# Why is the work done in moving a unit mass from infinity to a point (where gravitational field exists) negative?

Gravitational force acts towards the center. Here, while performing work, force is towards the center and displacement is also towards the center, then why is the work performed on the body considered negative? "As in work, $W= -GM/r$"

where $G$ is the gravitational constant, $M$ is the mass of the Earth, $r$ is the distance between the center of the Earth and the object.

I understand the mathematical derivation(https://en.wikipedia.org/wiki/Gravitational_potential) but don't know the physical significance of negative sign. Also,

"Work= -(Change in gravitational potential energy) = -(final gravitational PE - initial gravitational PE)"

Convention taken here is Gravitational PE at infinity to be zero, and nearer to the earth, gravitational PE is -x, we get above equation as,

Work = $-(-x -0) = x$ which is positive. I am confused. Please help me clearing my concepts.

Textbook Statement for Gravitational Potential:(Provided in the photo attached)

• Do you have a link to where this equation is shown? I don't remember ever seeing a work-formula like that. The potential energy formula on the other hand does have a minus sign, but work is, as you say, $W=\vec F\cdot\vec r$. – Steeven Feb 13 '17 at 17:44
• Work= change in Kinetic energy= final KE - initial KE = intial PE - final PE i.e. gain in KE = loss in PE =work done on the body – Abbas Miya Feb 13 '17 at 17:46
• Thank you, but I am asking to the other work formula you wrote: $W=-GM/r$. Where do you have that from? Would you mind quote the whole text-line that introduces this formula? – Steeven Feb 13 '17 at 18:09
• A very simple thing to remember: Each displacement of a particle in a force field requires a certain amount of work -- positive if the force opposes the displacement and negative if the force aids the displacement. No equations are needed in my opinion, just this simple rule. – K7PEH Feb 13 '17 at 18:25
• Gravitational force on a unit mass at x distance away (initial position at infinity, now at x); F = GM/x^2 , in moving the mass dx distance towards the earth small work done is dW=Fdx , integrating both sides we get total work done when the body is brought at r distance from earth (initially at infinity). This gives the result. @Steeven – Abbas Miya Feb 14 '17 at 1:41

Remember that work is a transition function - not a state function. You only do work over a distance, not at a point. So when using $W=F\Delta x$ in integral-form we must remember the edges of the integral: $$W=\int_{r_i}^{r_f} F\;dx$$

Let's choose the positive axis outwards from Earth (force is negative $-F_g$ and displacement is $r_f<r_i$) and calculate work done on a falling object:

\begin{align} W&=\int_{r_i}^{r_f} (-F_g)\; dx\\ &=\left[-\left(-\frac{GM}x\right)\right]_{r_i}^{r_f}=\left[\frac{GM}x\right]_{r_i}^{r_f}\\ &=\frac{GM}{r_f}-\frac{GM}{r_i} \end{align}

$r_f<r_i$, and because they are in the denominators, $\frac{GM}{r_f}>\frac{GM}{r_i}$ and thus work $W$ is positive.

Or we can try with another choice of axis, e.g. from the starting point of the falling object and inwards towards Earth (force is positive $+F_g$ and displacement is $r_f>r_i$):

\begin{align} W&=\int_{r_i}^{r_f} F_g\; dx\\ &\left[-\frac{GM}x\right]_{r_i}^{r_f}\\ &=\left(-\frac{GM}{r_f}\right)-\left(-\frac{GM}{r_i}\right)\\ &=\frac{GM}{r_i}-\frac{GM}{r_f} \end{align}

In this case $r_i<r_f$, so $\frac{GM}{r_i}>\frac{GM}{r_f}$ and work is still positive. Choice of axis doesn't matter - a choice just has to be made to get the signs clear.

This can also be looked at from pure energy.

Gravitational potential energy is:

$$U=\int F \;dr=-\frac{GM}r$$

Work done by Earth on something falling is:

\begin{align} W&=\Delta K=U_i-U_f\\ &=-\frac{GM}{r_i}-\left(-\frac{GM}{r_f}\right)\\ &=\frac{GM}{r_f}-\frac{GM}{r_i} \end{align}

Again work $W$ will be positive.

A sign rule-of-thumb is always the formula: $W=Fr$: Work is positive if force and displacement are in the same direction, and negative if opposite.

Gravity therefor always does positive work on a falling object. But was rising like a weatherballoon, the work done by gravity would be negative. Negative work just means that your efforts don't really work; the object still moves in another direction than in which you are pushing/pulling.

