Considering the -ve sign in Gravitational Potentail Energy

[hms.tech]

I believe that I have found a serious mistake in the wiki entry about Potential Energy.

notice the -ve sign in the formula on the web page.

According to the article, the definition of this term is : the gravitational potential at a location is equal to the work (energy transferred) per unit mass that is done by the force of gravity to move an object from infinity to the location.

The wording itself is mathematically inconsistent with the formula -$\frac{GMm}{r}$.

Here is the problem :

It is clear that gravity does positive work on the object when it is moved from $\infty$ to some distance "x" . (this is simply because Work = F.s ; Force and displacement are in the same direction).
Why is it that the Formula contains a -ve sign when gravity is doing positive work on the object ; hence transferring energy to the object (by the intuition of "work").
The energy of the object is increasing .

Say we take a point X= 2 m

By definition , Gravitational potential energy is the work done(by gravity) to move the object from infinity to X=2 m .We plug in the values and we get an answer which is -ve .

This -ve answer implies that the work done by gravity in moving the mass from infinity to X=2m is NEGATIVE but we have already established that gravity does +ve work on any object which moves closer to itself ( such as this object moving closer to the Earth) [since work = F.s, and force and displacement are in the same direction]

This is clearly a contradiction !

Please help, I am ecstatic about this conundrum.

 

[Dale]

The Wikipedia entry is correct. Let's say that the object is in free fall so all we need to worry about is the KE and the PE. It starts at infinity with 0 KE and 0 PE (according to that formula). The total energy is KE+PE=0, which is a conserved quantity. Then it falls to some finite radius r during which gravity does positive work on it. This increases the KE. Since the KE is positive and since KE+PE=0 we know that the PE must now be negative. Hence the sign.

Although this analysis is specific to an object with 0 total energy, it easily generalizes to other situations.

 

[hms.tech]

I think this problem looks convoluted but its really not.

The only solution I can find is to accept that the wording of the definition IS flawed, however considering it from a more open Point of view, I am willing to accept that the definition is the best which can be formulated.

I must not take the definition too literally and just limit it to being the defining statement of "Gravitational Potential energy" and not the "work done by gravity".

On a similar note, we can relate this definition to Work done by Gravity as ;

W(gravity) = - ΔP.E

 

[sophiecentaur]

I think this problem looks convoluted but its really not.

The only solution I can find is to accept that the wording of the definition IS flawed, however considering it from a more open Point of view, I am willing to accept that the definition is the best which can be formulated.

I must not take the definition too literally and just limit it to being the defining statement of "Gravitational Potential energy" and not the "work done by gravity".

On a similar note, we can relate this definition to Work done by Gravity as ;

W(gravity) = - ΔP.E

The definition is not flawed. It is arbitrary and the choice that's been made for the definition of potential energy is the Work Done in taking a standard mass to a certain position (by the experimenter - not by the Field). Bringing a mass from infinity involves Negative Work (You get energy out).

If the inverse definition had been chosen then all the signs would be opposite in all circumstances.

 

[apelling]

Wiki says:-In classical mechanics, the gravitational potential at a location is equal to the work (energy transferred) per unit mass that is done by the force of gravity to move an object to a fixed reference location.

By fixed reference location they mean infinity. It is the work done by gravity moving from r to infinity not from infinity to r. This is equivalent to the work done against gravity in moving from r to infinity. This latter definition is the more normal definition as given by hyperphysics

 

[sophiecentaur]

Wiki says:-In classical mechanics, the gravitational potential at a location is equal to the work (energy transferred) per unit mass that is done by the force of gravity to move an object to a fixed reference location.

By fixed reference location they mean infinity. It is the work done by gravity moving from r to infinity not from infinity to r. This is equivalent to the work done against gravity in moving from r to infinity. This latter definition is the more normal definition as given by hyperphysics

I think you have managed to add confusion to this. The standard expression for GPE (relative to infinity) has a minus sign and that implies that the work done, in question, must be done by the experimenter and not 'by gravity'. Why not just leave it at that (as the Hyperphysics page says). Once you open up both alternatives, you can only expect grief at some stage. Your Bank Balance is in the Negative, when you owe the Bank money and not when they owe you and even the most inventive accountant could hardly put it the other way round.

 

[jbriggs444]

Whether potential energy is the work that would have been done by the experimenter to take a test object from the defined reference point (e.g. at infinity) to its current position or whether it is the work that would be done by gravity on the test object as it moves from its current position to the defined reference point is a matter of no particular concern.

For an object moved slowly between the test point and the reference point, Newton's second law tells us that the force of gravity and the force exerted by the experimenter are equal and opposite. It follows that the work done by the experimenter traversing a path in one direction is equal to the work done by gravity traversing the same path in the opposite direction.

Of course, since [static] Newtonian gravity is a conservative field, the exact path doesn't matter; only the endpoints do.

Sophie, you seem to see a sign discrepancy here that I do not.

 

[jtbell]

The first paragraph of the Wikipedia page says:

In classical mechanics, the gravitational potential at a location is equal to the work (energy transferred) per unit mass that is done by the force of gravity to move an object to a fixed reference location. [...] By convention, the reference location is usually taken at infinity

I put the key phrase in boldface.

However, further along, the section Mathematical form starts out by saying:

The potential V at a distance x from a point mass of mass M can be defined as the work done by the gravitational field bringing a unit mass in from infinity to that point

Again, I put the key phrase in boldface. Compare it to the one in the first quote. They are inconsistent. The first version is correct, the second one is not.

 

[rcgldr]

Note that the definition of potential energy within a field from point A to point B is defined as the negative of the work done by the force from the field on an object moving from point A to point B.

For an attractive force like gravity, potential energy increases with distance from the source. If GPE at ∞ is defined to be zero, then GPE is negative at any finite distance. Note that in the case where g (acceleration from gravity) is assumed to be constant when near the surface of the earth, then GPE is commonly defined to be zero at h (height) = 0, and GPE is positive for h > 0. In both cases, GPE increases (becomes less negative or more positive) as distance from the source of gravity increases.

 

[sophiecentaur]

Whether potential energy is the work that would have been done by the experimenter to take a test object from the defined reference point (e.g. at infinity) to its current position or whether it is the work that would be done by gravity on the test object as it moves from its current position to the defined reference point is a matter of no particular concern.

For an object moved slowly between the test point and the reference point, Newton's second law tells us that the force of gravity and the force exerted by the experimenter are equal and opposite. It follows that the work done by the experimenter traversing a path in one direction is equal to the work done by gravity traversing the same path in the opposite direction.

Of course, since [static] Newtonian gravity is a conservative field, the exact path doesn't matter; only the endpoints do.

Sophie, you seem to see a sign discrepancy here that I do not.

I guess that, whichever way you look at it, the Work is a scalar so there's no real distinction between 'who does what' and that doesn't imply a sign discrepancy. I think my concern is to do with the 'share' of the energy that the planet and the unit mass get, as a result of the motion. When a unit mass is brought towards a planet, the force times distance moved by the planet is almost zero where the force times distance moved by the unit (vanishingly small) mass is huge. This isn't really relevant, though - as I now see.

 

[hms.tech]

G.P.E at a point is defined as the work done by "gravity" to move a mass from that point to infinity.