Why gravitational potential energy is a system property?

[rajen0201]

Spring has more potential energy when it is compressed or stretched from its initially balanced state. As external work is done, it stores energy in the form of potential energy. Here, we know energy is stored in spring but For the Earth-ball system, where the energy stored?

 

[PeroK]

Summary:: gravitational potential energy is described as a system property and it depends on the distance between two objects.

Spring has more potential energy when it is compressed or stretched from its initially balanced state. As external work is done, it stores energy in the form of potential energy. Here, we know energy is stored in spring but For the Earth-ball system, where the energy stored?

It's a good question. In principle, the question is the same as asking how the force of gravity can be exerted by one body on another across space, with no physical interaction. And that question also applies to electromagnetic forces.

The simplest answer is that in Newtonian gravity and classical electromagnetism, the fields can and do store energy.

PS see also post #2. Ultimately the spring is storing its energy in atomic and molecular electromagnetic fields. Hence, it's not conceptually very different if you look at it closely enough.

 

[weirdoguy]

PS see also post #2.

Since it was deleted as "unhelpful" (I guess it was to laconic) I'm going to restate what I said there, maybe in a more helpful way. What I was getting at is that you, @rajen0201, said with such an ease that "potential energy of spring is stored in the spring itself". Can you explain to us what do you think it means? Where exactly in the spring this energy resides? In the center, in the back? For me the question you ask is almost the same so it urges to have the same answer. And if not, why not?

 

[rajen0201]

Yes, Newtonian gravity and classical electromagnetism, the fields can store energy.
But how this phenomenon works for Earth consumes energy and ball stores it. For spring we can see it is twisted. But We don't see any effect on elevated ball.

 

[Ibix]

For spring we can see it is twisted.

You mean that you can see that the spring has changed configuration when compressed. You can see that the various parts are in different places.

But We don't see any effect on elevated ball.

Are you telling me that you can't see that the Earth-ball system has changed configuration when the ball is elevated? Or you can't see that the various parts are in different places?

 

[Adesh]

Summary:: gravitational potential energy is described as a system property and it depends on the distance between two objects.

Here, we know energy is stored in spring

Not exactly, in normal colloquial language we can say that but that’s not theoretically true.

Summary:: gravitational potential energy is described as a system property and it depends on the distance between two objects.

For the Earth-ball system, where the energy stored?

In the field. Assume the gravitational field as some kind of a rubber band connecting the Earth and the ball. If you move the ball, rubber band (field) will get stretched. And please remember rubber band doesn’t belong to Earth or to the object alone but it’s a shared entity.

 

[rajen0201]

You mean that you can see that the spring has changed configuration when compressed. You can see that the various parts are in different places.

Are you telling me that you can't see that the Earth-ball system has changed configuration when the ball is elevated? Or you can't see that the various parts are in different places?

I can't see that the Earth-ball system has changed configuration except they separated. this is not corelating with spring case where its intermolecular bonds are weakened. For Earth ball system, man pushes ball and Earth with same force but in opposite direction as per Newton's third law. So, man gives energy to both Earth & ball to counter gravitational force.

 

[Adesh]

I can't see that the Earth-ball system has changed configuration except they separated

That is what we call change of configuration, that is change in position which was specified to them.

this is not corelating with spring case where its intermolecular bonds are weakened.

Inter-molecular bond is just electrostatic force, if you disturb the configuration energy will be gained by the system.

For Earth ball system, man pushes ball and Earth with same force but in opposite direction as per Newton's third law. So, man gives energy to both Earth & ball to counter gravitational force.

That’s not correct.

 

[sophiecentaur]

So, man gives energy to both Earth & ball to counter gravitational force.

OK. Work is Force times Distance. Same magnitude of force on both ball and Earth. How do the distances moved compare? Not the same Work is involved. You need to be much more rigorous in your thinking.

 

[rajen0201]

Force = mass x gravity value on earth. now multiply with height h = mgh that is work for object. same is applicable to Earth case. Here, gravity of object on Earth to be considered. So, mgh Earth =mgh object.

 

[Ibix]

Force = mass x gravity value on earth. now multiply with height h = mgh that is work for object. same is applicable to Earth case. Here, gravity of object on Earth to be considered. So, mgh Earth =mgh object.

Are you observing that the gravitational potential energy change is the same whether you regard the Earth as moving or the object as moving? If so, yes. Or are you arguing that gravitational potential energy is a property of each object and hence that the gravitational potential energy change is twice the work done on the system, violating conservation of energy? If so, no.

 

[Ibix]

I appear to have missed an earlier reply to me - apologies.

I can't see that the Earth-ball system has changed configuration except they separated.

So you can see that the configuration has changed.

this is not corelating with spring case where its intermolecular bonds are weakened.

Why do you think intermolecular forces are weakened by compressing a spring? If that were the case you might expect it to get easier to compress a spring the more it is already compressed - at the very least you'd expect that the force from a spring was proportional to $x^f$ where $f<1$, but it's always modeled in the elastic regime as proportional to $x$.

