Equation for Gravitation potential energy

AI Thread Summary
The discussion centers on the concept of gravitational potential energy (U) and its equations, specifically U = -Gm1m2/r and PE = mgh. The confusion arises from the interpretation of potential energy at different heights, with the realization that potential energy is relative and depends on the chosen reference point. While U approaches zero as distance (r) approaches infinity, lifting an object increases its potential energy relative to a defined reference point, such as the ground. The key takeaway is that the difference in potential energy (ΔU) between two positions is what holds significance, rather than the absolute values of potential energy. Understanding this distinction clarifies the relationship between height, mass, and gravitational potential energy in various contexts.
member 392791
So I was looking at the equation for Gravitation potential energy

U = - Gm1m2/r

This implies the energy is 0 as r→∞

However, if I lift a book above my head, wouldn't it have more potential energy than if it was lying on the ground, meaning as r (h) increases, potential energy increases? Where is my misconception here?

PE = mgh

Wikipedia: ''The factors that affect an object's gravitational potential energy are its height relative to some reference point, its mass, and the strength of the gravitational field it is in. Thus, a book lying on a table has less gravitational potential energy than the same book on top of a taller cupboard, and less gravitational potential energy than a heavier book lying on the same table. An object at a certain height above the Moon's surface has less gravitational potential energy than at the same height above the Earth's surface because the Moon's gravity is weaker. Note that "height" in the common sense of the term cannot be used for gravitational potential energy calculations when gravity is not assumed to be a constant. The following sections provide more detail.''
 
Physics news on Phys.org
Woopydalan said:
So I was looking at the equation for Gravitation potential energy

U = -G \frac{m_1 m_2}{r}

This implies the energy is 0 as ''r''→∞

However, if I lift a book above my head, wouldn't it have more potential energy than if it was lying on the ground, meaning as r (h) increases, potential energy increases? Where is my misconception here?

PE = mgh

it does have more potential energy at ''r''→∞ than it has on the ground. 0 is bigger than a negative number.

Wikipedia: ''The factors that affect an object's gravitational potential energy are its height relative to some reference point, its mass, and the strength of the gravitational field it is in. Thus, a book lying on a table has less gravitational potential energy than the same book on top of a taller cupboard, and less gravitational potential energy than a heavier book lying on the same table. An object at a certain height above the Moon's surface has less gravitational potential energy than at the same height above the Earth's surface because the Moon's gravity is weaker. Note that "height" in the common sense of the term cannot be used for gravitational potential energy calculations when gravity is not assumed to be a constant. The following sections provide more detail.''

from a classical POV, the thing you're missing is that potential energy, by itself, is sort of meaningless. it's the difference in potential energy between the two locations that matters.
 
The formula U=-G \frac{m_1 m_2}{r} obeys the convention that U is zero at 'infinity'.

The formula U = mgh uses the convention that U is zero when h = 0. And we could say that h = 0 at floor level, at ground level, at table-top level, or wherever suits us for solving a particular problem. Alternatively (and, for me, preferably) we can write the formula as ΔU = mgΔh, in which case we never concern ourselves about where U is zero. It should be said that the mgΔh formula only applies over regions where the field is uniform, for example near the Earth's surface, for Δh << radius of Earth. So it's no use for, say, calculating escape velocity or elliptical orbits.
 
Last edited:
I agree completely with Philip Wood...it would be much better if we used ΔU=mgΔh
There would be less confusion.
 
And note we could do the same thing over longer scales, just say ΔU = Gm1m2*Δ(1/r). Note for small Δ(1/r), those two formulae become the same. So it is as has been said-- only ΔU is meaningful.
 
Hi there, im studying nanoscience at the university in Basel. Today I looked at the topic of intertial and non-inertial reference frames and the existence of fictitious forces. I understand that you call forces real in physics if they appear in interplay. Meaning that a force is real when there is the "actio" partner to the "reactio" partner. If this condition is not satisfied the force is not real. I also understand that if you specifically look at non-inertial reference frames you can...
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top