[jeffreyH (OP)]

An object moving away from a large mass has an initial kinetic energy. Gravity removes this positive kinetic energy until it reaches a value of zero. It then continues to remove kinetic energy so that the object now travels towards the centre of the force with increasing negative kinetic energy. Where does the kinetic energy go? Is it conserved?

[chiralSPO]

An object moving away from a large mass has an initial kinetic energy. Gravity removes this positive kinetic energy until it reaches a value of zero.

I'm with you so far...

It then continues to remove kinetic energy so that the object now travels towards the centre of the force with increasing negative kinetic energy.

I don't think this is right. The kinetic energy is still positive, even if the velocity is defined as negative (velocity is squared in the equation K.E. = 0.5*m*v², so the sign of the velocity is inconsequential, only the magnitude matters here).

Where does the kinetic energy go? Is it conserved?

This problem is traditionally thought of as kinetic energy being converted into potential energy (gravitational potential energy in this case), which is then converted back to kinetic energy. Energy overall is conserved, but neither kinetic energy nor potential energy is conserved.

[Janus]

Basically, the equation  Et = mv^2/2-GMm/r 
Where
Et is the total energy of mass m with respect to M
M is the gravitating mass
r is the center to center distance between m and M
gives a constant value of Et for mass m in a free fall state (including orbits)

For a closed trajectory (circular or elliptical orbits), Et will be negative.
For a parabolic trajectory (v= escape velocity at r) Et will be 0. (Et is set at zero for mass m when it is at rest with respect to M at an infinite distance from M)
For a hyperbolic trajectory (v> escape velocity at r) Et will be positive.

[jeffreyH (OP)]

The momentum of a body falling back towards the earth will be p = -m v. Do we distinguish which quantity 'owns' the minus sign? We can sensible assume that it is velocity since negative mass in this context makes no sense. If a force is attractive only it must remove energy. Potential energy balances the books but does it hide the underlying mechanism? The negative velocity squared does give a positive result but then information is lost since you cannot tell from the answer whether the kinetic energy was directed away from or towards the source. Since velocity, a vector, times mass, a scalar, is itself a vector momentum preserves directionality. Since energy is a scalar we cannot rely on it to tell us everything we need. This has to be the job of momentum.
So if the momentum is negative can we say the same about the energy? Do we need a concept of negative kinetic energy?

[jeffreyH (OP)]

Basically, the equation  Et = mv^2/2-GMm/r 
Where
Et is the total energy of mass m with respect to M
M is the gravitating mass
r is the center to center distance between m and M
gives a constant value of Et for mass m in a free fall state (including orbits)

For a closed trajectory (circular or elliptical orbits), Et will be negative.
For a parabolic trajectory (v= escape velocity at r) Et will be 0. (Et is set at zero for mass m when it is at rest with respect to M at an infinite distance from M)
For a hyperbolic trajectory (v> escape velocity at r) Et will be positive.

That is just a Lagrangian. That was not my point.

[jeffreyH (OP)]

Basically, the equation  Et = mv^2/2-GMm/r 
Where
Et is the total energy of mass m with respect to M
M is the gravitating mass
r is the center to center distance between m and M
gives a constant value of Et for mass m in a free fall state (including orbits)

For a closed trajectory (circular or elliptical orbits), Et will be negative.
For a parabolic trajectory (v= escape velocity at r) Et will be 0. (Et is set at zero for mass m when it is at rest with respect to M at an infinite distance from M)
For a hyperbolic trajectory (v> escape velocity at r) Et will be positive.

Thank you for your reply it has helped me to move in the right direction.

[alancalverd]

An object moving away from a large mass has an initial kinetic energy. Gravity removes this positive kinetic energy until it reaches a value of zero.

So far, so good

It then continues to remove kinetic energy so that the object now travels towards the centre of the force with increasing negative kinetic energy. Where does the kinetic energy go? Is it conserved?

And there's the mistake.As the body rises it loses kinetic energy and gains potential energy. As it descends it loses pe and gains ke.

Energy is conserved. Energy is the sum of ke and pe.

[timey]

If a body gains potential energy at h from M, then how can the equivalence principle be upheld?

A caesium atom is thought to only 'appear' to have a higher frequency at h from M from the perspective of the lower reference frame.

