[Bill S (OP)]

https://en.wikipedia.org/wiki/Negative_energy 

“The strength of the gravitational attraction between two objects represents the amount of gravitational energy in the field which attracts them towards each other. When they are infinitely far apart, the gravitational attraction and hence energy approach zero. As two such massive objects move towards each other, the motion accelerates under gravity causing an increase in the positive kinetic energy of the system. At the same time, the gravitational attraction - and hence energy - also increase in magnitude, but the law of energy conservation requires that the net energy of the system not change. This issue can only be resolved if the change in gravitational energy is negative, thus cancelling out the positive change in kinetic energy. Since the gravitational energy is getting stronger, this decrease can only mean that it is negative.”

Why is gravitational energy, rather than kinetic energy, seen as being negative?

[alancalverd]

It isn't.

The gravitational potential of a small test mass is defined as zero at infinite distance from a single large attractor, hence negative at the centre of said attractor. Thus the potential energy of a test mass is always positive with respect to the large attractor, and in free fall towards it, the loss of potential energy equals the gain in kinetic energy.

Obviously a very idealised model, but we use the same convention to define electric potential: the potential at a point is the amount of work needed to bring unit charge from infinity to that point.

Obviously this will start you worrying about the definition of infinity, but the greasyfingered engineer who resides in my head says "whatever distance is good enough" and the physicist says "thanks to the inverse square law, you don't have to go very far for a few volts plus or minus a millivolt".

[PmbPhy]

https://en.wikipedia.org/wiki/Negative_energy 

“The strength of the gravitational attraction between two objects represents the amount of gravitational energy in the field which attracts them towards each other. When they are infinitely far apart, the gravitational attraction and hence energy approach zero. As two such massive objects move towards each other, the motion accelerates under gravity causing an increase in the positive kinetic energy of the system. At the same time, the gravitational attraction - and hence energy - also increase in magnitude, but the law of energy conservation requires that the net energy of the system not change. This issue can only be resolved if the change in gravitational energy is negative, thus cancelling out the positive change in kinetic energy. Since the gravitational energy is getting stronger, this decrease can only mean that it is negative.”

Why is gravitational energy, rather than kinetic energy, seen as being negative?

It depends on the particular example. In this case the gravitational potential, U, is negative so that -grad U = F. This also holds for a positive source and a negative particle moving in the sources field. There are cases when U is negative in one region and positive in another. An infinitely long line of mass is an example.

[Bill S (OP)]

Obviously this will start you worrying about the definition of infinity

Absolutely not!  I have come to terms with the variety of uses of "infinity". I used manually focused cameras for about 60 yrs.  I just have to question the extrapolation of mathematical infinities to "prove" points that seem to be outside their scope.


The rest of your post, and Pete's will need some thought. Thanks.

[Bill S (OP)]

The gravitational potential of a small test mass is defined as zero at infinite distance from a single large attractor, hence negative at the centre of said attractor.

The GPE reduces from zero to minus infinity as the small mass approaches the large attractor. (?)

Was the decision to define GPE as zero at infinite distance, an arbitrary decision (mathematical convenience?), or is there another reason?   
Definitions such as +ve and -ve electrical terminals come to mind.

[pensador]

Wiki says it was Pascual Gordon who first suggested gravity can be viewed as negative energy.
Pascual Jordan first suggested that since the positive energy of a star's mass and the negative energy of its gravitational field together may have zero total energy, conservation of energy would not prevent a star being created by a quantum transition of the vacuum. George Gamow recounted putting this idea to Albert Einstein: "Einstein stopped in his tracks and, since we were crossing a street, several cars had to stop to avoid running us down".[3]
The zero-energy universe theory originated in 1973, when Edward Tryon proposed in the journal Nature that the universe emerged from a large-scale quantum fluctuation of vacuum energy, resulting in its positive mass-energy being exactly balanced by its negative gravitational potential energy.[4]

https://en.wikipedia.org/wiki/Zero-energy_universe

[Bill S (OP)]

. In this case the gravitational potential, U, is negative so that -grad U = F.

I read that, I understand each word, then I think: "could I explain that in simple terms to someone else?" The answer is “no”; so I guess that means I don’t really understand it, even if it feels as though I do.