Eventually you do get into a plastic regime, but then the spring force is non-conservative and you cannot use potential energy to describe it.

For Earth ball system, man pushes ball and Earth with same force but in opposite direction as per Newton's third law. So, man gives energy to both Earth & ball

See sophiecentaur's and my responses to this. Work done is force times distance; the Earth moves a distance $h_1$ and the ball moves a distance $h_2$, where the $h$ that you put into the gravitational potential energy change is $h=h_1+h_2$. The forces applied to both Earth and ball are equal and of magnitude $mg$, so the work done on the Earth is $mgh_1$ and the work done on the ball is $mgh_2$ for a total work of $mgh$. This is the correct version of your analysis in post #10.

(Note that I'm being a bit careless about signs here, but working properly with vector quantities would return the same answer.)

 

[jbriggs444]

Force = mass x gravity value on earth. now multiply with height h = mgh that is work for object. same is applicable to Earth case. Here, gravity of object on Earth to be considered. So, mgh Earth =mgh object.

That would be true if the Earth moved a distance h when you raised a small mass a distance h above it.

In practice, the object moves perceptibly and the Earth moves imperceptibly. It certainly does not move by a distance h.

 

[rajen0201]

Kindly note that, Earth & object both has gravity influence on each other in attached sketch. both floating in full vacuum universe without influence of other forces/conditions.

 

[Ibix]

Kindly note that, Earth & object both has gravity influence on each other in attached sketch.

Of course. In what way did you think @jbriggs444 and my responses didn't rely on that fact?

 

[sophiecentaur]

In practice, the object moves perceptibly and the Earth moves imperceptibly. It certainly does not move by a distance h.

It does depend on the reference frame that's used. Relative to the object, the Earth does move. If an experimenter on the object didn't already have some information about the relative masses then he/she could not be 'certain'.

We all suffer from the way that elementary physics is (necessarily) taught and PF is always picking up the pieces resulting from the fact that Reference Frames are not always grasped by people. And this is hundreds of years after Newton!

 

[rajen0201]

Of course. In what way did you think @jbriggs444 and my responses didn't rely on that fact?

Please see attachment.

 

[sophiecentaur]

I wasn't arguing with anyone. I was just restating the basic problem that most (all?) of us experience when stepping up from basic pre- 'Newtonian - relativity'. It's so 'obvious' to people that the Earth or the wall don't move. 'Common Sense' (haha) takes us on from that intuitive feeling.

 

[Ibix]

Please see attachment.

That is just restating your mistaken claim that both Earth and bullet move the same distance. That is not the case in the frame where they were initially at rest.

 

[sophiecentaur]

Please see attachment.

The attachment is fine as long as you realize that H and h are very different. The Earth, in this context is 'immoveable' and H is as near zero as makes no difference.

 

[jbriggs444]

It does depend on the reference frame that's used. Relative to the object, the Earth does move.

Indeed.

It is convenient to use a frame that is not moving around much -- one that is inertial. If we are going to speak about conservation of momentum, such is normally assumed.

It is also convenient to pick a frame where the Earth and the object are not whizzing past. One where the center of mass of the system is stationary is often used.

 

[Ibix]

The attachment is fine as long as you realize that H and h are very different.

It's only partially fine - it concludes that the work done on the two is equal, which is not the case. The forces on the two are equal, so the work done on one is $Fh$ and on the other $FH$ (with a few simplifying assumptions about $F$ not being time varying). But $h$ and $H$ are not equal, as you observe.

 

[rajen0201]

That is just restating your mistaken claim that both Earth and bullet move the same distance. That is not the case in the frame where they were initially at rest.

No. I am not claiming both covers same distance. All three parameters Mg(obj)H = mg(earth)h value at both side is different

 

[Ibix]

No. I am not claiming both covers same distance. All three parameters Mg(obj)H = mg(earth)h value at both side is different

Given that $Mg_{obj}=mg_{earth}$, which follows directly from Newton's law of gravity, your third sentence appears to be a claim that $h=H$. So which are you actually claiming? $h=H$ or $h\neq H$? The latter would be correct, but would undermine your claim about equal energies - as I pointed out in post #12, to which your diagram was a response.

 

[sophiecentaur]

with a few simplifying assumptions about FFF not being time varying

Variation with time is not relevant. N3 applies all the time and the rest follows.

No. I am not claiming both covers same distance. All three parameters Mg(obj)H = mg(earth)h value at both side is different

This appears to be total nonsense. The forces on Earth and ball are equal and, with the ball close to the surface they are constant with time. But that is not relevant - the Energy transferred to the ball as KE is much higher than the Energy transferred to the Earth.
I wish you would let this lie. I am not even sure what point you are trying to make as the problem was solved in Newton's time. I think your time would be better spent in trying to follow and understand the accepted approach which works extremely well for everyone else.