It is supposedly the gravity potential that causes this to occur, but if one is in the reference frame with the clock in the higher gravity potential, there is no extra gravity potential energy apparent for the atom.

The equivalence principle states that the atom will be the equivalent in all reference frames.

So how can it be said that a body gains potential energy at h from m?

[guest4091]

Begin with a test mass m on the surface of earth mass M and physicist Biff. The g-field from M is always on, i.e. m is accelerating downward but the ground gets in the way preventing any motion. Inserting a scale between the ground and m will easily prove this. To make the acceleration apparent m must be separated from the ground. The 3 masses define a closed system of energy. Biff tosses m vertically supplying the ke from his body. As m rises gravity imparts ke downward until m stops and reverses direction and returns to the ground. To label the acceleration of gravity as potential energy is misleading (what's new) since the g-field is a continuous dynamic process. The ke supplied by Biff returns to the ground with the accumulated ke from the g-field. If Biff picked up m and threw it back to the ground with the same effort, the result is obvious. A variation of the example would be to let m pass near the earth, gain momentum (ke) and continue into space. Since the g-field is produced by M, m has indirectly removed some energy/mass from M instead of returning it.

[Yahya]

mathematically there is something decreasing ( speed of object) which decreases kinetic energy , and there is something increasing at the same time speed decreases which is height, at each small decrements in kinetic energy there is an equal increment in potential energy( conservation of energy ). you can add small change in g.
theoretically ,  the object does work against gravity - force-,  it consumes this work done from its own kinetic energy, gradually , the object finally in fact LOSES its kinetic energy, but being moved again towards the opposite direction is an independent process, which is force(gravity ) acting on object and causing it to move , the object had lost  its kinetic energy  being given it again is another process.

[timey]

As m rises gravity imparts ke downward until m stops and reverses direction and returns to the ground.

...and by what mechanism is gravity imparting ke downwards?

[JohnDuffield]

An object moving away from a large mass has an initial kinetic energy. Gravity removes this positive kinetic energy until it reaches a value of zero.

At which point the mass of the body has increased. That external kinetic energy is now potential energy, which is mass-energy, which is actually internal kinetic energy. If some of that internal kinetic energy were released, what you'd have is a radiating body losing mass as described in Einstein's E=mc² paper.

It then continues to remove kinetic energy so that the object now travels towards the centre of the force with increasing negative kinetic energy. Where does the kinetic energy go? Is it conserved?

That kinetic energy isn't negative, it's positive. When our body falls back down gravity converts some of the potential energy or mass-energy or internal kinetic energy into external kinetic energy. After the body hits the ground this external kinetic energy is dissipated, and you're left with a mass deficit. See the Wikipedia binding energy article:

"Classically, a bound system is at a lower energy level than its unbound constituents. Its mass must be less than the total mass of its unbound constituents. For systems with low binding energies, this "lost" mass after binding may be fractionally small. For systems with high binding energies, however, the missing mass may be an easily measurable fraction. This missing mass may be lost during the process of binding as energy in the form of heat or light, with the removed energy corresponding to removed mass through Einstein's equation E = mc²".

Note that there is an effect on the large body, but whilst momentum is equal and opposite, kinetic energy is not. When you drop a brick there's a measureable change in the kinetic energy of the brick, but there's no measurable change in the kinetic energy of the Earth. It's so slight that we discount it. 

[JohnDuffield]

...and by what mechanism is gravity imparting ke downwards?

By making light curve downwards, because of the wave nature of matter. Think about pair production, where we made an electron and a positron out of light. Also remember the Einstein-de Haas effect, which "demonstrates that spin angular momentum is indeed of the same nature as the angular momentum of rotating bodies as conceived in classical mechanics". Then see Hans Ohanian’s 1984 paper what is spin? He says this: “the means for filling the gap have been at hand since 1939, when Belinfante established that the spin could be regarded as due to a circulating flow of energy”. Check out the Poynting vector, then think of the electron as light going round and round, then simplify it further to light going round a square path. Remember gravity makes light curve downward, so what happens to that electron? This:



The horizontals curve downwards a little, and the electron falls down. In doing so internal kinetic energy is converted into external kinetic energy. Because only the horizontals bend downwards, the deflection of light is twice the Newtonian deflection of matter. See Ned Wright's article for something on that. "Before Einstein developed the full theory of General Relativity he also  predicted a deflection of 0.875 arcseconds in 1913, and asked astronomers to look for it. But World War I intervened, and during the war Einstein changed his prediction to 1.75 arcseconds, which is twice the Newtonian deflection".

