A friend and I were discussing this, and couldn't decide. Since it takes energy to physically raise something up above the planet's surface, does it continue to require energy in order to keep it above the surface?

While gravity is extremely weak relative to atomic/molecular bonding forces, we were split on whether or not some tiny amount of energy needed to be expended - perhaps like radioactive decay? - to keep that bond from being eventually broken apart by gravity (ie, the table breaking and eventually flattening to an even thickness in being pulled to ground level by gravity).

UPDATE I: if no energy is being spent to keep the table from sinking down to ground level, how does this differ from, say, a ball of something magnetic being suspended in the air over an electromagnet? Energy is required to keep it up - if you cut power the the electromagnet, the ball falls.

UPDATE II: This question is a direct result of trying to learn more about it on my own :) I'll be taking intro physics as an MD program pre-req in Sept, and have started exploring snippets in a hit or miss way governed by what is of interest to me. So my knowledge of physics is extremely patchy right now :)

  • $\begingroup$ The energy loss in the magnetic levitation case occurs in the magnet and would happen even if nothing was being levitated. Use a superconducting magnet and it goes away. $\endgroup$ – dmckee --- ex-moderator kitten Jan 12 '14 at 18:00
  • $\begingroup$ Related to physics.stackexchange.com/questions/1984/… in the context of why it is tiring to hold heavy things up. I'm also sure we've done this before. $\endgroup$ – dmckee --- ex-moderator kitten Jan 12 '14 at 18:06
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    $\begingroup$ Do you understand the physics definition of energy? The work--energy theorem? If there is no motion, there can be no work and if there is no work there is no expenditure of energy. In the biological system (muscles) there is motion, so it requires energy to hold things up using muscles. In the case of the table there is no motion. $\endgroup$ – dmckee --- ex-moderator kitten Jan 13 '14 at 1:39
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    $\begingroup$ Intuitively I don't think that's correct, and @Mark Eichenlaub's answer seems to supports my intuition: dE=-dW+dQ "Heat is simply any energy transferred between two systems by means aside from work." So there can be a transfer of energy without work, if I'm understanding it correctly. $\endgroup$ – Logos Jan 13 '14 at 15:44
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    $\begingroup$ @dmckee: well, that's related to my question. If the energy comes from the atomic bonds, then eventually there wouldn't be enough energy there to keep the table upright against the pull of gravity, collapsing it. So that just leads round to my question, again, and makes it sound like yes, energy does need to be expended steadily in order to keep the table top from falling. $\endgroup$ – Logos Jan 13 '14 at 18:54

No, a table does not expend energy to keep its top from falling.

Energy is an example of a "state function". A "state" is just the particular way a system is set up - this atom is here, that atom is there, this is the electromagnetic field, etc etc. When you specify just where everything is and what the nature of the interactions is, you have the state.

For a system as big as a table, we do not actually specify where every atom is to describe the state. We just say "there is 20 kg of wood with this density at this temperature in this shape..." until we have given a good enough description.

Every possible state has some energy associated with it. Given a state, you can apply some formulas to calculate the energy. For example, for the table, there is thermal energy stored in the random motions of the molecules and there is elastic energy stored in the slight deformations in the table that counter the force of gravity pulling the table down. We can use this to find the energy in the table. However, this energy is not going anywhere or doing anything. It's just sitting there. Or really, it's just a label used to describe what the state is. I wrote more about what energy is in this answer.

As long as the state of the table does not change, its energy does not change. This is probably good enough to intuitively answer your question. The table is just sitting there. It stays in the same state all the time. Its energy therefore doesn't change, so no energy is expended. This is why a table does not "get tired" the way a person does.

Technically, though, that fact that the energy of the table doesn't change simply means that however much energy flows into the table also flows out of the table.

The energy flow in and out of the table (and energy flow in general) is divided into two categories: heat and work. Work is things like mechanical pushes on the table. Heat is things like conduction with the surrounding air molecules and radiation (i.e. infrared light given off and absorbed by everything all the time) cooling or heating the table.

Unless there's an earthquake or some such going on, the table is mechanically isolated - no work is being done on it. There are forces on it, but work is a product force*motion. Since there is no motion, there is no work. Therefore, the only energy going in or out is heat. The table gives off radiation, but unless it's colder or hotter than the surrounding room, it absorbs and emits radiation at the same rate. Likewise, the table loses heat through conduction, but gains it back at the same rate.

The electromagnet you mentioned is different from the table. If you were to measure its temperature, you'd find that the electromagnet is consistently hotter than the surrounding room. It is continually converting electrical energy into thermal energy and giving off this thermal energy as heat via conduction and radiation. This is not strictly necessary to levitate something. The levitating object is not gaining energy. All the energy is going into heating the electromagnet. This is an artifact of the way that the electromagnet works. Current runs through some wires, and the wires have resistance. If we could get rid of the resistance of the wires, we could levitate the object without an input of energy, as mentioned by dmckee in the comments.

If a human being were to hold the table up, they would have to expend energy to do so, and eventually would get tired. But this is similarly because humans are simply inefficient. The energy they expend is not going into holding the table up. It is again going into heat. The person is continually converting chemical energy stored in their body into heat.

If the person were in the process of lifting the table, so that the table were rising, then the state of the table would change. As it rose higher up, it would have more gravitational energy. Because its energy is increasing, we know energy must be coming in from somewhere. So while the table is rising, the person lifting it is doing work on it. But once the table is stationary, its energy is unchanging and no energy need be "spent", where "spent" energy means that the energy is converted from one form (e.g. chemical energy in the human body or electrical energy in a motor) into another form (gravitational potential energy of the table).

