For an object or force to do work, it needs energy. But, from where does the gravitational force get the energy to do work upon, say, a falling object? The gravitational force is doing work upon the object, isn't it?

I searched the internet and Physics SE, and found this. Caesar asked a similar question and udiboy1209 answered it.

udiboy1209 says:

Let's take the example of a ball dropped from some height. Gravity of the earth pulls it downward, doing work on the ball and giving it kinetic energy. The question you ask is where did it get this energy from? Go back a step and think about how this ball ended up at such a height? You lifted it up with your arms and put it on that height. Your arms did work against gravity, spent some energy to put that ball on that height. Where did that spent energy go? This was given to gravity!

When you do work against gravity, you store energy in the gravitational field as gravitational potential energy, which then gravity uses to do work on that object.

But isn't the work done by our arms stored on the ball? He says that the energy spent is stored in the gravitational field. Work is done upon the ball; shouldn't the energy be stored on the ball? If so, then where does gravitational force get its energy to do work upon the ball on the first place?

  • $\begingroup$ Relativity is the only acceptable answer here and I don't think you're going to get an answer that explains what you're looking for. $\endgroup$ Commented Nov 16, 2016 at 16:56
  • $\begingroup$ Potential energy is assigned to a system and is due to the specific configuration of that system. In case, system includes ball and earth! $\endgroup$
    – user104909
    Commented Nov 17, 2016 at 10:31
  • $\begingroup$ The conclusion that, "the energy was given to gravity" needs to be reconsidered. The gravitational field was there before the ball was introduced, and will be there after the ball leaves. Both masses have a gravitational field inherent to their mass. The presence of the mass in gravitational field results in a force on both masses. As long as the ball is in the gravity field, a force will be present, acting on the ball. When energy is expended to move the ball against the force, Gravitational Potential Energy is stored in the system of gravitational field and ball/mass. $\endgroup$ Commented Jun 25, 2018 at 23:20

6 Answers 6


You can think of gravitational energy being stored in a system of bodies, not just one body or the other. When you apply force over a distance (work) to the ball, it is being stored in the system of "the ball and the Earth." We can capture the concept of this energy stored in the system by saying its "stored in the gravitational field," but at the very minimum we should say that it's stored in the system.

Similar issues show up in electrostatics. In electrostatics, potential energy is almost always between two bodies, not in one or the other. If you choose to think of it as being in one body or the other, you end up in some really peculiar paradoxes.

What makes this tricky to understand intuitively is that we have many cases where one object is so astonishingly massive compared to the other that we can often handwave away this system-wide thinking, and pretend that the ball is the thing that actually has the gravitational potential energy. This is similar to how electrical engineers assume there is such a thing as a "ground" and that it can sink infinite electrical energy (there's a glorious pile of issues like ground loops which are associated with faulty assumptions regarding grounds). However, in many reasonable environments, these simplifications (such as assuming the earth doesn't move in response to us jumping upwards) are effective, so we keep using them.

There are also theories regarding what gravity "is" in general relativity and quantum mechanics. If one wishes, one can pursue those and come to a deeper answer. However, I don't believe they are necessary for everyone to learn.

  • $\begingroup$ The ball and the earth is a system. But the earth itself is part of the universe. So the ball is a part a bigger system, the universe. So, if we do work on any object, it is stored in the whole system. If this is so, then what does the concept of an isolated system imply ? $\endgroup$ Commented Nov 18, 2016 at 10:41
  • 1
    $\begingroup$ @Gokul The real answer to that would be that there are no truly isolated systems. However, there are cases where they are isolated enough that you can get good predictions out of pretending they were isolated. For example, in this case, the potential energy of the ball is also affected by Jupiter, or even Alpha Cenaturi! However, if you run the numbers for the isolated Earth-ball system and run the numbers for a system with all the stars and planets in it, you find that the stars and other planets have such an extraordinarily weak effect that you almost get the same answer with... $\endgroup$
    – Cort Ammon
    Commented Nov 18, 2016 at 15:32
  • 1
    $\begingroup$ .. the simplified Earth-ball system as you did with the full universe system (even though it took umpteen-billion more calculations to do the full universe... they just didn't matter all that much). You can, however, create cases where those little details do matter. An example is the Lagrange point between the Earth and the moon. A Lagrange point is where the pull of gravity from those two bodies cancel each-other out perfectly. If you were to put a ball at a Lagrange point for the earth and moon, and calculate it with an isolated Earth-ball-moon system, you'd find it should stay still. $\endgroup$
    – Cort Ammon
    Commented Nov 18, 2016 at 15:33
  • 1
    $\begingroup$ In reality, the ball would be ever so slightly disturbed from this perfect balance by the sun, or by Jupiter, and fall off of that special point. At the Lagrange point, all of the big effects from the Earth and moon cancel out such that those itty bitty effects from the sun or other planets can actually affect things. Of course, as the ball falls away from the Lagrange point, towards either the Earth or the moon (depending on which way it was perturbed), the two major gravitational effects will no longer perfectly cancel. Once the ball moves away from that point, it once again is... $\endgroup$
    – Cort Ammon
    Commented Nov 18, 2016 at 15:37
  • $\begingroup$ ... effective to just model the simplified Earth-ball-moon system. The effect of "the rest of the universe" will once again be small enough that you can get away with the simplification. $\endgroup$
    – Cort Ammon
    Commented Nov 18, 2016 at 15:38