[timey]

Nope, I'm sorry John, but - according to the equivalence principle you cannot add that potential energy to the body because this means that the atom isn't the equivalent in each reference frame.

Also - Pmbphy has mentioned in another thread elsewhere that potential energy does not affect relativistic mass.  He states that it is only kinetic energy that affects relativistic mass.

[jeffreyH (OP)]

Ok. Negative kinetic energy requires an imaginary velocity, complex in other words. So that wasn't really an option. However the velocity itself can become negative due to choice of coordinates. Therefore negative momentum is also due to choice of coordinates. If we turn the tables and view the upward motion of an object to be due to a negative acceleration. Bear with me. Then Newtons laws are violated since a decreasing force should still cause a positive velocity. So gravity sucks big time. If I put a straw into a glass of milk and suck the amount of milk decreases and the force causing it, me, gains something from the milk. Potential milk anyone?

[jeffreyH (OP)]

Nope, I'm sorry John, but - according to the equivalence principle you cannot add that potential energy to the body because this means that the atom isn't the equivalent in each reference frame.

Also - Pmbphy has mentioned in another thread elsewhere that potential energy does not affect relativistic mass.  He states that it is only kinetic energy that affects relativistic mass.

You are certainly on fire at the moment. Keep it up!

[timey]

Remember gravity makes light curve downward,

...obviously, but by what mechanism?

"Before Einstein developed the full theory of General Relativity he also  predicted a deflection of 0.875 arcseconds in 1913, and asked astronomers to look for it. But World War I intervened, and during the war Einstein changed his prediction to 1.75 arcseconds, which is twice the Newtonian deflection".

Interestingly though John, 1.75 is the so far dimensionless gravitational coupling constant, negating the power 10 consideration...

https://en.m.wikipedia.org/wiki/Gravitational_coupling_constant

(Jeff, thanks for comment!)

[JohnDuffield]

Nope, I'm sorry John, but - according to the equivalence principle you cannot add that potential energy to the body because this means that the atom isn't the equivalent in each reference frame.

You do. There is no magical mechanism by which the very real kinetic energy of a bullet fired upwards somehow disappears or zips across space to some other place. The bullet has that kinetic energy, and it stays with the bullet, as internal kinetic energy aka potential energy aka mass-energy. Conservation of energy applies. The mass deficit is not something I made up. A body at rest at a low elevation has less mass-energy than the same body at rest at the higher elevation.   

Also - Pmbphy has mentioned in another thread elsewhere that potential energy does not affect relativistic mass.  He states that it is only kinetic energy that affects relativistic mass.

Relativistic mass is not rest mass.

...obviously, but by what mechanism?

It's akin to refraction. We don't call it gravitational lensing for nothing. Check out what Einstein said: "the curvature of light rays occurs only in spaces where the speed of light is spatially variable". It's rather like sonar actually:

[timey]

what Einstein said: "the curvature of light rays occurs only in spaces where the speed of light is spatially variable".

It can work just as equally if one says:

"The curvature of light rays occurs only in spaces where the speed of light is 'temporally' variable"

To far better and more sensible results!

[jeffreyH (OP)]

Nope, I'm sorry John, but - according to the equivalence principle you cannot add that potential energy to the body because this means that the atom isn't the equivalent in each reference frame.

You do. There is no magical mechanism by which the very real kinetic energy of a bullet fired upwards somehow disappears or zips across space to some other place. The bullet has that kinetic energy, and it stays with the bullet, as internal kinetic energy aka potential energy aka mass-energy. Conservation of energy applies. The mass deficit is not something I made up. A body at rest at a low elevation has less mass-energy than the same body at rest at the higher elevation.   

Also - Pmbphy has mentioned in another thread elsewhere that potential energy does not affect relativistic mass.  He states that it is only kinetic energy that affects relativistic mass.

Relativistic mass is not rest mass.

...obviously, but by what mechanism?

It's akin to refraction. We don't call it gravitational lensing for nothing. Check out what Einstein said: "the curvature of light rays occurs only in spaces where the speed of light is spatially variable". It's rather like sonar actually:

Do you actually understand what potential energy actually is?