# Signs in proof of gravitation potential energy (GPE)

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.

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

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

• 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. Commented Oct 13, 2013 at 19:17
• 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 $\Delta PE=-W$.
– user4552
Commented Oct 13, 2013 at 20:15
• 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 Commented Oct 14, 2013 at 16:52
• 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.
– user4552
Commented Oct 14, 2013 at 18:51

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).