If the table eventually breaks down and falls, its gravitational and elastic energy will go down. They will be converted into heat - the table, the floor it lands on, and the air it travels through will all heat up a little bit. Usually, we don't notice this heating very much because atoms are generally whizzing around at hundreds to thousands of miles per hour, so the small speeds of everyday life are not very significant. However, if you put a lot of mechanical energy into a small system, you can notice this heating. For example, by banging on the head of a nail with a hammer, you can heat the nail head, or by rubbing your hands against each other vigorously you can heat up the surface of your skin.

The second law of thermodynamics says that entropy in an isolated system can only increase. A world in which the table has collapsed and released its gravitational and elastic potential energy as heat has higher entropy than one where the table that has not done so. This means that if you leave the table alone long enough, it will eventually collapse and will not rebuild itself. In order to reduce the entropy of the system by rebuilding the table, you would need to do work on the system, thus "expending energy" on it. So ultimately, it takes some small energy input to keep the table up because without some sort of maintenance, the table will eventually fall apart. However, the second law doesn't place a limit on how long this has to take; the energy input can be arbitrarily small.

The study of the way that energy can be "spent" by converting it from one form to another is called thermodynamics.

  • $\begingroup$ A very thorough explanation, thank you much! I do have a couple other questions tho: (1) Does the effects of gravity on the table only count "once"? ie, when you set up the table, and then for all time after that it doesn't count until the table's structure changes due to entropic energy losses? (2) Does gravity have any effect on the entropic energy loss of the table? (I'm assuming that energy is lost as heat radiated away or something like that) $\endgroup$ – Logos Jan 13 '14 at 15:29
  • $\begingroup$ The energy of the table changes whenever its state changes. If the table breaks and collapses all at once, then that's how it happens. If it slowly sags with age, then that's how it happens. I don't understand the second question. $\endgroup$ – Mark Eichenlaub Jan 13 '14 at 19:35
  • $\begingroup$ I presume I'm probably not phrasing the question very well then, as my familiarity with the terminology is pretty much a layman's as of yet :( You said "...it takes some small energy input to keep the table up because without some sort of maintenance, the table will eventually fall apart." I assume that that means there is some small, regular energy drain over time. Does gravity have any effect on the rate of that energy drain? ie, without gravity would the process take longer? $\endgroup$ – Logos Jan 14 '14 at 1:16
  • $\begingroup$ That is very difficult to calculate. All I am saying is that things decay. Old buildings fall down, old statues rust, etc. Presumably a table will decay a bit more quickly with gravity than without it. There is not necessarily a constant energy input required. It is just the case that some non-zero amount of energy is required over long times in order to counter the table's gradual decay. $\endgroup$ – Mark Eichenlaub Jan 14 '14 at 1:26
  • $\begingroup$ Oh, I don't need to calculate it - I just want to understand what, exactly, is going on. Being able to quantify it would just be a nice bonus at this point :) $\endgroup$ – Logos Jan 14 '14 at 13:47

The key point to consider is how to apply the equation for calculating work: the integral of force over displacement. Since there is no displacement there will be no net work.


If you leave a table say for a billion years it would not be smashed to the floor only if it were perfectly aligned ( at a molecular scale) to not be pulled by gravity. But nothing is that perfect. Look at old glass panels, they get heavier in the bottom. So yes energy would be needed to re arrange it to a stable position everytime it goes a little wrong.

Think of it as sort of layers ( microscopic like the ones in graphite.) start to pull down. Even a diamond would give up but it would take an infinite amount of time. Try to think on a grander scale about it, what is happening to the earth on which the table is standing.

  • $\begingroup$ That is my point to my friend, actually, but he insists that no energy transfer takes place. Would another way to look at it would be to swap gravity and duration amounts - say, millions of gravities over a short duration rather than one gee over millions of years? $\endgroup$ – Logos Jan 12 '14 at 15:31
  • $\begingroup$ That table is in for a rough time. Millions of gs would be a black hole. Energy is a number associated with something. Energy transfer is not something like a physical object transfering between the body and another body. His understanding of energy is not very good I might say. Nobody really understands it that well, so confustion is natural. $\endgroup$ – Greg Jan 12 '14 at 16:59
  • $\begingroup$ Hmm... wish there were more answers going, I'd like to hear other opinions :( $\endgroup$ – Logos Jan 12 '14 at 17:37
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    $\begingroup$ This answer is more wrong than right. The one place it is right is in the importance of time scales. Things change and the table falls in time, to be sure, but not because it is expending energy in the mean time. $\endgroup$ – dmckee --- ex-moderator kitten Jan 12 '14 at 18:01
  • $\begingroup$ @dmckee: so where does it go wrong, and where right then? doesn't much help to say it's wrong, but give no details :( $\endgroup$ – Logos Jan 13 '14 at 1:24

No, it does not continue to consume energy to keep something above cerrtain height, you are comparing the pain you feel in hand for raising something for some time, that is not due to the expense of energy to keep the ball there. Also in magnetic levitation, the energy is wasted whether you levitate something or not, so that too is a bad example. Also if you talk about rarioactive decay, every body is known to radiate energy in this form for different time periods, nearly all bodies have carbon and uranium footprints so expense of energy in this form has nothing to do with raising something.

Take this as an example, you place something on a giant spring, now energy is stored as the spring potential energy of spring but once the forces are in euilibria no more energy is spent. If you want to check you can release the spring and you will get the energy back in same amount. Similarly in the case of raising something above a massive object against gravity, the energy is stored as potential energy and the system remains in equilibrium without expense of additional energy.


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