# Contradiction on gravitational potential energy

I was reading the derivation of the gravitational energy of a point mass and I seem to have found a contradiction.

The derivation in my textbook is given as follows:-

Let there me a large fixed mass 'M' and a small mass 'm' placed at a distance of $r_1$ from mass M.

Now let there be some external force $F_{ext}$ that displaces mass m from $r_1$ to $r_2$.

Now according to the work energy theorem, we will have the following equation:- $$K_2 - K_1 = W_g + W_{ext}$$ If we make sure that the kinetic energy of the system does not increase:- $$0 = W_g + W_{ext}$$ $$W_g = -W_{ext}$$ The change in gravitational potential energy of the system is equal to the negative of the work done by the gravitational force. $$U(r_2) - U(r_1) = -W_g$$ Now let $r_1 = r$ and $r_2 = r + dr$ where Dr is an infinitesimally small distance. $$U(r_2) - U(r_1) = -\int F_g.dr$$ Since the gravitational force and the displacement are in opposite directions:- $$U(r_2) - U(r_1) = -\int F_gdrcos(\pi)$$ $$U(r_2) - U(r_1) = \int F_gdr$$ $$U(r_2) - U(r_1) = \int \frac{GMm}{r^2}dr$$ $$U(r_2) - U(r_1) = -\frac{GMm}{r}$$ Now let $r_1$ = $\infty$ and $r_2 = r$ $$U(r) - U(\infty) = GMm(\frac{1}{\infty}-\frac{1}{r})$$ $$U(r) - U(\infty) = GMm(0-\frac{1}{r})$$ Now we will assume that the potential at infinity is zero. $$U(r) = -\frac{GMm}{r}$$

My query:-

But if we let $r_1 = \infty$ and $r_2$ = r , $r_1$ will be greater than $r_2$(since $\infty>r$ ). Our original assumption was that the mass m was moving away from the mass M and that's why the dot product of gravitational force and displacement was negative. But now we are assuming it is moving towards mass M (from infinity to a separation of r). This looks like a contradiction to me.

## 1 Answer

But if we let $r_1 = \infty$ and $r_2$ = r , $r_1$ will be greater than $r_2$(since $\infty>r$ ). Our original assumption was that the mass m was moving away from the mass M and that's why the dot product of gravitational force and displacement was negative. But now we are assuming it is moving towards mass M (from infinity to a separation of r). This looks like a contradiction to me.

I assume that you are referring to this statement

Since the gravitational force and the displacement are in opposite directions

and the equation which follows it??

$$U(r_2) - U(r_1) = -\int F_g\, dr\cos(\pi)$$

The gravitational force is $\vec F_{\rm g} = F_{\rm g} (-\hat r)$ and the displacement is $dr \,(+\hat r)$ so that the work done by the gravitational force is

$\vec F_{\rm g} \cdot dr (+\hat i) = F_{\rm g} \,dr (-\hat i)\cdot (+\hat i) = F_{\rm g} \,dr \cos \pi = - F_{\rm g} \,dr$

To find the change in the gravitational potential energy you have to evaluate minus the work done by the gravitational force

$$U(r_2) - U(r_1) = +\int^{r_2}_{r_1} F_g \, dr= \dfrac {GMm}{r_1} - \dfrac {GMm}{r_2}$$

At the end you then say, hang on if I go the other way ie from $r=r2$ to $r=r1$ things go wrong because now my path and the gravitational force are in the same direction so I do not need the $\cos π$ in my equation should it not be $\cos 0$ ?

The answer is no because that is all taken care of when you set your limits of integration.
So in your integral if the lower limit is smaller than the upper limit then the increments $dr$ are positive whereas if the lower limit is larger than the upper limit then the increments $dr$ are negative.

$$+\int^{r_2}_{r_1} F_g \, dr = - \int^{r_1}_{r_2} F_g \, dr$$

which is illustrated below. 