It is pretty intuitive that increasing the height or distance of an object and also increasing the mass of it increases the gravitational potential energy (GPE).

Basically the question is asking, with regards to this equation $$U=-\frac{GMm}{r}$$ if you increase mass, this makes the $U$ more negative whilst increasing the distance makes it less negative (closer to 0). This is confusing because even though both increasing distance and mass will increase "GPE", it's not reflected in the output number of the formula, $U$.

Could someone help me understand that although both increase the stored energy, they change in $U$ differently?

No, my question is not a duplicate, as the flagged question asks why $U$ is a negative value, of which I completely understand why it is negative. My question is why the $U$ changes as it does, becoming more negative with more mass which is odd, as it would seem to suggest a decrease in the GPE of the system.


3 Answers 3


It is pretty intuitive that increasing the height or distance of an object and also increasing the mass of it increases the gravitational potential energy (GPE).

This is true for the casual calculation where $U = mgh$ and where you are considering $h > 0$.

Let's instead look at a case where we might want to use $h < 0$.

We have a 1kg bucket and a 10kg toolchest to be lowered below a construction site. Sitting on the ground, both have potential energy equal to zero, because we set the ground to be $h = 0$. With the bottom of the pit 10m down, the gravitational potential energy of each will become:

$$U_{bucket} \sim -100 J$$ $$U_{toolchest} \sim -1000 J$$

In this case increasing the mass of the object didn't increase the potential energy, it decreased it. What it really does is increase the change as we depart from the zero point. It's more positive or more negative for a given distance.

So the difference is that most uses of $U = mgh$ tend to be at locations above the zero point, while all uses of $U = -\frac{GMm}{r}$ are at locations below the zero point.

  • $\begingroup$ This is already explained in the answers to the flagged question. $\endgroup$ Commented Jan 17, 2019 at 0:25

I think that the confusion is arising due to two different formulas of GPE that are $mgh$ and $\frac{−GMm}{r}$

$mgh$ is actually a derived and slightly "edited" from of the latter.

See the following SE question : Derivation

So it is important to note that $mgh$ is change in energy and not energy itself. The latter formula is the true formula of GPE.

Regarding your question : As one increases distance, GPE would become less negative i.e. change in energy is positive as predicted by the $mgh$.

As we increase mass, GPE would become more negative so change in GPE should be negative but we can't use $mgh$ to get the change now. While deriving the formula, we assume mass to be same and distance varying. If we want to study change in GPE as per change in mass we will need to consider distance constant and then derive a new formula.


The law of gravitation is


Where $F$ is the force exerted by $M$ on $m$ and vice versa. The gravitational potential energy function can be derived from the law of gravitation.

Lets assume $M$ is the mass of the earth and $m$ is your mass. The force the earth exerts on you will equal your mass, $m$, times the acceleration due to gravity at the surface of the earth, $g$. We can calculate $g$ from the following



  • $G$ is the universal gravitational constant $= 6.674\times10^{11}$ N kg$^{-2}$ m$^2$
  • $M$ is the mass of the earth $= 5.9742 \times 10^{24}$ kg
  • $r$ is the radius of the earth $= 6378$ km at the equator.

Which gives us, using proper units, $g=-9.8 \frac{\text{m}}{\text{s}^2}$ at the surface of the earth.

Now, when we apply the equation $GPE=mgh$, it is the difference in the gravitational potential energy of $m$ between the surface of the earth and the height $h$. $g$ can be assumed to be constant provided $h$ is negligibly small compared to $r$. If it isn’t, then you would need to redo the above calculation to determine a new $g$ using a new value of $r$

Bottom line, the $h$ in $mgh$ is not the same as $r$ in the gravitational potential energy equation. Consequently an increase in either $m$ or $h$ results in an increase in gravitational potential energy near the surface of the earth. This does not contradict the gravitational potential energy function.

Hope this helps.


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