 

[Ibix]

Variation with time is not relevant. N3 applies all the time and the rest

It's relevant if you are trying to apply W=Fd without expressing it as an integral, which the OP is doing as well as applying Newton's third law.

On another thread I tried to get him to analyse a simpler case that only required momentum conservation, but have got a response (with the same diagram from this thread attached for some reason) that suggests he doesn't understand that.

 

[sophiecentaur]

It's relevant if you are trying to apply W=Fd without expressing it as an integral, which the OP is doing as well as applying Newton's third law.

On another thread I tried to get him to analyse a simpler case that only required momentum conservation, but have got a response (with the same diagram from this thread attached for some reason) that suggests he doesn't understand that.

I sympathise. I never can understand how people will argue without attempting to take on board what’s in all textbooks. At this level they can’t really believe there’s a realistic alternative view. Being wrong carries no stigma, once one admits it.

 

[rajen0201]

Variation with time is not relevant. N3 applies all the time and the rest follows.This appears to be total nonsense. The forces on Earth and ball are equal and, with the ball close to the surface they are constant with time. But that is not relevant - the Energy transferred to the ball as KE is much higher than the Energy transferred to the Earth.
I wish you would let this lie. I am not even sure what point you are trying to make as the problem was solved in Newton's time. I think your time would be better spent in trying to follow and understand the accepted approach which works extremely well for everyone else.

the energy transferred to the ball as KE is much higher than the Energy transferred to the Earth.
Our sense doesn't allow us to think differently for the case. Note that for Earth 1/2MV2, M is too much so, we don't feel Earth velocity. you didn't see the attachment. Let me what is wrong in the sketch? so, I can correct my side.

 

[rajen0201]

Given that $Mg_{obj}=mg_{earth}$, which follows directly from Newton's law of gravity, your third sentence appears to be a claim that $h=H$. So which are you actually claiming? $h=H$ or $h\neq H$? The latter would be correct, but would undermine your claim about equal energies - as I pointed out in post #12, to which your diagram was a response.

h≠H? The latter would be correct, but would undermine your claim about equal energies - as I pointed out in post #14, to which your diagram was a response.
Yes, h≠H, But, I don't interprete it is undermining the equal energy distriution between Earth and Bullet.
Force F = P x A is the same for both Earth and bullet, Now h=H applicable when both items mass is same, and Fxh = FxH. But here, Earth mass is too much So, Fxh ≠ FxH. See the attachment, At rest(against gravity) Mg(obj)xH = mg(earth)xh will be satisfied.

 

[Ibix]

you didn't see the attachment

We saw the attachment twice on two different threads. The problem is that you keep applying the wrong mathematics to solve this problem, just as in the thread I linked in my post above. You need to apply conservation of momentum to derive the initial velocity ratio of the Earth and the ball/bullet, and only then can you apply the conservation of energy to work out their relative movements.

 

[rajen0201]

Variation with time is not relevant. N3 applies all the time and the rest follows.This appears to be total nonsense. The forces on Earth and ball are equal and, with the ball close to the surface they are constant with time. But that is not relevant - the Energy transferred to the ball as KE is much higher than the Energy transferred to the Earth.
I wish you would let this lie. I am not even sure what point you are trying to make as the problem was solved in Newton's time. I think your time would be better spent in trying to follow and understand the accepted approach which works extremely well for everyone else.

See attached sketch in #14, Force F = P x A is the same for both Earth and bullet, Now h=H applicable when both items mass is same, and Fxh = FxH. But here, Earth mass is too much So, Fxh ≠ FxH. See the attachment, At rest(against gravity) Mg(obj)xH = mg(earth)xh will be satisfied.

 

[Ibix]

But here, Earth mass is too much So, Fxh ≠ FxH. See the attachment, At rest(against gravity) Mg(obj)xH = mg(earth)xh will be satisfied.

This is just nonsense. Why don't you apply the conservation of momentum? This will actually get you to the right answer.

 

[rajen0201]

This is just nonsense. Why don't you apply the conservation of momentum? This will actually get you to the right answer.

Tell me what is wrong with it.

 

[Ibix]

Tell me what is wrong with it.

You didn't use the conservation of momentum and your result violates it.

 

[rajen0201]

You didn't use the conservation of momentum and your result violates it.

conservation of momentum is not violated. refer below explanation.
The magnitudes of the forces on the objects are the same, but accelerations will not, that is because the masses are too different. Therefore, the changes in the velocity of an object will be higher than the earth.

Now introduce masses M earth, and m, the products of the mass and the change of velocity will be equal in magnitude.
MdV/dt(earth) = mdv/dt(object).
Now, we know the masses of the Earth & object do not change during the interaction.
d/dt(MV)(earth) = d/dt(mv)(object).
d/dt(P)(earth) = d/dt(p)(object). So, both Earth & oject momentum is same at every location & every time.
Also, it will be 0 on Earth surface & 0 at the highest elevations when no force is applied. In the object's return condition(due to gravity), same is also true.