[PmbPhy]

The gravitational potential of a small test mass is defined as zero at infinite distance from a single large attractor, hence negative at the centre of said attractor. Thus the potential energy of a test mass is always positive with respect to the large attractor, and in free fall towards it, the loss of potential energy equals the gain in kinetic energy.

Bill  - Sorry I wasn't able to explain it better. I think you do understand it. Explaining to others is troublesome due to the wide variety of each of the "other" people. To answer your question one has to use math because your question is math oriented and thus about the math of Newtonian gravity itself.

Not so. The potential energy of a particle in such a field is zero having the value U = -GMm/r

[alancalverd]

Now you've got me worried. I have a mass m at a large distance from M. As it hurtles earthwards it exchanges mgh for ½mv², which is always positive, so its potential energy must have been positive before it started moving.

Thank goodness for time zones. I've had a hard day evaluating collision avoidance systems (successfully, as you can see) and I look forward to Pete explaining it all while I get some sleep.

[Janus]

Now you've got me worried. I have a mass m at a large distance from M. As it hurtles earthwards it exchanges mgh for ½mv², which is always positive, so its potential energy must have been positive before it started moving.

Thank goodness for time zones. I've had a hard day evaluating collision avoidance systems (successfully, as you can see) and I look forward to Pete explaining it all while I get some sleep.

mgh only gives a good answer if h is small compared to r, the distance from the center of the Earth.  So for example, if you are on the surface of the Earth, r is 6378000 m, and g = ~9.80153837 m/s^2 at the surface.  mgh gives an energy of 98.01153837 joules.   But this assumes that g is a constant during the entire fall. It is not, it is very slighty weaker 10 m up than it is at the surface.    To figure out how much the GPE changes due to this you take the antiderivative of how of how much g changes.   Gravitational force is F = GMm/r^2 and the antiderivative of this is -GMm/r.   
If r = 6378000 at the surface of the Earth, then   the GPE difference between there and 10 m up is -GMm/6378000+10)-(-GMm/6378000) = GMm(1/6378000-1/6378000) = 98.01138438 joules or 0.00015399 joules less than what you got using mgh.  Not much of a difference.   However, if we were make h equal to the radius of the Earth, then mgh gives as answer of 62511759.17 joules, while the correct answer is 31255879.59.   The first answer would give you a impact velocity of 11.18139161 km/s*  which is equal to the escape velocity of the Earth, while that correct answer is  7.906 km/s

Going from -GMm/r to -GMm/(r+h), is going from a negative value to a less negative value, which represents a positive increase, just like going from mg(0) to mgh ( where h is postive) does.

*This happens to be equal to the escape velocity from the surface. 

[PmbPhy]

Now you've got me worried. I have a mass m at a large distance from M. As it hurtles earthwards it exchanges mgh for ½mv², which is always positive, so its potential energy must have been positive before it started moving.

The potential is U = -GMm/r just like I said earlier and Janus confirms. Suppose the kinetic energy is positive in the beginning at a large distance from r = 0. Then the potential is negative and the kinetic energy is positive while the total mechanical energy, E, is positive. The closer it gets to r = 0 the greater the kinetic energy and the less U is where E = U+ K = constant.

[Petrochemicals]

On a mathmatical point, negative in relation to physics always ends up being something we dont understand , electrically negative charge has a over burdance of electrons, whilst the positive charge (negative in the circuit) has a deficiency, a mathmatically inferior number, when both are subtracted gives a negative number. Mathmatically we understand gravity, physically we dont.

Seems what the wiki article says is you get increaced attraction. True, but you have lost some potential and gained energy too. Being as we dont understand gravity, therefore the answer is I love lamp.

[Bill S (OP)]

I’ve wrestled with the maths and tried to manipulate the logic, and I think Petrochemicals might have hit the nail on the head.

Mathmatically we understand gravity, physically we dont.

The problem is not so much paucity of explanations, as the way I think; but it’s probably worth a final shot at clarity.

The gravitational potential of a small test mass is defined as zero at infinite distance from a single large attractor

If the GPE is zero, how does that differ from having no energy?

…hence negative at the centre of said attractor………
Thus the potential energy of a test mass is always positive with respect to the large attractor.

