Imagine two locations with different amounts of gravity. I carry up a weight in low gravity, move it on this height over to the other place, and let it fall down there with higher gravity.

experiment setup

Wouldn't falling down release more energy than lifting up hast cost? If so, is it theoretically possible to generate such a transition between different levels of gravity near to each other?

  • $\begingroup$ No, the amount of energy needed to lift up the object is exactly the same as the energy released when it falls back down again (ignoring air friction etc). This is because gravity is a conservative force. $\endgroup$ Jan 21 '14 at 7:45
  • $\begingroup$ @JohnRennie So lifting the ball up in a less gravity location would take the same amount of work I could get out of the falling ball in the high gravity location? $\endgroup$
    – danijar
    Jan 21 '14 at 11:01
  • $\begingroup$ Yes, but moving the ball from the low to the high gravity locations would take work, and that would make up the difference. $\endgroup$ Jan 21 '14 at 11:14
  • $\begingroup$ Oops, sorry I misread your comment. Lifting the ball in the low gravity location takes less work than you get back from dropping it in the high gravity location. However moving it from the low to high gravity location and back takes work. The amount of work lifting the ball plus the work moving it adds up to the same as the work you get back by dropping the ball. $\endgroup$ Jan 21 '14 at 15:33
  • $\begingroup$ @JohnRennie couldn't I use two ramps to let the ball roll from one side to another? Their slope could be minimal to not waste heights. $\endgroup$
    – danijar
    Jan 21 '14 at 20:38

The main point is that Newtonian gravity fields are conservative. What that means is that it is impossible to have a configuration like the one you drew without there being gravitational fields pointing to the left and to the right in the regions where you want to do the 'horizontal' transfer.

For example, you might try to achieve this on Earth by taking the usual uniform gravitational field and locating a very heavy mass just under the foot of the conveyor belt on the left. This will mean, though, that as you move your mass from the foot of that conveyor belt you will be fighting against the attraction of that very same mass, as shown with the red arrows:

enter image description here

The net result is that doing both of those horizontal transfers takes work, and in fact it must take exactly the same amount of work as what you've gained from lifting the object in the weaker field. There are, of course, many possible ways to achieve the fields you want, apart from the one in my image, but because all gravitational fields are the sum of attractive forces to a bunch of point masses, and the field of each point mass is conservative, you will always, necessarily, have cross-pointing fields like the one I pictured that will do away with any perpetual motion engine.

  • $\begingroup$ Interesting, so if we could generate both fields just pointing downwards, it would work, right? But we can't do that of course. Edit: One more though, the higher we take to ball, the higher the energy gain. But the energy needed to move the ball horizontally stays constant, doesn't it? $\endgroup$
    – danijar
    Jan 22 '14 at 22:15
  • $\begingroup$ @danijar If you could generate the field it would work, but the field generator would be putting energy in. This happens with other types of fields. An electron can move around in a circle against friction and continue indefinitely, but only because humans are powering a machine that makes those fields, using electricity. Moving the ball to the side takes no energy. You could put it on a string and let it go in a circle... until the string breaks. $\endgroup$ Jan 22 '14 at 23:10
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    $\begingroup$ @danijar That's correct: if you could build it, it would work. But then it wouldn't be a gravity field as we know them. And no, it doesn't stay constant. You can probably do it yourself, and it will be a good exercise: calculate the work needed to move a unit mass around a rectangular loop with a point mass below one corner, including the dot products and all. You'll see it sums to zero. $\endgroup$ Jan 22 '14 at 23:19
  • $\begingroup$ @AlanSE your assertion that moving the ball to the side takes no energy is incorrect, at least in any nonfictional gravity field that's consistent with Newton's law. $\endgroup$ Jan 22 '14 at 23:21
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    $\begingroup$ Yes. Which is why I restrincted my answer to nonfictional physics. The real question is what happens with perpetual motion questions: the only way for them to not deserve immediate closure, in my opinion is if they can be phrased as "why wouldn't this work", and answers explain why they wouldn't in the real world. For me this is very simple, and here it boils down to why the proposed gravity field won't work. You can, of course, post according to your own interpretation. Good night! $\endgroup$ Jan 23 '14 at 0:39

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