Suppose we have a system with Two point masses of mass $M$ and mass $m$. And we want to derive Work done. Lets say M is fixed or $M>>m$. Initially assume mass m is at rest at a distance of $a$ from $M$ and after some time it reaches $b$. And we want to calculate work done. Since magnitude of Gravitational Force is $\frac{GMm}{r^2}$ and displacement of $m$ is towards the big mass so Force and dissplacement are in same direction so work done should we positive. But if we calculate $\int_a^b F\cos(\theta) \,dx$ you get a different answer its negative. Since $b<a$ (Gravity is a repulsive force and after some time distance decreases between masses final distance is b and initial is a by definition) and gravitational force is just $\int_a^b \frac{GMm}{r^2} \,dr$. This has to be right because both displacement and Force are in same direction so Force multiplied by dr is as I showed above. Now when you calculate this integral you get $GMm(\frac{1}{a}-\frac{1}{b})$ which is clearly negative. What am I doing wrong. If you continue with this working to calculate potential energy you get it as positive which is also wrong. whats the mistake. The same reasoning I did for spring force and other phenonmenon its correct but gravity something wrong is happening what is it.

  • $\begingroup$ $b < a$ is confusing. Maybe make it $r_1$ and $r_2$, with $r_1 < r_2$. It would make your reasoning easier to follow. $\endgroup$
    – JEB
    Commented Mar 19 at 15:01
  • $\begingroup$ Your integral is simply bounded incorrectly. The integral from b to a is the negative of the integral from a to b. It's convention to have the lower bound less than the upper bound. This would fix your issue. $\endgroup$ Commented Mar 19 at 15:19

1 Answer 1


One way to look at what is going on is that when you do an integral like $$ \int_a^b f(r) dr $$ with $a>b$, then $dr$ is negative. More precisely, if you convert this integral into a Riemann sum, $$ \sum_{i=1}^N f(r_i) \Delta r $$ with $r_1=a>r_N=b$, then $\Delta r=\frac{b-a}{N}<0$.

So let's go back to how we would do the dot product to get to the integral for work. To do this carefully, we need a vector expression for the force (not just the magnitude). We introduce a unit vector $\hat{e}_r$ that points radially outward. Then $$ \vec{F} = -\frac{GMm}{r^2}\hat{e}_r $$ since the force points radially inward.

Now we come to the displacement vector -- this is the tricky part. What you would like to say is that the displacement is radially inward, so that $$ d\vec{r} = -\hat{e}_r dr \ \ \ {\rm (WRONG)} $$ However, this is wrong, because:

  • We know $-\hat{e}_r$ points radially inward
  • We know that $dr$ is negative, by the argument above.
  • Therefore, combining the above two points, $-\hat{e}_r dr$ points radially outward, but we know that the actual displacement $d\vec{r}$ must point radially inward.

Therefore, the correct equation here is $$ d\vec{r} = {\color{red} +} \hat{e}_r dr $$ With this substituion, we get \begin{eqnarray} W = \int \vec{F} \cdot d\vec{r} = - (\hat{e}_r \cdot \hat{e}_r) \int_a^b \frac{GMm}{r^2} dr = {\color{red}-} \int_a^b \frac{GMm}{r^2} dr = GMm \left(\frac{1}{b}-\frac{1}{a}\right) > 0 \end{eqnarray} as expected.

Another way to say this, which is perhaps less error prone, is that the displacement is $$ d\vec{r} = \hat{e}_r |dr| $$ You can avoid having to worry about the sign, if you remember that $|dr|$ means you should always choose the limits of integration to go from the smaller $r$ value to the larger $r$ value, so that $dr>0$ and $|dr|=dr$.


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