Thought Experiment

The energy comes from your imagining a world where you can arbitrarily change universal constants.

The earth's gravity varies across the surface of the planet due to variations in the underlying mass, not in G.

You might as well ask, "If a bullet traveling at velocity, V, has energy 1/2 mV² and I suddenly double the mass, where does the extra energy come from?"

If you "suddenly" double the mass of a speeding bullet, why would you assume that the velocity, V, would not change?

Heck of a back wind...

Welcome, Guest,

Seems to me that the change in energy would come about from the act of changing g (the acceleration due to Earth's gravity, as opposed to G - aka, Big G - the fundamental, universal gravitational constant). The only way to do that, as far as I know, is to change the mass of the Earth. As this is a thought experiment, we needn't be concerned about what technology might be necessary to pull this off; we simply assume that we can (I'm writing this to forestall any spurious discussions along these lines by others).

Say we remove a chunk of mass from the Earth. This will lower g reducing the potential energy of M (and everything else on Earth in similar circumstances - including the potential energy of the mass we remove because of the reduction of g), but the energy of the Earth-chunk system overall won't change, because we'll get it back when we restore the chunk. The energy we put into removing this piece includes the reduction of M's potential energy. What we're doing is simply rearranging the distribution of energy, but the total system energy remains constant.

If we add mass, we'll have to take it from somewhere else. That Somewhere Else chunk will have a certain potential energy with respect to Earth, and the energy of that Earth-chunk system will also remain constant. Again, we're merely redistributing the energy. The total energy doesn't change.

My two pesos' worth, at any rate.

By having two continental plates slide across each other will change the diameter of the earth and rotational speed and therefore g, The difference will be small but it is possible.

Energy does not just appear or leave or is sitting on a bench to be called into the game by the coach.

with conservation of energy it may change states, one cannot help getting into this without getting into physics, by separating G out or talk about changing it with out M have something to do with it.

This topic is too late in the day for me....

Actually you may be on to something. Jesse Beams at the University of Virginia (he of the magnetic centrifuges for enriching uranium) believed that G (yes, that's big G) had a time variance in the 14th decimal. So, what you do is hoist fat old Aunt Maud up to the top of the barn on the hay hook, wait about a millenium or two and her 375 pounds will be changed.

Where will it go? Dare I say entropy?

Entropy, Thats the word I wanted to use, but I ran out of entropy to remember what that word was.

Don't think entropy. Think 'margarita-on-the-beach.'

You can rise from the ashes tomorrow, Phoenix. Get some rest, dude!

What do you mean that "Energy does not just appear or leave or is setting on a bench"??? I have a teenage daughter, she will be two sick to go to school for two days running. Let a friend call, and she has the energy to get up and go and party for the next ten hours strait. If that was not energy sitting on the bench and waiting to be called into play, well I'll just have to eat "Cat food" of something?

I dun-know where the energy comes from, do you?

You have potential energy, not actual energy. And as its referred too as potential energy, any of the variables can change with out effecting the concept of it being potential only. It is only real energy when it enters the real world and the potential becomes kinetic? So whether its on the earth or on the moon you are only referring to its potential.

JD.

JD is on the right track. The potential energy exists only because of the gravitational attraction between the object and the earth, and the fact that something is holding the object immobile relative to the earth. (Uh, oh! I mentioned the R word! Can Jorrie be far away?) Remove the restraining something, and gravity takes over.

Now, let's say we have several objects, all of the same mass, aboard an aircraft that is flying at a constant altitude above the surface. The potential energy of the objects will vary as the plane flies over areas of varying densities, due to different densities of different types of soil, rock, vegetation, and water. However the changes in potential energy will be miniscule due to the total mass of the earth and the short distance between the objects and the earth. If someone aboard the plane were to drop the objects one at a time as the plane passed over varying terrain, the objects would appear to fall with the same acceleration. We would have to have an extremely precise and sensitive instrument to measure their falls and determine any difference.

hi 3 Doug. Reading your post brought to mind a further analogy, if we have a plane with several object of equal weight aboard a plane flying at a constant altitude, ( maintaining the same radial distance from the earth centre), not only will the potential energy vary with gravity, but also with the terrain, so when flying over a mounting range and dropping an object it will have a different potential to a object dropped into a valley, etc. yes further food for thought.

