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So in my textbook I am told:

two masses have gravitational potential energy because work had to be done in order to move one the masses form a position very far away (lets say infinity) to the position near the other mass.

Additionally, I found on a physics forum:

"zero potential energy is at an infinite distance from the centre of a spherical object."

I am lost. My intuition suggests that you have zero gravitational potential energy when the two masses are right next to each other! As gravity is attractive, you need to do work in fact get away, not to get closer! So when you are "infinitely" far away from another mass m (there exists nothing else other than m and you), you actually have very high gravitational potential energy (I know I am over-simplifying).

What am I getting wrong?

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  • $\begingroup$ Did you consider writing down the formula for gravitational potential and staring at it for a few minutes. Try this with the Newtonian gravitational potential energy equation and see what happens for very a small distance. Then compare this to the case of very large distance. A plot my help you. As I recall this is discussed nicely in Landau Vol. I early on, with several graphs. I do not have a copy to hand to give you a page reference, but the index should help you find it. Finally note that a negative potential energy is an attractive force. $\endgroup$ – Flint72 May 14 '14 at 22:08
  • $\begingroup$ @Flint72 Thank you for your comment. I have looked at some graphs, but my confusion, I doubt stems from how the graphs seem to behave. The problem for me is that, I cannot see why we define something very close to the earth to have more gravitational potential energy (of course in the negative) than something very far away. Surely if 2 asteroids were dropped onto the earth, one 3km above the surface and one from 3000km, the one further away would have had more gravitational potential energy, as this would account for its much higher destructive power. $\endgroup$ – Just_a_fool May 14 '14 at 22:19
  • $\begingroup$ @Flint72 In my mind, if gravity were to be repulsive, it would make sense why work would have to be put into bringing a mass close to the earth. $\endgroup$ – Just_a_fool May 14 '14 at 22:22
  • $\begingroup$ You are correct to suspect that $\Phi(\infty)=0$ is arbitrary, because it is. You are free to define it using whatever you want as a reference point for zero, but conventionally for gravitational potentials it's common to use $\Phi(\infty)=0$ for convenience, ie, to define zero energy as when everything is completely separated. $\endgroup$ – DumpsterDoofus May 14 '14 at 23:42
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    $\begingroup$ Also, the textbook wording is sort of misleading. While it's technically correct to say that "work had to be done in order to move one the masses from a position very far away (lets say infinity) to the position near the other mass", it's negative work, because the masses are attracting each other. So you are correct to suspect that "As gravity is attractive, you need to do work in fact get away, not to get closer", but the discrepancy is resolved because the textbook is assuming the amount of work done on the system is negative. $\endgroup$ – DumpsterDoofus May 14 '14 at 23:47
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You never "need" the absolute value of potential energy, since you always look at its difference. Therefore you are free to choose any reference energy as the zero of the potential. When looking at a potential which is defined by the following force: $$ F=\frac{\alpha}{\|\vec{x}\|^3}\vec{x} $$ You get an infinitely large force when $\|\vec{x}\|$ becomes zero. Which means an infinite amount of work would have to be done to get away from zero. So this would be a bad choice for a reference energy.

But I think people doe find it easier to set the reference potential energy at a "boundary". And the only remaining "boundary" would be at infinity.

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  • $\begingroup$ Nitpick: an divergent force near $0$ does not imply that it would take an infinite amount of work to get away from there. The exponent in the Newtonian gravitational law, however, does imply it. $\endgroup$ – Stan Liou May 14 '14 at 23:48

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