1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why is gravitational potential energy defined at infinity?

  1. Feb 21, 2014 #1
    Why is gravitational potential energy defined at infinity???

    Like here on earth there is Zero potential energy at the center of the earth (if you could theoroeetically go there) so why not define it as 0 at zero distance from the force supplier instead of at infinity??? I understand why the potential energy is 0 at infinity but I am struggling with the reason why mgh is positive yet gravitational is negative??? Can't mgh be defined the same way (yes I realize the force varies exponentially so the "height" would be erroneous) but I am still unsure. Help

    Sent from my iPhone using Physics Forums
  2. jcsd
  3. Feb 21, 2014 #2


    User Avatar
    Homework Helper

    For point masses or non-overlapping spherical masses of uniform density

    GPE = - G m1 m2 / r + K

    Where K is some arbitrary constant. Setting K = 0 means that GPE = 0 for r = ∞. It's just a matter of convenience.

    In the case of an infinitely large plane with some fixed amount of mass per unit area, then

    GPE = g m h + K

    where g is the acceleration produced on any object in the gravitational field produced by the infinite plane. Setting K = 0, means that GPE = 0 for h = 0. For objects close to the earths surface, treating the earth's surface as an approximation to an infinite plane is common and for this approximation, GPE = m g h.

    In the case of a point mass outside of an infinitely long cylinder with fixed amount of mass per unit length:

    GPE = g m ln(r/K) = g m (ln(r) - ln(K))

    Where K is the reference radial distance from the center of the cylinder, and g an acceleration factor related to the mass density in the cylinder. Setting K = 1 results in

    GPE = g m ln(r)

    In this case the choice of K = 1 would depend on the units.

    Note that in all cases, GPE increases (becomes less negative or becomes more positive) as r (or h) increases.
  4. Feb 21, 2014 #3
    So when you conserve energy in an equation then how come you could have -gpe + kinetic energy = ... It would appear as though gpe were subtracting energy and hence yield a wrong result if gpe wasn't on the right side of the equation as well.

    Sent from my iPhone using Physics Forums
  5. Feb 21, 2014 #4


    User Avatar

    Staff: Mentor

    How do you know that? All that ever matters physically is the difference in potential energy, U2 - U1 = ΔU, between two locations. It's purely a matter of mathematical convenience which location we choose to have potential energy U = 0.

    To see the connection between U = mgh and U = -GMm/r, maybe this post will help:


    This analysis works only for r > R, i.e. at the earth's surface and beyond. It doesn't work for r < R, inside the earth, because F ≠ GMm/r2 and U ≠ -GMm/r + K inside the earth.
    Last edited: Feb 21, 2014
  6. Feb 22, 2014 #5
    I think you need to clarify your question. The mechanical energy is defined as E = K + U,
    where U is the potential energy. U may be a negative quantity, but the formula is E = K + U.
  7. Feb 22, 2014 #6

    Yeah wouldn't a negative potential energy subtract from the total mechanical energy??? So how can that be right?
  8. Feb 22, 2014 #7
    Yes, it would subtract from the mechanical energy, but that's not a problem.
    The potential energy itself has no physical significance. Only changes to the energy are meaningful.
    You can always add a constant to the total potential energy (which also adds a constant to the mechanical energy) with no effect on the physics. That constant can be positive or negative.
    Even if you chose the surface of the earth as your reference level, you would still have to deal with negative potential energy for points inside of the earth.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook