $\endgroup$ Oct 11, 2017 at 23:40
  • $\begingroup$ @BenjaminRogers-Newsome $\lvert \vec{F} \rvert = \lvert -\nabla \phi \rvert $. So you find the force as usual, but it's a vector. To get a useful 1d quantity out of that vector, we find its magnitude. Essentially what we're plotting is the "length" of the force vector. $\endgroup$
    – CDCM
    Oct 11, 2017 at 23:48

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