Ball, Field or Force

When you do work against gravity, you store energy in the gravitational field as gravitational potential energy, which then gravity uses to do work on that object.

But isn't the work done by our arms stored on the ball?
He says that the energy spent is stored in the gravitational field.
Work is done upon the ball; shouldn't the energy be stored on the ball?

(my emphasis)

No - it doesn't follow that energy should be stored in or on the ball.

Forget gravity for a moment. Imagine you are floating far away in space and have two large objects next to each other connected by a spring. If you put your feet on one object and use your arms to move the other object away, you feel as if you are doing work on the object but the energy is stored in the spring.

Origin of energy

where does gravitational force get it's energy to do work upon the ball

As Virgo and Cort Ammon's answers explain, when you move the ball away from the Earth you are storing energy in the gravitational field, it is a function of the configuration of objects within that field.

Storage mechanism

or - what the heck is this gravity thing anyway?

As Feynmann once explained, sometimes you can't satisfactorily explain phenomenon other than through mathematics.

  • 1
    $\begingroup$ The energy stored in a spring is explained by elastic deformation and energies between molecular bonding being stretched etc. etc. The OP wants to know where the gravitational energy comes from. $\endgroup$ Commented Nov 16, 2016 at 19:42
  • $\begingroup$ The OP is not wondering where the energy is stored he's wondering where it comes from. $\endgroup$ Commented Nov 17, 2016 at 10:05

In Newtonian gravity, the potential energy of the ball does not get "stored" anywhere. It is just a function of the configuration of the ball and the earth system.

In general relativity, the concept of gravitational energy is not always well defined. But it is meaningful in the Newtonian limit which requires a weak gravitational field and velocities much less than light. In that case, the mass of the system needs to take into account the Newtonian binding energy.

Another case energy is well defined in is if you have an isolated system in asymptotically flat spacetime, such as two orbiting neutron stars. Then one can evaluate how much energy is carried away by the gravitational waves. This causes the neutron stars to inspiral and has been measured very precisely.


Space acts as reservoir of energy.

For simplicity, consider space as invisible spring connecting "each mass/energy unit of one body" to "each mass/energy unit of another body". The spring is one way, it only shrinks and it does so per inverse square law. You can never press it enough to make it a pushing spring.

That invisible spring can be considered as the energy pool (potential).

  • $\begingroup$ The energy stored in a spring is explained by elastic deformation and energies between molecular bonding being stretched etc. etc. The OP wants to know where gravitational energy comes from. $\endgroup$ Commented Nov 16, 2016 at 21:10
  • $\begingroup$ @BillAlsept: From the reservoir, which is space, the shrinking spring, invisible. $\endgroup$
    – kpv
    Commented Nov 16, 2016 at 22:49
  • $\begingroup$ You're just saying shrinking Springs. You're not saying what your springs are or how they shrink etc. etc. You're not giving an answer $\endgroup$ Commented Nov 16, 2016 at 23:03
  • $\begingroup$ @BillAlsept: I wish I could answer where it gets its energy from. Unfortunately, we do not have verifiable answer in that manner. Only quantitative verification is possible. Per GR, as the bodies move closer the space curves more and more, thereby imparting its energy to the moving bodies as KE. In other words, shrinking of the spring imparts energy which the OP is asking. Natural state of space (in presence of mass/energy) is to be curved as much as it can. While falling to that natural state, it imparts energy to mass in the form of KE. $\endgroup$
    – kpv
    Commented Nov 17, 2016 at 0:33

Let's split Earth in 2 equal objects of mass

m1 = m2 = Mearth/2

Let's take a look at the potential energy between each of the 2 objects and a 3rd object of mass mobject.

