10
$\begingroup$

When an object falls through vacuum, gravitational potential energy is converted to kinetic energy. Is there some way to get electrical energy out of the equation by itself (i.e. somehow convert the gravitational potential energy to electrical energy)? Is this physically possible? If so, what properties must this object have?

By by itself, I mean without using any external (possibly stationary) "reference object" (e.g. a stationary coil), so a magnet falling through a coil does not count, i.e. the electricity is generated solely by the object that is falling. Note that the object itself can be arbitrarily complex internally, just that whatever mechanism it has inside must also be falling along with the object.

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ I just want to point out that it may not generate electricity, but a charged particle in free fall over the earth does radiate electromagnetic waves , physics.stackexchange.com/questions/13513/… $\endgroup$
    – anna v
    Commented Feb 23, 2017 at 13:35
  • 1
    $\begingroup$ Generally the answer is no (though there are creative possibilities as mentioned in some answers) but an object falling means that there is a massive gravity well (been watching too much Expanse) that it is falling into. If it's a planet or a star then there is often a magnetic field surrounding the body. A wire moving through such field will generate electricity. We've tried to harness this several times but each time we tried the wire broke $\endgroup$
    – slebetman
    Commented Feb 24, 2017 at 0:03
  • $\begingroup$ @annav your comment made me start thinking about a charged particle static and in free fall in a gravitational field. I did more research than was necessary and wound up finding out that not only does a charged particle NOT radiate in free fall, it also does not radiate in uniform acceleration. We know it falls at the same rate as an uncharged particle (meaning all gravitational potential turns to kinetic). If it were to also radiate, where would that energy come from? Feynman showed it doesn't radiate. $\endgroup$
    – Jim
    Commented Feb 24, 2017 at 12:45
  • $\begingroup$ @Jim well, Feynman is not the pope. The energy would come from the gravitational field , similar to the synchrotron radiation of an electron in a magnetic field, imo, but I also saw a number of preprints with similar propositions, and no, in these preprints the charged does not fall the same as the neutral. It is a very weak effect because the gravitational constant is very very small $\endgroup$
    – anna v
    Commented Feb 24, 2017 at 13:11
  • $\begingroup$ @annav FEYNMAN IS NOT THE POPE!??? This changes everything! In all seriousness, I'm not taking his word as law. The classical approach contradicts the well observed Bremsstrahlung effect but Feynman's solution agrees with it. That said, I'm now going to also go look more into specifically a free-falling case because most of what I read about was focused on the case of a charge static in a gravitational field $\endgroup$
    – Jim
    Commented Feb 24, 2017 at 13:16

3 Answers 3

Reset to default
20
$\begingroup$

A sufficiently large object will experience differential gravity ("tidal force") - this could be converted into a small amount of electrical energy by having two heavy spheres separated by a long rope; as they fall there will be a tension on the rope and you could let that tension do work on a generator / dynamo ("complex but internal to the object")

The concept here is that a ball closer to the earth will experience greater force and so fall a little bit faster - in the extreme case of falling to a black hole this leads to "spaghettification " but on a more normal scale it could give you a little bit of electricity. But without en external electric or magnetic field I can think of no way to convert most of the kinetic energy into electrical - the ability to do so would be a first step to an antigravity system. Let me know when you get there!

Improve this answer
$\endgroup$
8
  • $\begingroup$ +1 your post reminds me of a space tether system. en.wikipedia.org/wiki/Space_tether_missions $\endgroup$
    – user146020
    Commented Feb 23, 2017 at 13:04
  • $\begingroup$ You could also use a large object made of piezoelectric material instead of the spheres/rope/dynamo system. $\endgroup$
    – Diracology
    Commented Feb 23, 2017 at 17:30
  • $\begingroup$ @Diracology sure - I pointed out that a small amount of energy can be extracted from the gradient in the gravitational field. Any number of methods could be used to do the conversion from mechanical to electrical energy. $\endgroup$
    – Floris
    Commented Feb 23, 2017 at 17:38
  • $\begingroup$ In additon for this excelent answer you can just "release" the second "piece" of your "device" just little after the first "piece". That way the first will be falling faster than the second and will pull it and you don't need more a many kilometers rope nor a blackhole $\endgroup$
    – jean
    Commented Feb 23, 2017 at 17:44
  • 1
    $\begingroup$ @DrunkenCodeMonkey You can't reset the spring and wheel for less energy than it produced. Not possible. That means the energy has to come from somewhere else. The most likely place is the spacecraft (it's kinetic or potential energy). You could use solar, but now that defeats the purpose of having the system. Taking energy from the s/c means you'll decay its orbit and it would be unsustainable. $\endgroup$
    – Jim
    Commented Feb 24, 2017 at 12:47
19
$\begingroup$

Floris posted an answer that assumes the object is large enough for different parts of it to experience noticeably different gravitational forces. This is one way to accomplish it and written rather well, so I won't discuss that case further.

If you require that the object is small enough that all parts of it would be at approximately the same gravitational potential as every other part at any given time, then the answer is "no". Free fall is a geodesic, which means in its frame, the internal components would experience no real difference from a situation where it is not falling, so there wouldn't be a change that would allow it to produce energy for itself.

Looking at it a slightly different way, converting the gravitational potential energy into electrical energy would mean not all is converted into kinetic energy, which means you'd effectively be slowing the fall of the object compared to something not producing electricity. You already said we can't have it interacting with the massive body through anything but gravity, so you can't have the fall slowed by anything and, thus, all energy must transform to kinetic.

Improve this answer
$\endgroup$
1
  • 4
    $\begingroup$ Your last paragraph nicely expands on my "first step to an antigravity system". $\endgroup$
    – Floris
    Commented Feb 23, 2017 at 16:11
1
$\begingroup$

If by 'electrical energy' you mean 'an electric current', then no, that won't work. The stationary coil you are excluding is there to provide the supply of electrons to be moved by the magnet - no electrons, no current.

And no, nothing can be generated within the falling object. Current is generated by the relative motion of a magnet and an electron source; an object in free-fall is never in motion relative to itself.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.