What is the equation for gravitational and electrical potential energy?

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The discussion focuses on deriving gravitational potential energy (GPE) and electrical potential energy (EPE) using the work-energy principle. For GPE, the work done is calculated using the gravitational force, leading to the expression W = GmM(h/(R(R+h))) for small heights compared to the Earth's radius. In the case of EPE, a similar approach is taken with the electric force, resulting in W = (e^2/k)(h/(r(r+h))) for small h relative to the electron's orbital radius. The conversation highlights the importance of defining limits in integrals and the approximation of potential energy formulas under specific conditions. Overall, both forms of potential energy increase with distance, and the discussion emphasizes the nuances in their derivations and interpretations.
  • #31
rogerk8 said:
Ep is arbitrary chosen, right?

It is not fixed like Ek, right?

It is based on a reference point.
Umm, KE is also not fixed. It is based on a reference frame. I am sitting on my chair, so in the Earth's reference frame my KE is 0, but in the Sun's reference frame I am moving so I have a substantial KE.

Energy is conserved regardless of which reference frame you use for KE and what reference point you use for PE, but different reference frames will give different values for KE and different reference points will give different values for PE.

rogerk8 said:
But if Ep=0 at infinity, Ep might as well be negative, right?
Yes, if you choose the common convention of setting gravitational PE to 0 at infinity then at all finite distances the PE is negative. There is nothing wrong with that.
rogerk8 said:
The interesting thing here is that if we compare FGPE with EPE the major difference is the sign.

Both has the exact same structure but differs in the sign.
I haven't been following your electrical PE stuff. Just like there are two masses in the gravitational equation there should also be two charges in the electrical one. If they have the same sign then the overall expression (using PE = 0 at infinity) is positive, indicating that it is repulsive. If they have opposite signs then the expression is negative, indicating that it is attractive. So the sign is folded into the charge. When the force is attractive, like gravity, you automatically get a negative sign for the PE, like gravity.
 
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  • #32
'And if we set

GM/R=g'

You can see this setting is incorrect; the units do not balance. Consequently, your result for W also doesn't give the right equation.
In the limits of integration, if you put (R+a) and (R+h) in place of a and h, you get:
W= mgR2 [(R+a)-1 - (R+h)-1]
Neglecting R(a+h) and (ah) in comparison with R2, we get:
W= mg(h - a).

I hope this would be helpful.
P. Radhakrishnamurty
 

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