From universal law of gravitation, gravitational force exerted on a body of mass m by another body of mass M is $$ \mathbf F = \frac{GMm}{x^2} $$ where x is the distance between the centres of both the objects.

So, work done by gravitational force in bringing the object of mass m from infinity to a distance r from the centre of body of mass M is $$ W = \int_\infty^r \vec{F(x)}.\vec{dx}$$ $$=\int_\infty^r \frac{GMm}{x^2}\hat x.\vec{dx}$$ (where $\hat x$ is the unit vector in the direction in which the body of mass M is attracting the body of mass m, i.e. the direction of $\vec{dx}$ which results the angle between both vectors $0$) $$ =\int_\infty^r \frac{GMm}{x^2} {dx}\ cos0$$

$$ = - GMm\left(\frac{1}{r}-\frac{1}{\infty}\right) $$ $$= -\frac{GMm}{r}$$

Now, we know that $$W=-(∆U)$$ $$-\frac{GMm}{r} = -(U_r - U_\infty)$$ $$-\frac{GMm}{r} = (U_\infty - U_r)$$ Since, Zero of potential energy is at infinity by convention, so $U_\infty$ = 0 $$-\frac{GMm}{r} = -U_r$$ $$\frac{GMm}{r} = U_r$$

I get potential energy at a distance r as positive, then why is it that gravitational potential energy is $$-\frac{GMm}{r}$$

What is wrong in my derivation?

  • $\begingroup$ Please see this: physics.stackexchange.com/q/64674 $\endgroup$ Commented Jan 23, 2018 at 16:22
  • 2
    $\begingroup$ For starters, in your first formula (v3), you equate a vector with a scalar. $\endgroup$
    – Qmechanic
    Commented Jan 23, 2018 at 16:47
  • $\begingroup$ I don't agree $U_\infty = 0 $. I think $U_r = 0 $ $\endgroup$
    – paparazzo
    Commented Jan 23, 2018 at 17:43
  • $\begingroup$ $U_\infty=0$ is the convention most commonly used. $\endgroup$ Commented Jan 23, 2018 at 18:27
  • $\begingroup$ Have a look at this derivation where it is shown that the limits of integration determine the sign of the work done. physics.stackexchange.com/a/302728/104696 $\endgroup$
    – Farcher
    Commented Jan 23, 2018 at 20:21

3 Answers 3


The definition of potential energy is the work done by external force in moving a mass from a distance r to infinity without changing its kinetic energy. Or you can say potential energy is the negative of the work done by the gravitational force in bringing a mass from a distance r to infinity.

Potential energy can be seen as that energy which on application to a mass would free it from the influence from the other mass(here Earth). So we integrate the expression from r to $\infty$.

Now writing the expression for the work done,

$$dW=\vec F.d\vec x$$ where $\vec F$ is the external force which is in the direction of displacement($\theta=0, \cos \theta=1$).So now the expression becomes $$dW=\int^\infty_r F.dx$$ integrate the expression and you may get the desired result.

  • $\begingroup$ $$W=-\int^r_\infty F.dx$$ $$W=-\int^r_\infty \frac{GMm}{x^2}dx$$ $$= -(( \frac{-GMm}{r}) -( \frac{-GMm}{\infty}))$$ $$\frac{GMm}{r} $$ $\endgroup$ Commented Jan 24, 2018 at 3:02
  • $\begingroup$ How can I get negative potential energy? $\endgroup$ Commented Jan 24, 2018 at 3:06
  • $\begingroup$ I got work done by external force as positive. Now since work done by external force is equal to potential energy at that point, so potential energy is equal to $$\frac{GMm}{r}$$. How will I get this expression negative? $\endgroup$ Commented Jan 24, 2018 at 3:23

Signs in these calculations are usually challenging. However, you pretty much know that your first definition of W is wrong. When gravitation grabs a mass from infinity and brings it close by, it obviously performs positive work. From M point of view, it added kinetic energy to m hence positive work. Your equation for W results in negative work.

The crux is in your definitions of directions along these lines: You are defining $\hat{x}$ as the direction from far to near. But the integral limits suggest $\hat{x}$ direction from near to far.

  • $\begingroup$ If I am integrating from infinity to r, then how is it that $\hat{x}$ points from near to far? $\endgroup$ Commented Jan 23, 2018 at 17:45
  • $\begingroup$ And further, $\hat{x}$ is the unit vector for force of gravitation, so i think that it should always point from far to near. Am i right? $\endgroup$ Commented Jan 23, 2018 at 17:55
  • 2
    $\begingroup$ Assume origin is at M. X axis direction towards m. dx direction is towards m. Integral limits are OK. But force is towards M and should be negative. $\endgroup$
    – npojo
    Commented Jan 23, 2018 at 18:17
  • $\begingroup$ Since I am bringing the mass from infinity to r, then how can dx be towards m, I think it should be towards M. $\endgroup$ Commented Jan 24, 2018 at 2:49

This is a mathematical issue purely. Your thinking and physical derivation is fine, you just perform a "backwards" integration. The issue appears in this step of yours:

$$ =\int_\infty^r \frac{GMm}{x^2} {dx}\ cos0$$

You integrate from a larger to a smaller value with $\int_\infty^r$.

Think of an integral in a geometric way: $\int_1^2$ is the area under the graph from point $1$ to $2$. So what is the flipped version $\int_2^1$? It is the "flipped" area from $1$ to $2$. The "negative" area, if you will. It will be added a minus sign mathematically. You must add a minus sign manually to "flip" it back if you wish to make "backwards" integration.

  • $\begingroup$ Okay, now I understand it. I should write it as $$-\int_\infty^r \frac{GMm}{x^2} {dx}\ cos0$$ or $$\int_r^\infty \frac{GMm}{x^2} {dx}\ cos0$$. $\endgroup$ Commented Jan 24, 2018 at 12:37

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