E1 = G m1 mobject / r1,object
E2 = G m2 mobject / r2,object

so if m1 and m2 are very close together r1,object = r2,object = r so the total potential energy is

Etotal = E1 + E2
= G m1 mobject / r1,object + G m2 mobject / r2,object
= G (m1+m2)mobject / r
= G Mearth mobject /r

The potential energy is actually obtained from integrating over ρdV and summing all contributions from all other objects:
E = Σi ( G mobject ρidV /ri,object )

The more the mass the more the Potential Energy, the closer the harder to escape from it.

So, if you would want to move Earth from Sun's orbit to another solar system you would need to give enough energy to Earth to escape from its orbit around the sun.

If you were to quantize this energy that you need to give to Earth as E = mc2 you will find it on the order of mass of our Moon.

G = 6.674×10−11 N(m/kg)2
Mearth = 5.9737 x 1024 kg
Msun = 1.989 x 1030 kg
rearth, sun = 149597890 103 m

Eearth, sun = G mearth m sun / rearth, sun

so Eearth,sun = 5.2387488887711 1033 Jules
and if we convert to mass (E=mc2) the equivalent mass is
mE = 5.8208320986346 1016 kg

So to move Earth form our solar system to another solar system we need energy to break this potential energy + some energy to "make" it arrive at its new destination... etc. The rest you can deduce by yourself.
As per total some 5.82 1016 kg from Earth (and Sun) are from our potential energy.

Gravity gets its energy from mass

Now if one wants to move to a deeper interpretation, like in the view of General Relativity, mass or energy curves space which means that gravity is but curved space around a mass, like denoted by Schwarzschild spherically symmetric metric
2 = (1 - 2GM/r)*dt2 - (1/c2)(dr2/(1 - 2GM/r) + r2(dθ2 + sin2θdφ2))

  • $\begingroup$ So that was an example of calculation of a mass defect. And a falling ball gets energy from the increasing mass defect. There's a subscript error in the formula of potential energy of earth. $\endgroup$
    – stuffu
    Commented Nov 18, 2016 at 0:00
  • $\begingroup$ Yes and yes, there were some typos sorry ( I've corrected it - I think...) $\endgroup$
    – Mihai B.
    Commented Nov 18, 2016 at 10:49
  • $\begingroup$ And I suppose we could have 2 objects send them to collide near a high gravitating mass. Like 2 electrons accelerated by black hole's gravity and which then collide near it. We'll get more energy from the collision. So we got energy "into" the electrons by accelerating them with a high gravitational field; and since they are in direct collision path ... pam pam pam ... more energy shows up in the collision. $\endgroup$
    – Mihai B.
    Commented Nov 18, 2016 at 11:03
  • $\begingroup$ Two electrons are sent to collide in a gravity well. In the collision for example 22 low-mass electrons and 20 low-mass positrons might be created, because there's the mass defect thing. $\endgroup$
    – stuffu
    Commented Nov 21, 2016 at 8:46
  • $\begingroup$ Let gamma = 1/sqr(1 -v^2/c^2). Let gamma1 = 2 and let gamma2 = 2. If the electrons are in direct collision course more stuff shows up on the map, whether it is in the form of leptons or EM waves or GW or whatever else. As per total as to the energy conservation rule. Yes maybe there is going to be less electrons/positrons but some other kind of energy is expected to show up too (EM/GW/etc). I was referring to the total energy of the "things" implicated in the collision meaning sum(E(before)) = sum(E(after)). $\endgroup$
    – Mihai B.
    Commented Nov 21, 2016 at 13:12

I am not going to try to tell any truths about where a falling ball is getting energy. Instead I just say something that maybe makes some sense:

An arm lifting a ball gives energy to the ball. Yes that is a reasonable idea.

Then the energy either stays in the ball, or it moves from the ball to somewhere else.

As we have learned about such idea that the energy spent lifting a ball is stored in the gravity field, we can make a guess that the energy moves from the ball to the gravity field, if the energy moves from the ball to somewhere else.

  • $\begingroup$ Yes, when lifting a ball, it certainly acquires Gravitational Potential Energy, but to attribute energy storage to either ball or field alone is not reasonable. I can move the same ball in a zero gravitational thought experiment, and there will be no GPE stored by that displacement. In other words, the ball and the gravitational field must both be present for a ball displacement to store GPE. Thus, a reasonable conclusion about the location of the energy storage is that the ball and gravity field form a system, and energy expended to displace the ball stores energy in that system. $\endgroup$ Commented Jun 25, 2018 at 19:53
  • $\begingroup$ in that case, is the energy split in some ratio between the field and the ball? $\endgroup$
    – Aveer
    Commented Jun 10, 2022 at 17:18

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