• Please examine my comment written above, it gives the equation W=-GM/r. Sorry for late reply. It was mid night yesterday when I posted this question. – Abbas Miya Feb 14 '17 at 1:49
• And yes, I have mixed mechanical work formula with potential energy formula. – Abbas Miya Feb 14 '17 at 2:46
• Am I not supposed to do that? I thought relating concepts would make my understanding broader. – Abbas Miya Feb 14 '17 at 2:47
• @Nature Okay, now I finally see the root of the confusion. See the update to my answer - you forget to set up a clear axis-direction, and because of that you are mixing some signs. – Steeven Feb 14 '17 at 9:12
• and i am too young to deserve "sir". – Abbas Miya Feb 14 '17 at 9:36

You have to think about what force is providing the work over a given distance. The gravitational force is an attractive force so that a heavy object will pull a lighter object towards itself (the heavier object). The work done on a point mass is done by a heavier gravitational force. It is common physics notation to denote work as negative when a system does work on itself and to denote work as positive when a an external force does work on the system. Also, point masses move objects from an higher potential to lower potential energy.

• Please be specific – Abbas Miya Feb 13 '17 at 17:49
• There isn't a heavier gravitational force. The force of the heavy object on the lighter is exactly the same magnitude as the force of the lighter on the heavier. Both forces do work. The distinction comes in when you discuss what happens to one of the objects. And please give me a reference for the notation you mention in your bold sentence. What you say implies by the work-energy theorem that if a "system does work on itself" then the kinetic energy of the system will decrease. That certainly wouldn't be the case for a system of 2 objects attracting each other gravitationally. – Bill N Feb 13 '17 at 21:43

Assume that $W$=Work is the energy that a source must put in the system to obtain a final configuration from a starting configuration. Let $K$ be the kinetic energy: you have $$W=K_f-K_i.$$ This means that, to speed up a particle, some force must act on it by doing positive work: this work is added up to the energy. In conservative systems, you have potential energy $V$ such as, for any initial and final configuration, you have $$W=K_f-K_i=V_i-V_f.$$ Let us take as system a gravitational system: we take the kinetic and gravitational energies to be $$K=\frac12 mv^2,\quad V=-G\frac{Mm}r,$$ with obvious definitions. We start from an initial configuration that is often taken as the zero of energy: we start with a motionless particle $(v_i=0)$ at $(r_i\to\infty)$, very far from the attractor: this means $K_i=0$ and $V_i=0$. As a final position, we take the particle to be at some radius $r_f$ with velocity $v_f$. The work done by the gravitational field is $$W=-V_f=G\frac{Mm}{r_f}.$$ This is a positive quantity, so this means that $K_f=W$ will be a positive kinetical energy: the particle speeds up when coming near to the center.

As you can see, negative difference of potential energies means positive work done on the system by the forces that you describe through your potential. This means that the gravitational force will attract the particle towards the center.

If we started with "repulsive gravity", $V=G\frac{Mm}r$, we would have had the opposite situation: we would have gotten $W=-V_f=-G\frac{Mm}{r_f}$, that is negative. Since the kinetic energy cannot be negative, this means that the particle will never get nearer to the origin in this situation, and an energy cost must be paid to make the particle become closer to the origin.

To close up everything, if work is positive you will have that the final configuration may be reachable from the starting position without having to give energy to the system.

• this is the same conclusion i have. My text book states that, bringing a unit mass from infinity to some point nearer to earth does negative work on the body. But up above in your answer, there is positive work shown. – Abbas Miya Feb 13 '17 at 17:56
• Remember that you should see the difference of potentials as the work done by the forces, in this case the gravitational force: as they do a negative work, you can say that some energy got transferred from the gravitational field (even if we're not modelling it) to the kinetic energy of particles. Notice that negative work done by the forces means positive work done on the particles. An energy of $-W$ leaves the gravitational field, an energy $W$ is absorbed by the particle that speeds up. – Salvatore Baldino Feb 13 '17 at 18:50

It's the work done by us such that it starts from infinity (at zero speed) and ends at that "certain point" (also at zero speed).

If you do no work at all, then the mass will "fall" from infinity to that point, and have (potentially very significant) kinetic energy when it reaches the "end" point. But, that's not what we want; we want it to end at rest. So, we have to slow it down, which involves doing negative work on the mass. Credit: Erick Anson from Quora

Logic for work-energy theorem: Since both initial and final velocity of the mass is 0, there is no change in kinetic energy which prevents us from applying the theorem. The work done by us gets stored in the body in the form of Gravitational Potential Energy.

Here, we do not want to change the mass' kinetic energy. So in order to do that we have to apply a force opposite to the attractive gravitational force.

Therefore, in this case, the work done by us will be $-ve$ because the angle between the force that we are applying and the displacement is $180$ ($\cos 180 = -1$).

But the work done by the body with mass $M$ (which is pulling the body on which we are applying the force) will be $+ve$ because the angle between the force applied by this body and the displacement is $0$ ($cos0 = 1$).