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Proof of gravitational potential energy.

Work done by gravity in bringing mass from infinity to a distance of $r$ between masses.

When we use the integration formula and arrive at the answer we get $-GMm/r$ taking lower limit as infinity and upper as $r$.

But this work should be positive as force and displacement are in same direction.

Please explain.

If my proof was wrong, then tell any other satisfying proof for GPE.

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The work done by gravity on the infalling mass $m$ is $\int_\infty^r F(r') dr'=-\int_r^\infty F(r') dr'=GMm/r$.

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  • $\begingroup$ Ben Crowell : according to my knowledge this work done by gravity on the infalling mass m is known as the potential energy at that point where it is at a distance of r from the source mass M. now this has come positive but we define it as -GMm/r. please correct if wrong. $\endgroup$ – Chris Oct 13 '13 at 19:17
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    $\begingroup$ The work done by the gravitational force equals the increase in the object's kinetic energy, $W=\Delta KE$. By conservation of energy, $\Delta KE+\Delta PE=0$, so $\Delta PE=-W$. $\endgroup$ – Ben Crowell Oct 13 '13 at 20:15
  • $\begingroup$ why this : ΔKE+ΔPE=0 our goal is to introduce potential energy,so can you please elaborate from when we got work done by gravity +ve and how we chosee potential energy zero at infinity and what would happenn if any other point is chosen as 0.thank you $\endgroup$ – Chris Oct 14 '13 at 16:52
  • $\begingroup$ The equation $\Delta KE+\Delta PE=0$ is expressed entirely in terms of differences, so it's valid regardless of where you choose as your zero level of PE. $\endgroup$ – Ben Crowell Oct 14 '13 at 18:51
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Maybe the confusion arises from the fact that the potential energy in a point $P$ can be interpreted as the work needed to bring a particle from a reference point $O$ to $P$, without altering its kinetic energy. Due to the fact that $\Delta K = W$ this is exactly minus the work done by the conservative forces.

In this case $O$ is “the infinity” and the work needed to bring it to the point $P$ is negative (in fact it goes there spontaneously).

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