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An outcome of my high school syllabus is to "define gravitational potential energy as the work done to move an object from a very large distance away to a point in a gravitational field: $-GMm/r$".

It makes sense that GPE is always negative since a positive force against gravity and hence positive work must be done to move an object very far away, where it has 0 GPE since the field is not present at all.

But I don't understand why moving an object from an infinite distance to a point in the field is negative work. The object would experience a net negative force, and be moved a negative displacement. Won't $W=Fs$ then be positive? What does negative work even mean?

Thank you very much for helping.

Diracology
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houston
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    This post talks about negative potential energy. http://physics.stackexchange.com/q/94281/37364. Work causes a change in energy. Positive work increases energy, negative work decreases it. If F is in the opposite direction of s, which is the direction of motion, then F slows the object down. – mmesser314 Jul 06 '16 at 14:03
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    Possibly useful : Sign of Work and potential energy in electrostatics http://physics.stackexchange.com/q/99086 – sammy gerbil Jul 06 '16 at 19:45

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The work done by an external agency in bringing the body to a point is needed. Note that the process must be quasistatic, otherwise kinetic energy terms will be needed. Now gravity will tend to attract a mass. Thus to keep the process quasistatic, one must oppose this gravitational attraction. In other words, F is directed oposite to the gravitational attraction to keep the process quasistatic. Hence the negative sign.

Lelouch
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  • Thank you but I still don't understand because unfortunately I don't know what quasistatic means. But intuitively, won't the force be in the same direction as gravity if the object is moving closer, away from infinity? – houston Jul 06 '16 at 10:59
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    You can think of moving quasistatically to mean moving infinitely slowly(i.e with a non zero infinitesimal velocity), but with a gravitational field only actin on the body it will accelerate. This would require the inclusion of kinetic energy terms in the work done. So to avoid it we must counter gravity eith just enough force to prevent the body from accelerating, which in the limit becomes same as the force of gravity in magnitude. – Lelouch Jul 06 '16 at 11:41
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The object would experience a net negative force, and be moved a negative displacement.

Potential is defined in terms of the work done by an external force.
The object has a negative force acting on it due to the gravitational attraction so the external force acting on the object must be in the positive direction to have a net zero force on the object.
It is this positive external force which is multiplied by a displacement in the opposite direction giving negative work done which means that work is done on whatever system is applying the external force.

Perhaps it will be easier to understand what is going on if one considers moving one positive charge towards another positive charge.
The force on the charge to be moved is outwards so to move the charge inwards the external force of equal magnitude must be inwards in the same direction as the displacement of the external force so positive work is done.
The external force does work pushing the two charges together whereas the masses subjected to an attractive force want to come together and the whatever is providing the external force is being pulled by this force of attraction.

Another way is to consider the "test" mass which is being moved to find the work done on it to find the potential energy as the system.
That mass has a force due to the gravitational field and the external force acting on it and those two forces are equal and opposite.
This means that the net force on the test mass is zero and so the net work done on the test mass is zero.
This means that if the mass was moving at the start it had an amount of kinetic energy which would be the same as it had at the end - this is the work-energy theorem in action.
Now the gravitational field does positive work as the direction of the gravitational force (inwards) is the same as that of the displacement (inwards) whereas the external force does an equal amount of negative work as the external force has the same magnitude as the gravitational attraction as the gravitational attraction but is opposite in direction.
Note that this means that the total work done on the test mass is zero.

Farcher
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Saying that work done is +ve or -ve is a mathematical convention used for calculating energy transfers. It is +ve when it is done by us on the system, and -ve when it is done by the system on us.

Positive work is done in pushing against a force to reach a configuration - eg pushing a car uphill. Negative work is done if a force pulls in the direction we want to go - eg lowering a car downhill. In keeping the car from accelerating out of control, we can extract useful work, store it, and perhaps use it to push the car back uphill to where it started from. When we get there, the car has the same GPE as when it started, so the net work done on it is zero. We have done +ve work pushing it back uphill, so we must have done -ve work on it (it has done +ve work on us, or for us) when we lowered it downhill.

If the force were not there (eg if the hill were not there) we would have to do no work at all to move the car from one place to another.

When talking about the gravitational attraction between two objects, by convention their mutual GPE is zero when there is no interaction between them - ie when the separation is infinite. Moving one from infinity towards the other is like rolling the car downhill. The force of attraction does work for or on us. When we have got them to the required separation, we have to do +ve work to separate them again. If the mutual GPE is zero again after doing +ve work, we must have done -ve work in moving them closer.

sammy gerbil
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Image object $m$ is at some point $a$ and you were to supply a force opposite to the gravitational force caused by $M$. This force is equal to $F_{stop}=-F_g$ so that $m$ hovers completely still at point $a$.

M ------------------- a
                      ↑
                      m

Obviously this force can't do any work because the object doesn't move and so the net displacement is zero. Now imagine you want to move $m$ from $a$ to either $b$ (further away from $M$) or $c$ (closer to $M$) like this:

M ---------- c ------ a ------ b

Work is defined more specifically as $$W=\int_{a}^{b} {\vec F\cdot d\vec s}$$ where $d\vec s$ is a really small vector pointing from $a$ to $b$ (because our path is a straight line) and $\cdot $ is the vector dot product. Note that the dot product is positive if two vectors are pointing in the same direction and negative otherwise.

If you want to move $m$ from $a$ to $b$, you will have to supply some force which I will call $F_b$ in addition to $F_{stop}$. Since $F_b$ points in the same direction as $d\vec s$, the work done will be positive. Simimlarly, you would have to supply a force $F_c$ to move from $a$ to $c$. This time the force is opposite to $d\vec s$ (remember $d\vec s$ is always from start to end) so the work done is negative.

AccidentalTaylorExpansion
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