Everyone knows kinetic energy is converted to potential energy in the body when it moves up against the earth's opposing gravitational force. But I am facing some problem with this.

What I learnt about Newton's third law of motion and ... :

From the answer and comment of my another question written by @Floris, I have learnt that Newton's 3rd law of motion is a direct consequence of law of conservation of energy. When a body moves in a certain direction and an opposing force acts on it , it does exert a reacting force (by Newton's law) on that agent doing positive work on the agent and by law of conservation of energy the agent gains that energy which the moving body spends while doing work on the agent. Hence, the body loses its kinetic energy.

Problem in the case of the body moving up against the earth's gravitational force:

When a body starts moving up from the earth, it immediately faces the opposing gravitational force . So the body will also exert reactive force on the earth,the agent of the gravitational force. Thus using the concept what I have learnt , the body must do work on earth which will eventually make it to lose energy and by the law of conservation of energy,the lost energy will be taken by the agent upon which the body exerts reactive force ie. earth .

But nothing does happen like that. During its ascent,kinetic energy is converted to potential energy which is stored in the body,not in the earth.

I have some loophole in my understanding. But where have I done the mistake? Where is my intuition incorrect? Please clear my confusion.

  • $\begingroup$ FYI. When a body accelerates in a certain direction and an opposing force acts on it [...] It doesn't have to accelerate, it just has to be moving to have an opposing force acting on it and for all this to take place. $\endgroup$
    – Steeven
    Commented Oct 15, 2014 at 17:14
  • 1
    $\begingroup$ The potential energy is not stored in the body, it is stored in the relative positions of the body and Earth. If we removed Earth from the picture, the body would no longer have gravitational potential energy. I suspect that that is your primary misunderstanding. $\endgroup$ Commented Oct 15, 2014 at 17:20

5 Answers 5


If I understand you correctly, your mistake is in using friction as an analog to gravity.

Because friction is a non-conservative force the work done is dependent on the path taken. Furthermore the energy "lost" due to friction is stored in a way that is not spontaneously reversible within the system (e.g. heat, plastic deformations, etc.). Gravity on the other hand is conservative so the work done by gravity is not dependent on the path taken, and is dependent only on the distance between the two objects.

To say an object does work on another object is also slightly incorrect because the work is done by a force. In this respect gravity does work on both objects, because the force applies to both of them as understood through Newton's Third Law,and Newton's Law of Gravitation.

Alternatively looking at different reference frames might help. e.g. In an "earth-centric" reference frame I would push off the earth with my feet, and lift the ball relative to the earth with my hand. In a "ball-centric" reference frame I would push off the ball with my hand, and lift the the earth relative to the ball with my feet.


Here are a few points to keep in mind:

  1. Potential energy is always described as the potential energy of the system. For example, the gravitational potential energy of the Earth-Moon system, belongs to the system as a whole, not the Earth or the Moon individually. So for your example, if you are for instance throwing a brick upwards, it would be the potential of the brick-Earth system.

  2. The Work-Energy Theorem can be written (in terms of conservative, non-conservative and other, external forces) as: \begin{align*} W_{total} &= \Delta K = K_{f} - K_{i}\\ W_{cons} + W_{non-cons} + W_{other} &= K_{f} - K_{i} \end{align*} But for a conservative force, the definition of the associated potential energy is $$W_{cons} \equiv -\Delta U = -(U_{f}-U_{i}),$$ and so our previous equation becomes: \begin{align*} W_{cons} + W_{non-cons} + W_{other} &= K_{f} - K_{i}\\ -(U_{f}-U_{i}) + W_{non-cons} + W_{other} &= K_{f} - K_{i}\\ U_{i} + K_{i} + W_{non-cons} + W_{other} &= K_{f} + U_{f}\\ \end{align*} If there are no external forces or non-conservative forces, then: $$U_{i} + K_{i} = U_{f} + K_{f}.$$ So we see that we can either use the concept of the work done by gravity, OR we can use the concept of gravitational potential energy. But we don't want to do both at the same time, as then we would count the influence of gravity twice.


