Equation for Gravitation potential energy

1. May 20, 2012

Woopydalan

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.''

2. May 20, 2012

rbj

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

from a classical POV, the thing you're missing is that potential energy, by itself, is sorta meaningless. it's the difference in potential energy between the two locations that matters.

3. May 21, 2012

Philip Wood

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: May 21, 2012
4. May 21, 2012

truesearch

I agree completely with Philip Wood....it would be much better if we used ΔU=mgΔh
There would be less confusion.

5. May 21, 2012

Ken G

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.

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