# The sign of Gravitational Potential Energy

1. Dec 19, 2008

### ritwik06

1. The problem statement, all variables and given/known data
"The gravitational potential energy of a particle at a point with respect to another particle (kept at rest) is equal to the work done by the force which brings them from infinite separation at that point, without gain in kinetic energy."

Then I tried to mathematically calculate th value of gravitational potential energy when the particles are at a distance "R". Here is what I did.

3. The attempt at a solution
As there is no gain or loss in kinetic energy, I take my force exactly equal to the gravitational attraction between th particles, but opposite in direction.

The work done when th body is displaced by a distance dr=
$$\frac{-Gm_{1}m_{2}.dr}{r^{2}}$$

This must be negative as th force and the displacement are in opposite directions.

Integrating it with limits :
$$^{r}_{\infty}\int\frac{-Gm_{1}m_{2}.dr}{r^{2}}$$
=Gm1m2/r

But the answer given in my book is negative, with the same magnitude. Where have I committed a mistake?

2. Dec 19, 2008

### alphysicist

Hi ritwik06,

I think you may be misreading this statement. The "force" that this statement is referring to is another applied force, not gravity. For example, if I am required to do 50 J of work to push a particle from infinite separation to a point R (with no kinetic energy gain), then the particle has a PE of 50 J at that point. So this statement is the conservation of mechanical energy equation for the case of no change in kinetic energy and no initial potential energy, and with work done by a non-conservative force:

$$W_{\rm NC} = PE_f$$

(which follows from Wnc=(change in kinetic energy)+(change in potential energy)).

In your case, what I think what you want to calculate with is the definition of potential energy, which is that the potential energy difference (associated with some conservative force) is equal to the negative of the work done by that conservative force:

$$\Delta {\rm PE} = -\ W_{\rm C}$$

You'll set PEi=0 since it is at infinity for your integral, and then that should clear up the sign difference.

Last edited: Dec 19, 2008
3. Dec 20, 2008

### Staff: Mentor

Here's how I would look at it:
Good. Note that the force acts in the direction of positive r, thus it is:
$$\frac{Gm_{1}m_{2}}{r^{2}}$$

Note that dr is intrinsically in the direction of positive r, so an element of work is:
$$dW = \frac{Gm_{1}m_{2}}{r^{2}}dr$$

The work done, and thus the gravitational PE at r = R, will be negative when your limits of integration go from ∞ to R.

4. Dec 24, 2008

### ritwik06

Ok! Positive r is in the direction from the fixed mass to the moving mass, right? Force is +ve.

Will the element of work done for a small displacement +ve? the force is positive but the displacement is in opposite direction, isnt it?????

and,
Merry Christmas!

Last edited: Dec 24, 2008
5. Dec 25, 2008

### Staff: Mentor

It depends on which direction you go in. If your displacement is in the direction of positive r (you are separating the masses) then the work is positive and the change in PE is positive.

We define gravitational PE with respect to infinite separation, so in that case the displacement is opposite to the force (we are bringing the masses together) and the work done (and change in PE) is negative.

$$\text{GPE} \equiv \int_{\infty}^R dW = \int_{\infty}^R\frac{Gm_{1}m_{2}}{r^{2}}dr = -\frac{Gm_{1}m_{2}}{R}$$

It's Newton's birthday!

6. Dec 25, 2008

### HallsofIvy

Staff Emeritus
At the time Newton was born, England was still using the Julian calendar and had not yet switched to the Gregorian Calendar. The day of Newton's birth was indeed Dec 25 on the Julian Calendar (though Christmas was not considered a major holiday then) but is normally now given as Jan. 4 on the Gregorian Calendar.

7. Dec 25, 2008