Nothing is actually stored. (You will not find anything "in" the body :) )

The increase of potential energy means in this case that there is a force (of gravity) acting on a body, and the body's movement away from the source of this force increases the distance the body can p o t e n t i a l l y travel under the influence of this force. So if the body is eventually allowed to free fall, with every moment of the accelerated motion of this body toward the source of gravity its energy will be increasing according to the equation


Acceleration means the velocity will be increasing with every moment of the body's travel. Therefore, the further the body gets from the source of gravity, the longer its movement will be, so the greater the velocity it will attain, and therefore the greater the kinetic energy it will actually acquire.

In short, moving away from the source of acceleration increases the potential velocity a body can finally achieve on its way back (to the source of acceleration). That's all.

(Also, the movement away from the source of gravitation requires a force to counteract the force of gravity, which means expenditure of energy. Letting the body go allows it to acquire energy (back) - from its movement toward the source of gravity.)

To maintain consistence with the other question you referred to:

A body must do work against an opposing force to continue motion.

This is a statement that I have found many times. But what is the reason behind it ? Suppose $F_1$ is acting on a body to accelerate it (to increase the K.E). Then another force $F_2$ less than the former acts on the body in the opposite direction. So, according to the above statement, the body must have to lose energy. But why will the body lose energy?

I will give you a different answer than people did there, but perhaps a more straightforward one.

Why will the body loose energy? OK, how was its energy expressed? Before another force begun to oppose, the body's energy was constantly increasing as its kinetic energy equals to $mv^2/2$, and $v$ was constantly increasing, because according to this law by Newton: $F=ma$ the force was causing an acceleration. Now, when the opposing force appeared, the resultant force working on the body was $F_1-F_2$. This means that the acceleration of the body must have decreased, which means its velocity must have decreased, and which means its kinetic energy must have decreased. Hence the loss of energy.

OK, now let's go back to your initial statement: "A body must do work against opposing force to maintain motion". According to Newton, again, a body is in (uniform) motion when there is no (resultant) force working on it. You do need a force as an impetus to body's motion from a stationary state, but in order to maintain (uniform) motion you do not need it anymore. Now, if this body in uniform motion encounters an opposing force ($F_2$), this force will cause an acceleration of the body in the opposite direction - so first it will decelerate the body to a halt, and then accelerate it in the opposite direction. If you, however, want to counteract it and maintain the initial motion you need to apply another force ($F_2$) to the body. Obviously, this (and any) force needs time to exert its influence, which translates into distance ($s$), so given the equation for work: $W=Fs$ we can see there must be work done to maintain the motion.

  • $\begingroup$ Thank you, sir. But can you tell where my understanding is flawed ?...Is not my approach to Newton's third law right? Why does the body going upward not follow the same rule as the body at @Floris' answer? $\endgroup$
    – user36790
    Commented Oct 15, 2014 at 17:42
  • $\begingroup$ @user36790: I have expanded my answer to incorporate your other problem. $\endgroup$ Commented Oct 16, 2014 at 9:47

Paily's answer is the correct one. Mechanical energy is defined as the sum of kinetic and potential energy and equal to the work done by non-conservative forces. When these are absent (as when they act perpendicular to displacement) they do no work, therefore
ΔK+ΔU = 0 and K+U = constant. In short, the principle of conservation of mechanical energy: any change in kinetic energy K is always compensated by an equal but opposite change in potential energy U.

BTW, a conservative force describes the force that, like gravity, "returns" work that has been done against it. Friction, for example, is a non-conservative force.

  • $\begingroup$ >Gravity returns the work done against it. Sir,can you please explain how the body does work on the gravity which it returns later? It will be grateful if you explain. $\endgroup$
    – user36790
    Commented Oct 16, 2014 at 16:44

What you say is exaclty true, until:

But nothing does happen like that.

Oh yes it does!

During its ascent, kinetic energy is converted to potential energy which is stored in the body, not in the earth.

You can't generally say that potential energy is "stored" somewhere. Instead of correcting you, I'm rather investigating what led you there. I believe you have not included the fact that the body indeed applies force to the earth and that force indeed applies work to the earth! If some magic removed everything from the universe except the earth, the earth would move a liiiiitle differently than in the magic removed everything except the earth AND your body.


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