forces - Integrating a dot product gives wrong sign for work done

Consider a point mass which creates a gravitational field. The gravitational force pulls a 'test mass' towards the point mass. Since the displacement and gravitational force are in the same direction, the work done by the gravity is positive. I would like to show this mathematically, but my answer comes out negative:

I think the source of this error is how I deal with the displacement (highlighted). The displacement is in the negative direction (towards the point mass), but because you take its magnitude when doing the dot product, I didn't think its direction mattered. I therefore proceeded to swap $ds$ for $dr$ (not really knowing what I was doing, but it seemed to make sense, because $s$ and $r$ are both displacements, albeit in opposite directions).

How should I correctly integrate this dot product to get the right sign of work done?

Answer

Where did you go wrong?

According to you the work done must be positive because you are evaluating $|\vec F| \cos 0 |d\vec s|$ and all three terms in that expression are positive.

The magnitude of $\vec F$, $|\vec F|$, must be equal to the positive quantity $\dfrac{GMm}{r^2}$ as $r>0$ so $\vec F = \dfrac{GMm}{r^2} (- \hat r)$

$d\vec s = |d\vec s| (-\hat r) = ds (-\hat r)$

This gives $\vec F \cdot d\vec s = \dfrac{GMm}{r^2} (-\hat r) \cdot ds (-\hat r) = \dfrac{GMm}{r^2}\, ds$ with $ds$ positive as it is the magnitude of $d\vec s$.

So to use this expression you must have $ds$ positive ie any integration must be one where $r$ increases unless you substitute $(-dr)$ for $ds$.

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If you deal with components rather than magnitudes you do not run into the same problem.

Let us start with $\vec F \cdot d\vec s$ which in your case is the work done by the gravitational force when the mass $m$ is displaced by $d \vec s$.

Let $\hat r$ be the unit vector in the positive r-direction.

$\vec F = F \,\hat r$ and $d \vec s = dr\, \hat r$ where $F$ and $dr$ are components of $\vec F$ and $d \vec s$ in the $\hat r$ direction.

$\vec F \cdot d\vec s = F \,dr $

In your example $\vec F = -\dfrac{GMm}{r^2}$ and $dr$ is yet to be determined.

So what is the sign of $\vec F \cdot d\vec s = F \,dr = -\dfrac{GMm}{r^2} \, dr$?

That all depends on the sign of $dr$ which can either be positive or negative.

What determines the sign of $dr$?

The limits of integration will determine the sign of $dr$ during the process of integration.

In your example you move the particle from $r_1 \hat r$ to $r_2 \hat r$.

It does not matter at this stage whether $r_1 > r_2$ or $r_1 < r_2$ what you need to do is the integration.

$\displaystyle \int _{r_1}^{r_2} -\dfrac{GMm}{r^2} \, dr = GMm\ \left (\dfrac {1}{r_2} -\dfrac {1}{r_1} \right )$ and this is a positive quantity if $r_1 > r_2$.