JD.

Over complexity....

If we change G.

The energy is used in changing G.

You could have said exactly the same question but said H in place of G....

The answer is then obvious.... the work done in increasing H gives you increase in potential energy.

Next...

I always understood that that was why it was potential energy. You haven't got it yet, only when you drop it?

The potential energy (of an object) is real energy but you can consider it as a kind of "stored energy" due to gravity... It doesn't produce work but it's always there... And it's always ready to produce work if we let the object to fall (the potential energy is changing to kinetic energy)...

Complexity? Yes I would like life to be simple, but have not found it to be so. The energy used in changing G, and the energy used in changing H, can be equated to the changes in potential, I agree with this. But the original question is what happens to forces extraneous to the object taken in isolation, what happen to forces that vary due to no fault of the object that finds itself in that situation. I have answered the question by saying that potential energy is an abstract concept, which is possibly true in part, but if one considers several objects loaded on to a plane, the and energy used to take them to a fixed altitude, and compare that to the energy recovered, to that which was used to put it in to a position of potential. The answer is possible tied up with H, an object dropped on top of a mountain range has not fulfilled its potential, and an object dropped into a valley has exceeded it. But potential must be equated to the point of origin, so an object raised to a potential and moved to a different area of G will not be fully realised till returned to its point of origin, Over complex, yes, I agree with the simple answer.

JD.

If the gravity was lowered the energy was never expended to lift the mass to its present height. If gravity is higher more energy was use to lift it and that energy came from you or the device used to lift it. If we could make a device that could change the gravity of a selected mass. The energy would be added or subtracted from that mass not any mass suspended above it. Where would the energy come from and go to. My bet is the device thats changing the gravity. I ASSUME that is the only logical place.

Some excellent discussions, both of the practical and the philosophical.

My original post of this question, I think I was leaning more towards just looking at the variables on paper and manipulating them without regard for real world application, and then thinking about how the answer would apply in the real world.

Let me post the original question another way: If I used 10 units of energy to put an object at rest at a given PE (thus, the PE would be 10 units), and while it was at rest I decreased g, and thus decreased PE, and then set the object falling, the KE out would not equal the PE in. So, again, where did that energy go. One can not simply state that that the energy no longer exisits.

The energy did'nt go anywhere, Because it was never there to begin with.

when you decreased g, it also effects PE, and one can not look at what it took.........wait.

the energy went with the "g" that was removed.

For the increment, for example, of the g of the Earth you add energy to the system "Earth & object" (from your work,i.e. in order to increase the mass of the Earth)... This increment of the energy is translated as an increment of the PE of the object... This incresed PE remains in the object (as a kind of "stored energy") and is always ready to produce work (e.g. transformation to KE)...

Velocity is defined as distance/time. When you square any velocity you get an acceleration, since you have to square the denominator (in this case, the time).

Hence, the speed of light squared is an acceleration, not a velocity,,,

Hence, E=mc^2 is a specific case of Newton's F=mA based on a specific acceleration,,,

Hence, Force = Energy,,, hence energy is an illusion and does not really exist, there are only the effects of different masses upon one another

Hence,,,, a 10 pound ball at rest on the ground is in a constant state of negative acceleration relative to the gravitons colliding with it. For if there were no acceleration, any mass multiplied by zero would yield a force of zero and the ball would be weightless when at rest. And any mass of zero multiplied by an acceleration would still yield a force of zero, hence gravitons have mass.

hence there is no energy, only force......

Spock is hovering in the Earth's atmosphere in serious danger of falling. Scotty beams him over to another planet with a G of 16 ft./sec^2. No energy is used since the enterprise is using the latest "free energy" magnetic plasma generator. Provided he is not at an altitude which will allow him to reach terminal velocity and that the altitude is the same as his previous altitude above the Earth, the Vulcan will now only hit the planet at 1/4 the velocity and with 1/4 the force. Two thing are accomplished here. Spock will no longer die upon impact and Scotty is smiling.

Hence, the speed of light squared is an acceleration, not a velocity

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Acceleration is not quite the velocity squared, else you'd get distance²/time², which is something else entirely. As velocity is the change in position/time, acceleration is the change in velocity/time. Only the denominator is squared. c² is not an acceleration, nor is force equivalent to energy.

