I understand the general derivation of the shell theorem by feynman but I don't get why he denotes the potential energy with W, the symbol for work. I would understand this if they would be the same but the work done in taking a particle from infinity to earth should be positive (wether we provide the force or gravity provides the force) so why is it negative in this case?
Thanks in advance.
Here is the link to the webpage: feynmanlectures.caltech.edu/I_13.html
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1$\begingroup$ This is Newton's shell theorem. $\endgroup$– Qmechanic ♦Commented Aug 21, 2018 at 12:06
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$\begingroup$ Link to webpage? $\endgroup$– Qmechanic ♦Commented Aug 21, 2018 at 12:09
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$\begingroup$ feynmanlectures.caltech.edu/I_13.html $\endgroup$– delivosaCommented Aug 21, 2018 at 12:10
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1 Answer
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The gravitational force is attractive so the work done by an external force to bring the mass $m’$ from infinity to the shell (which is the change in the gravitational potential energy of mass $m’$) is negative as the direction of the external force $-\hat r$ is in the opposite direction of the displacement $+\hat r$.
This means that to reverse the process (positive) external work has to be done to move the mass $m’$ from the shell to infinity and the gravitational potential energy of the mass $m’$ increases.
Taking the zero of gravitational potential energy as when the mass $m’$ is at infinity moving it to the shell reduces its gravitational potential energy ie the gravitational potential of the mass $m’$ is negative.
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$\begingroup$ We're talking about the potential energy of a point P outside the shell. Also, I don't get why youy say the force is opposite to the displacement. $\endgroup$– delivosaCommented Aug 21, 2018 at 12:29
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1$\begingroup$ @delivosa I have amended my answer although the same principles are involved. $\endgroup$– FarcherCommented Aug 21, 2018 at 12:31
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$\begingroup$ why is the displacement in opposite direction to the force? $\endgroup$– delivosaCommented Aug 21, 2018 at 12:41
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1$\begingroup$ @delivosa The external force has to hold the mass $m’$ back to prevent it accelerating towards the shell. $\endgroup$– FarcherCommented Aug 21, 2018 at 12:44
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$\begingroup$ Thanks I understand now, It's just the same as with an object near the earth. $\endgroup$– delivosaCommented Aug 21, 2018 at 13:07