Let’s start by avoiding the assumption that these statements are contradictory.  Does this mean that the centre of the attractor is considered as an infinitesimally small point, so, like infinity, the test mass can never actually reach it; thus, the energy of the test mass never becomes negative?

…and in free fall towards it, the loss of potential energy equals the gain in kinetic energy.

Loss of potential energy (+ve) equals gain in kinetic energy (+ve) (?) There seems to be no need to invoke negative energy. 

The only thing related to GPE that seems to increase as the small object approaches the attractor is “gravitational attraction”.  Does this represent negative energy?

[jeffreyH]

Have a read of this.
https://www.s-cool.co.uk/a-level/physics/gravitational-potential-energy/revise-it/gravitational-potential
See how potential is simply a negative scalar. Basically E/m where E is energy and m is mass. Read it through a few times.

[alancalverd]

The potential is U = -GMm/r just like I said earlier and Janus confirms. Suppose the kinetic energy is positive in the beginning at a large distance from r = 0. Then the potential is negative and the kinetic energy is positive while the total mechanical energy, E, is positive. The closer it gets to r = 0 the greater the kinetic energy and the less U is where E = U+ K = constant.

Hold on! "Suppose the ke is positive in the beginning"? No, I'm releasing a small object m from rest at distance r from the barycentre of m and M. The ke of m is initially zero, and increases as it moves towards M, so its initial pe must be positive.

[PmbPhy]

The potential is U = -GMm/r just like I said earlier and Janus confirms. Suppose the kinetic energy is positive in the beginning at a large distance from r = 0. Then the potential is negative and the kinetic energy is positive while the total mechanical energy, E, is positive. The closer it gets to r = 0 the greater the kinetic energy and the less U is where E = U+ K = constant.

Hold on! "Suppose the ke is positive in the beginning"? No, I'm releasing a small object m from rest at distance r from the barycentre of m and M. The ke of m is initially zero, and increases as it moves towards M, so its initial pe must be positive.

If you're releasing a particle "from rest" then there is a choice of what to set U to at that point. There always is. If you add a constant to U then the potential will still be negative somewhere. That follows from the fact that F = -grad(U + constant) = -grad U. I could easily set U such that U(r) = 0 rather than U(inf) = 0.

I gotta say that I'm suprised you don't know all of this. You're much better at physics than the comments in this thread,

[PmbPhy]

Have a read of this.
https://www.s-cool.co.uk/a-level/physics/gravitational-potential-energy/revise-it/gravitational-potential
See how potential is simply a negative scalar. Basically E/m where E is energy and m is mass. Read it through a few times.

Bill asked about potential energy, not potential. U = potential energy whereas U/m = V = potential. Sometimes U is called just 'potential.'

Note: E is usually used to refer to total energy, not potential energy.

[Bill S (OP)]

Thanks for the link, Jeffrey. I’ve managed to read it twice so far.  Seems to make sense, generally. I think I will have a couple of questions, but I want to see if I can work them out first.

[Bill S (OP)]

In looking at the linked article; if I seem to be ignoring relevant comments elsewhere in the thread, it is because I’m trying to extract understanding from the article at this point.

Pete commented that I “….asked about potential energy, not potential”. This leads into the questions.

Am I right in interpreting GPE as the energy, due to its position in a gravitational field, of a mass of any size?
 
Am I right in interpreting potential, as used in this article, as energy per Kg, of this mass; due to its gravitational position?

Just like potential energy, the biggest value of potential you can get is zero

Have I missed an explanation, in the article, as to why this should be? 

Do 10 MJ kg-1 of work on raising an object from the Earth's surface and it will move up to a point where it's potential is - 53MJ kg-1. That's 10 MJ kg-1 greater than on the surface because it is 10MJ kg-1 closer to zero.
Confusing! Look at the following diagram. It shows how potential drops as you move further from the surface of the Earth.

That, and the diagram, make good sense, but only if the potential is set at zero, at infinity.  This touches on the crux (or rather, the cruces) of my original question.
   
What does being “set at zero” actually mean?
Is it a mathematical “convenience”?
Wouldn’t it make as much sense to set the kinetic energy at zero, and define that as negative, in the same scenario?
As an object falls under gravity its GPE decreases and its kinetic energy increases.  If both of these are considered as positive, doesn’t that satisfactorily balance the equation?