Why gravitational potential energy is a system property?

[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.

 

[Ibix]

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.

Ok. But in that case $Mg_{obj}H\neq mg_{earth}h$. So are you now retracting your claim that they are equal?

 

[Ibix]

Let's do this calculation properly.

Let Earth and the bullet have masses $M$ and $m$ respectively, and initial velocities of zero. Propellant detonates, releasing energy $E$ which is absorbed by Earth and the bullet, accelerating them to velocities $V$ and $v$ respectively. They move distances $H$ and $h$ respectively before coming to rest due to their mutual gravity slowing them down.

The initial total momentum is zero, and by the conservation of momentum, so must be the total momentum immediately after the propellant detonates. Thus $0=mv+MV$, which is to say that $v=-MV/m$. The kinetic energy of the bullet is $mv^2/2$, and substituting the previous result tells you that this is $(M/m)MV^2/2$, so the kinetic energy of the bullet is $(M/m)$ times that of the Earth.

Since the maximum height that the objects reach (note that height, in this case, would actually be better described as displacement from their initial positions) is when their kinetic energies are completely converted into gravitational potential we can simply equate the initial kinetic energies to the final potential energy gains. That is, $$\begin{eqnarray*}mg_{earth}h&=&\frac Mm\frac 12 MV^2\\
Mg_{obj}H&=&\frac 12 MV^2
\end{eqnarray*}$$These two things are clearly not equal - they differ by a factor of $M/m$. To insist that they are equal would require that $m=M$ (clearly not the case in this scenario), or else require you to violate conservation of momentum by having $mv+MV\neq 0$.

Edit: if you want to determine $v$, $V$, $h$, or $H$ this could be done by noting that the initial kinetic energies of Earth and bullet must sum to $E$.

 

[rajen0201]

Here, it is not the case energy transfers from Earth to bullet where you interpretation is applicabble.
Note that, Initially they pushed in opposite direction by external source of energy, So, total momentum is divided in two parts (to Earth and bullet) that is mv(bullet)+MV(earth) - (MV)Momentum provided by Detonator) = 0.

 

[Ibix]

So, total momentum is divided in two parts (to Earth and bullet) that is mv(bullet)+MV(earth) - (MV)Momentum provided by Detonator) = 0.

No. The detonator provides zero net momentum. This can be seen in your own earlier maths - the pressure on the top of the detonation chamber is the same as the pressure on the bottom. That means that the forces are equal and act for equal time and hence the net momentum change is zero. So your equation reduces to $mv+MV=0$, as I said.

 

[Ibix]

Furthermore, your $mg_{earth}h=Mg_{obj}H$ conclusion leads directly to $h=H$, which even you seem to recognise is absurd. Your own maths tells you in multiple ways that you are wrong. You should listen to it.

 

[sophiecentaur]

So, total momentum is divided in two parts

You keep making nonsense statements. You seem to be unaware of the difference between Momentum and Energy - a pre - Newtonian problem that has long since been sorted out. Try a bit of 'Physics Input', rather than 'non-Science fiction'.

 

[rajen0201]

You keep making nonsense statements. You seem to be unaware of the difference between Momentum and Energy - a pre - Newtonian problem that has long since been sorted out. Try a bit of 'Physics Input', rather than 'non-Science fiction'.

I understand from the conversation that kE of the bullet is higher than Earth.
let me know.
is that the efficiency (=work/energy) of a system directly depends on the mass ratio? as the bullet has too much energy against the earth?

 

[sophiecentaur]

I understand from the conversation that kE of the bullet is higher than Earth.
let me know.
is that the efficiency (=work/energy) of a system directly depends on the mass ratio? as the bullet has too much energy against the earth?

I wouldn't use the term 'efficiency' there because both the Energy quantities are the 'output' of the operation. Efficiency would have to relate to things like the chemical energy into the muscles or the explosion etc. What you say certainly doesn't apply to "a system". Have a look at the definition of Efficiency of machines etc..

 

[jbriggs444]

I understand from the conversation that kE of the bullet is higher than Earth.
let me know.
is that the efficiency (=work/energy) of a system directly depends on the mass ratio? as the bullet has too much energy against the earth?

It depends (as @sophiecentaur suggests) on what you mean by "efficiency".

If you are looking at the percentage of the input energy (from the explosives) that shows up as kinetic energy in the bullet then 100% efficiency is approached in the limit when the mass (m) of the bullet is negligible compared to the mass (M) of the earth.

This assumes that we are using a frame of reference in which both bullet and Earth start at rest. If we are using a frame where bullet and Earth are not initially motionless then the notion of "efficiency" becomes more slippery and results exceeding 100% are possible.

 

[sophiecentaur]

100% efficiency is approached in the limit when the mass (m) of the bullet is negligible compared to the mass (M) of the earth.

I haven't come across the word 'efficiency' being used to describe the share of the output of mechanical Energy from a process. Isn't it usually out/in? I guess you could argue that a system where there are no other significant losses in the propulsion system then the 'share' of output Energy when the two masses are comparable would dominate in any calculation of Efficiency but "I wouldn't encourage it".
We need to be careful that @rajen doesn't run away with the idea that he's using the term in a way that would make sense to most people - without a lot of further qualification.

 

[jbriggs444]

I haven't come across the word 'efficiency' being used to describe the share of the output of mechanical Energy from a process. Isn't it usually out/in? I guess you could argue that a system where there are no other significant losses in the propulsion system then the 'share' of output Energy when the two masses are comparable would dominate in any calculation of Efficiency but "I wouldn't encourage it".
We need to be careful that @rajen doesn't run away with the idea that he's using the term in a way that would make sense to most people - without a lot of further qualification.

I do not disagree, though the set of things which I would label as "efficiency" is clearly more broad than the set you would agree to so label.

A more realistic accounting would likely find that much of the chemical energy in the explosives had been dissipated into thermal energy and rapidly expanding combustion residue and that the explosion had not been very efficient at all in converting chemical energy to kinetic energy.

Edit: Wikipedia has some ballpark numbers. [And the xkcd in the references is priceless]