I have two questions.

1) The first one has to do with the formula for deriving Gravitational Potential Energy. I learned that, for the derivation of Gravitational Potential Energy given large distances, we have to use the mathematical analytic way to derive an expression for it at a given distance.

To do so, you need integrate F dot dr from r to infinity. However, what I don't understand are as follows:

  • Why do we need to take the zero point at r = infinity? Why can't I take it from any arbitrary point to get a general expression for its GPE when I integrate?

  • Why is the work required to push an object to that height equal to the force due to gravity times the distance? Don't I need to apply a force that overcomes the force due to gravity to even raise it to begin with? Fg * h is definitely greater in magnitude than fg, but if I'm applying work to an object equal to Fg * h in the opposite direction of where it wants to move (towards the dominant object's COM) how do I know the magnitude of that work is sufficient to do so?

If I was trying to figure out how much energy I need to give an object to raise it from one point in space to another relative to, say the Earth, I could take the change in energy from the two points. If its energy at its initial point is 2 and the energy at the point I want it to be at is 8, I need to supply 6 joules to it. But how do I reconcile that with the derivation from the above paragraph?

2) The second question has to do with zero points for potential energy. Is this allowed because, as long as the distance from each object relative to another is the same no matter where I place a zero point, everything resolves? If at point A, object 1 is 2 units from point A and object 2 is 5 units from point A (all in, say, the x axis), then I'm not cheating by taking point B to be at object 1's position and saying object 2 is now 3 units from point B, right? Wouldn't its potential energy then change here though? That's okay because it's all relative, right? But the magnitude changes. that's okay?


2 Answers 2


1) point 1: We do not need to put the $0$ point at infinity. Since it is potential energy, we can set it to $0$ anywhere we want. This is because adding a constant to potential energy does not change the force involved (since $F=-\frac{dU}{dr}$). The reason so many people choose to do this is because what we are usually interested in is a change in potential energy, rather than an absolute value of it. If we set $U = 0$ at infinity then things work out nicely. For example, in the case of gravity at large scales, $U=-\frac{Gm_1m_2}{r}$. This expression goes to $0$ as $r$ goes to infinity. Therefore we can look at $U(r)$ as a change in potential energy from infinity, and we don't need to keep track of some arbitrary constant. The change of potential energy between two points in space becomes $$\Delta U=U(r_2)-U(r_1)=-\frac{Gm_1m_2}{r_2}+\frac{Gm_1m_2}{r_1}$$

If we did want to set $U=0$ somewhere else, then we always have an arbitrary constant following us around, but then it goes away when we find changes in potential energy. Let's say $U=0$ at some point $r = R_0$. Then our function of the potential energy is $U(r)=-\frac{Gm_1m_2}{r}+\frac{Gm_1m_2}{R_0}$. This is perfectly valid physically, but we get the same result as before for the change in potential energy between two points in space:$$\Delta U=U(r_2)-U(r_1)=-\frac{Gm_1m_2}{r_2}+\frac{Gm_1m_2}{R_0}+\frac{Gm_1m_2}{r_1}-\frac{Gm_1m_2}{R_0}$$$$\Delta U=-\frac{Gm_1m_2}{r_2}+\frac{Gm_1m_2}{r_1}$$ Therefore, we just set $U=0$ at infinity since it is the simplest thing to do (similar to why we put potential energy to $0$ when a spring is at its resting state).

1) point 2: This is not how you find the work done to push something against gravity in general, but we can make some assumptions to get to where it seems like you need to be. In general the work done by any force is $\int \vec F\cdot d \vec r$. So if you push on the object with some force, then this is how you determine the work done by yourself if you do not know anything else. If you know that the only forces acting on the object is yours and gravity, then you can use energy conservation as another way to get the work you have done: $$W_{tot}=\Delta K=W_{me}+W_{grav}=W_{me}-\Delta U$$ where K is the kinetic energy. If the object starts and stops at the same speed, then we can go further:$$W_{me}=\Delta U$$ And this is probably where you getting confused. In the case where we are close to the Earth, then $W_{me}=mg \Delta h$. In the case where we are farther from the Earth, $W_{me}=-\frac{Gm_1m_2}{r_2}+\frac{Gm_1m_2}{r_1}$. So your issue might be from trying to apply the first (an approximation) when you really need to be applying the second.

2) I am kind of confused on the wording here, but if I understand what you are asking, you are just wondering if you can change where $U=0$ is defined. As already stated, this is perfectly fine! You just have to make sure you stay consistent. The absolute values of your potential energies can change, as long as the relative values between points remains constant.

(Side note: You can run into issues if you try to put $U=0$ at infinity if your mass distribution is nonzero at infinity as well, but since we are just looking at gravity around the Earth we should be fine)

  • $\begingroup$ In Newtonian physics you can choose the potential energy to be zero wherever you like. But in relativistic physics, the potential energy contributes to the total energy and thus affects the invariant mass. Relativistically the potential energy must be taken to be zero at infinity. Otherwise, for example, you don’t get the right mass for a hydrogen atom when considering the effect of the electrostatic potential energy. $\endgroup$
    – G. Smith
    Nov 12, 2018 at 4:44
  • $\begingroup$ @GSmith Yes thank you. I will edit if the OP indicates they are interested in relativity and not Newtonian physics. $\endgroup$ Nov 12, 2018 at 4:49

1a) You have to take it from infinity to calculate all the GPE. If you only take it from an arbitrary point the energy required to move from that point to infinty will not be included.
1b) Work done is equal to net force time distance. When a onject falls the net force is g. When an object is raised the net force will include a term for -g, but nevertheless can be any magnitude depending upon the mechanism being used to lift the object. If a large force is used this will overcome GPE and also induce kinetic energy KE.
If you subtract GPE at two positions, the integration to infinity cancel each other out - so in fact then you can jsut integrate from one to the other to get a change in GPE, a $\delta GPE$
2) You have to integrate radially in relation to the center of mass point, and with GPE being higher further away from the mass, so radially outward. As you guessed it's not OK to change this unless the center of mass moved.
Strictly speaking (for a non-extended point-like mass) the center of mass point is not zero GPE, zero GPE is a point at infinity; rather the center of mass point is a negative minimum.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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