Why is gravitational potential energy defined at infinity?

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SUMMARY

The discussion centers on the definition of gravitational potential energy (GPE) at infinity, particularly in relation to the equations GPE = -G m1 m2 / r + K and GPE = mgh. It is established that setting K = 0 allows GPE to equal zero at infinity, which is a matter of mathematical convenience. The conversation also clarifies that while GPE can be negative, it is the difference in potential energy that holds physical significance, not the absolute values. The relationship between potential energy and mechanical energy is emphasized, highlighting that negative potential energy does not detract from total mechanical energy.

PREREQUISITES
  • Understanding of gravitational potential energy equations
  • Familiarity with concepts of mechanical energy (E = K + U)
  • Knowledge of gravitational force equations, particularly F = GMm/r²
  • Basic grasp of logarithmic functions in physics (e.g., GPE = g m ln(r/K))
NEXT STEPS
  • Study the derivation of gravitational potential energy equations, focusing on GPE = -G m1 m2 / r
  • Explore the implications of setting reference points in potential energy calculations
  • Learn about the differences in gravitational potential energy inside and outside of celestial bodies
  • Investigate the role of constants in energy equations and their physical significance
USEFUL FOR

Students of physics, educators teaching gravitational concepts, and anyone interested in understanding the mathematical foundations of gravitational potential energy.

jaredvert
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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. HelpSent from my iPhone using Physics Forums
 
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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 Earth's 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.
 
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
 
jaredvert said:
Like here on Earth there is Zero potential energy at the center of the earth

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.

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?

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

https://www.physicsforums.com/showthread.php?p=4651831#post4651831

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:
jaredvert said:
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

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.
 
dauto said:
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.
Yeah wouldn't a negative potential energy subtract from the total mechanical energy? So how can that be right?
 
jaredvert said:
Yeah wouldn't a negative potential energy subtract from the total mechanical energy? So how can that be right?

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.
 

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