Hrrrm,,,,

change in position or distance over change in time. Yes. Delta X / Delta t = velocity

acceleration = Delta velocity / Delta Time = change in velocity/change in time. Yes.

However, accelerations are always expressed in time squared provided the units of time are identical. 32.2 ft. / sec / sec = 32.2 ft. / sec^2. As you noted,velocities can not be expressed in units of time squared.

a velocity of:

10 meters / sec

when squared is shown as:

100 meters^2 / sec^2

A velocity of 1 meter/sec when squared is 1 meter^2 /sec^2

Agreed distance^2/time^2 is not the same as distance/time^2.

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If a numerator is divided by 2 or more denominators, you may multiply the denominators and divide.

100 / 2 / 5 is the same as 100/10

If the denominators are identical, multiply a denominator by itself is the same as squaring it

100 / 5 / 5 is the same as 100 / 5^2 or 100 / 25

units of measurement such as time or distance, when multiplied are shown as powers.

Hence the question is did Einy mean:

(299 792 458 meters / second)^2

or:

299 792 458 meters / second^2

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3 focused beams of particles/photon/whatever are sent out from a single source simultaneously. Beam 1 travels at 100,000 miles/sec below C, beam 2 travels at C, beam 3 travels at 100,000 miles/sec above C

If you are the target and are stationary relative to the source, beam 2 will be visible while beams 1 and 3 are not. If you travel at 100,000 miles/sec toward the source, beam 1 appears to be moving at the speed of light and may appear visible while the other two are not. When moving away at 100,000 miles/sec beam 3 is visible or appears to be traveling at the speed of light.

When light hits you, what is the rate of negative acceleration?

It's all about the relative acceleration man!

BTW - it's doctor McCoy who is actually smiling, Scotty just runs the transporter.

It's all about the relative acceleration man!

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Who is the relative acceleration man? Isn't he the guy who pushed your mother-in-law out the airlock?

Can I borrow him?

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Seriously, back to your question: c² means just that: (3e8 m/s)².

Decide whether you want your beams made of photons or of particles. You know, thingies having a rest mass.

If photons, beam 1 cannot travel at less than c if we're doing this experiment in a vacuum. If the medium is glass, well, that's different. If the medium is a Bose-Einstein condensate, well, that's really different. Let's stick to vacuum. Simplicity is a virtue. Especially when it comes to TV remotes.

If particles, beam 2 cannot exist. (Why not?) If photons, no problem.

If either particles or photons, beam 3 cannot exist. (Why not?) Even Spock'll tell you that.

LOL !!!

Now your cookin!

except, Spock can do warp speed!

This reminds me of another puzzle that leads one away from the basic concept that left and right sides of an equation must remain in balance.

PE = M*H*g therefore PE/M*H = g

now if we vary the value g on the right hand side of the equation then we must also do so to the left hand side.

PE/M*H*2 = g/2 gravity has been varied but the out come remains unchanged.

JD.

So in this hypothetical case, a hypothetical frictionless hack-saw is used to cut a homogeneous Earth into two identical homogeneous hemispheres, then both hemispheres are made spherical again by some unknown zero-work process and then a hypothetical jack is used separate them by an infinite distance (phew - that's hard work!). Now, the mass of the body to which M is attracted is now half what it was, so the acceleration due to gravity is now half what it was.

Newton said, "F = G x m₁ x m₂ / r²".

Archimedes said, "Eureka!", or is that just a rumour?

'Stinky Pete' said, "My arm hurts from all that jacking" [no comments, please, he isn't here to be offended; he's just gone off to scrounge a cup of tea].

Hypothetical frictionless hack saw, ****, had not thought of that, and I did not realise the mass would all end up on one side, I should have know? All that work and stress to create parrallel worlds, now we will never know, if they cant agree what happen to g. Perhaps if we gave half the mass to each and brought them together, and if they agree we might end up where we started, then again, if not you might be right.****.

Regards JD.

That was quick, not mean to be personaly, just an obsevation of some of my weaknesses, as compared to others. Regards JD.

"If you don't know me by now,

You will never never never know me.

Doooo -diddle-e-dooooo."

Its a learning curve, pleased to make your acquaintance.

Regards JD.

Mutual!