The simple explanation that textbooks and the internet say is that "gravitional potential is a scalar quantity hence can be added algebraically".

However, I'm not sure if it is that simple. Take for example, at point O where it is on the horizontal axis connecting the centre of the Moon and the centre of the Earth such that the gravitational field strength at that point is zero. Any point above or below point O will have a net gravitational field strength that is the vector sum of both Earth's (gearth) and Moon's gravitational field strength (gmoon). This vector sum will be smaller than (gearth)+(gmoon) but larger than (gearth) or (gmoon) alone.

Now back to the definition of gravitational potential at a point in a gravitational field, which is the work done (by an external agent) per unit mass in moving a mass from infinity to that point. This work done will be the integral of the gravitational field strength with respect to the distance away from the source of the field.

Therefore, since the gravitational field strength at point O due to the Moon and the Earth is zero, and the fact that gravitational field strength above and below point O isn't (gearth)+(gmoon), hence the work done per unit mass in moving a mass from infinity to point O will not be as simple as adding the gravitational potential due to the Moon and Earth algebriaically.

Am I correct?

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    $\begingroup$ I'm not sure if I fully get your reasoning. You are saying since the magnitude of the vector sum near point $O$ is less than the sums of the magnitudes of each force, this means you can't add the potential energy contributions from each force? Am I understanding you correctly? $\endgroup$ Aug 25, 2018 at 3:28
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    $\begingroup$ @AaronStevens Right that is what I meant, please correct me if there is any flawed reasoning. $\endgroup$ Aug 25, 2018 at 3:30
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    $\begingroup$ You mention explicitly that: "since the gravitational field strength at point O due to the Moon and the Earth is zero" I don't know if this relates to the flaw in your thinking, but let me emphasize: the gravitational potential at point O is not the same as at infinity. Sure, the gravitational potental at point O is higher than at either Earth or Moon, because if you move from point O to either the Earth or the Moon you go to a somewhat lower potential. Still, moving from infinity to point O is already the bulk of the potential drop. $\endgroup$
    – Cleonis
    Aug 25, 2018 at 7:20
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    $\begingroup$ @Cleonis Right, I've also thought about that part of my reasoning (gravitational field strength at point O is zero, which may suggest the gravitational potential at that point to be zero). However I've reasoned that part by imagining moving a mass from infinity to point O. In the process of moving that mass to point O, work must have been done against the gravitational field strength around point O, hence there will be a potential at point O. $\endgroup$ Aug 25, 2018 at 9:15
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    $\begingroup$ It is a postulation. (one way to look at it) it is same as why we can superposite forces. $\endgroup$
    – Shing
    Aug 25, 2018 at 10:20

2 Answers 2


It looks like you are confusing vector magnitudes with potentials.

The potentials are indeed additive because the forces are additive, which is confirmed experientially.

If $\vec F=\vec F_1+\vec F_2$, and we know that for a conservative force $\vec F=-\frac{dU}{dx} \hat x$ then we have

$$-\frac{dU}{dx} \hat x=-\frac{dU_1}{dx} \hat x-\frac{dU_2}{dx} \hat x$$

Then we can integrate both sides with respect to $x$ to get


Where $U_0$ is some constant. So you see the potentials do add given that the forces add, which it seems like you agree is true. This argument works in more than one dimension as well. Notice also that none of this depends on how the magnitudes of $F_1$ and $F_2$ compare to each other or their sum.

We can also use what you reference about doing work from infinity:

$$U=-\int_{\infty}^O \vec F \cdot d \vec x=-\int_{\infty}^O \vec F_1 \cdot d \vec x-\int_{\infty}^O \vec F_2 \cdot d \vec x=U_1+U_2$$

If you think about it these two methods aren't that different. The latter one just determines what $U_0$ is by setting the potential to $0$ at infinity.

Point O is a point of unstable equilibrium. This means that the potential energy is actually at a local maximum at point O. Forces just determine the slope/gradient of the potential energy, not values of the potential energy. This is evident from the above work, and knowing that you can always add a constant to the potential energy without changing the physics.

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    $\begingroup$ Thanks for the answer. But is there an explanation as to why the forces are additive? I'm imagining that an object placed above point O experiences a force A by the Earth and force B by the Moon, and hence the net force, C, that point O experiences is the vector sum of force A and B (which isn't A+B). $\endgroup$ Aug 25, 2018 at 3:55
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    $\begingroup$ Forces are additive because that's how our universe works. As for your example, you are mixing up adding vectors and adding their magnitudes. When adding forces it's always a vector sum. It doesn't matter what the magnitudes sum to. $\endgroup$ Aug 25, 2018 at 4:00
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    $\begingroup$ @BobbyLeung $\vec A +\vec B=\vec 0$ doesn't tell us anything about what $|\vec A| +|\vec B|=A+B$ is. One sum doesn't determine anything about the other in general. $\endgroup$ Aug 25, 2018 at 4:04
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    $\begingroup$ The parallelogram law of the addition of forces was discovered experimentally and it was that discovery that showed the amazing connection between Geometry and mechanics. So maybe the question is: Why is it that forces act this way in so many situations, and with so many different phenomena, why do gravitational force or electic force obey the same simple law? $\endgroup$ Aug 25, 2018 at 13:45
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    $\begingroup$ Read Ernst Mach: The Science of Mechanics; The chapter on the parallelogram law of forces. Many of your questions will be answered. $\endgroup$ Aug 25, 2018 at 13:47

The physical reason for why we can successfully add gravitational potentials in Newtonian gravity is because it possesses no graviton self-interactions. Since (Newtonian) gravity does not interact with itself, the resultant theory is linear: so adding $2$ solutions gives us another solution to the differential equation of motion.

But Mercury's observed perihelion precession cannot be accounted for using Newtonian gravity precisely because of this absence of non-linearities. On the other hand, general relativity (GR) has gravitational self-interactions and is therefore non-linear; it correctly describes the precession. The idea that gravity interacts with itself in GR could be traced back to the idea of upholding the weak equivalence principle.

Deriving Newtonian gravity from GR involves studying the approximate version of GR called linearized general relativity where the metric is broken into a small perturbation around flat space: $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$. The equations of motion for $h$ are linear, and $h_{00} = -2\phi$, where $\phi$ is the Newtonian gravitational potential.

In a similar vein, electromagnetism is linear (assuming all sources and currents are held fixed, so there are no back reaction effects).


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