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I am reading "The Feynman lectures on physics volume 1".

Feynman says

"The total work done in going around a complete cycle should be zero for gravity forces"

When gravity forces are given, I think we cannot move an object around an arbitrary complete cycle.

I cannot understand what Feynman is saying.

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  • $\begingroup$ Why can't you? You can apply external force to do that. $\endgroup$
    – user243267
    Dec 21, 2019 at 13:04
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    $\begingroup$ What about Earth moving around the Sun? $\endgroup$
    – my2cts
    Dec 21, 2019 at 13:31
  • $\begingroup$ By the way Feynman's statement should be true only for static fields. For time dependent gravitomagnetic fields there should be an anologon of induction. $\endgroup$
    – my2cts
    Dec 21, 2019 at 13:31
  • $\begingroup$ What he is saying is simply that for conservative forces the work around a loop is zero. Because such forces can be written as the gradient of a potential and $dW$ is an exact differential (assuming space is simply connected). The converse is even more true: if the work around any loop is zero the force field is conservative. $\endgroup$
    – lcv
    Dec 21, 2019 at 13:48

2 Answers 2

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When gravity forces are given, I think we cannot move an object around an arbitrary complete cycle.

What about round-trips, roll-coaster, merry-go-round, planet orbits?

Yes, there may be some external forces to oppose frictions and air resistance, but in the Feynman's statement only gravity forces are taken into account for computing total work.

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Well, we may consider, for instance, orbits when only 2 objects are considered and interact with gravity. If you consider a closed loop, the final point is equal to the initial point and since it's a cycle, the kinetic energy will also be the same. Hence, there is no variation of the gravitational potential, no variation of the kinetic energy and, from the conservation law, the work done by the gravitational forces will be zero.

If you consider a different case where other forces are also present and you may have periodic movement that is not a purely gravitational orbit, the reasoning will be similar. It's a consequence of gravity being a conservative force, or more generally, since we have gravitational potential energy, that determines the gravitational force, we can apply Stokes's Theorem.

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