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I want to find the work done by the force of gravity to move an object of mass $m$ from infinity to a point $P$ at distance $r_p$ from a body of mass $M$ (that I assume fixed). The formula should be \begin{equation*} W = \int_{\infty}^{P} \vec{F}(r) \cdot d\vec{r} \end{equation*} The force is conservative, so I can assume a straight path, and since the force on $m$ is directed toward the mass $M$, it is parallel to the displacement and the dot product is positive, equal to magnitude of the force. \begin{equation*} \vec{F}(r) \cdot d\vec{r} = G\frac{Mm}{r^2}dr \end{equation*} But doing so result in a negative work \begin{equation*} \int_{\infty}^{P} G\frac{Mm}{r^2} dr = -G\frac{mM}{r_p} \end{equation*} and this is wrong since it should be positive. What accounts for the discrepancy?

Note that Halliday and Resnick in their book "Physics" (chapter 13), do almost the same thing, because they compute the work from the point $P$ to infinity. In their case the math checks out because the force is antiparallel to the displacement and the dot product is negative, resulting in a negative work (which is ok because in that case the movement is against the force). Here it should be positive, but the problem is that not only the integral bounds are switched, but also the sign of the integrand, and so the overall sign is not changed.

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  • $\begingroup$ This might be related: Why is gravitational potential energy negative and what does that mean? $\endgroup$
    – Philip
    Jun 19 '20 at 10:55
  • $\begingroup$ @Philip I don't know about that. It's related, for sure, but that question seems to be asking about an energy formula (pure algebra) while this one is asking about an integral. $\endgroup$
    – David Z
    Jun 19 '20 at 10:57
  • $\begingroup$ @DavidZ Ah, you have a point; I've modified my comment. $\endgroup$
    – Philip
    Jun 19 '20 at 11:00
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This is kind of a weird one, so let's follow the logic step by step:

First of all, you're right that $\vec{F}\cdot\mathrm{d}\vec{r}$ should be positive. It has to be, because as you noticed, the force is always acting in the same direction as the path of integration, so each infinitesimal contribution to the integral is going to be positive. If you expressed it as a Riemann sum, you'd be summing up a bunch of positive quantities.

Given that $\vec{F}\cdot\mathrm{d}\vec{r} > 0$, and $\vec{F}$ obviously points in the negative $\hat{r}$ direction, it must also be the case that $\mathrm{d}\vec{r}$ points in the negative $\hat{r}$ direction. You might say $$\mathrm{d}\vec{r} = -\lvert\mathrm{d}r\rvert\hat{r}$$

But here's the tricky part. That last equation actually involves two different variables. The $\mathrm{d}\vec{r}$ on the left side is a differential that represents an infinitesimal progression along the path of integration (what might be otherwise denoted $\mathrm{d}\vec{\ell}$ or $\mathrm{d}\vec{s}$), while the $r$ on the right side is a coordinate which measures distance from the origin (what might otherwise be denoted $\rho$, or $x$ if you're integrating along the $x$ axis, or so on).

a small coordinate axes showing rho increasing to the right and l pointing to the left

So to avoid confusion, you might want to write $$\mathrm{d}\vec{\ell} = -\lvert\mathrm{d}\rho\rvert\hat{\rho}$$ Now, as you step along the line of integration from $\infty$ towards $P$, $\rho$ decreases. So if you're going to do a change of variables to replace $\mathrm{d}\vec{\ell}$ with its expression in terms of $\rho$, you're going to need $\mathrm{d}\rho < 0$, which in turn means that $\mathrm{d}\vec{\ell} = \mathrm{d}\rho\,\hat{\rho}$. Therefore, $$\int_{\infty}^{P}\vec{F}\cdot\mathrm{d}\vec{\ell} = \int_{\rho=\infty}^{\rho=r_P}\biggl(-G\frac{Mm}{\rho^2}\hat{\rho}\biggr)\cdot(\mathrm{d}\rho\,\hat{\rho})$$ The difference with what you were trying to do was that you assumed $\mathrm{d}r$ (non-vector, after the dot product) would be positive. That $\mathrm{d}r$ actually corresponds to $\mathrm{d}\rho$, and is negative. Roughly speaking, you implicitly did a change of variables from $\ell$ to $\rho$, which introduces a sign change, but it was hidden from you because you used the same letter ($r$) for both variables.

A precise formulation

Here's a fully precise way to do this: take the definition of work as a line integral of the force vector field, $$W = \int_{0}^{1} \vec{F}\bigl(\vec{r}(s)\bigr)\cdot\underbrace{\vec{r}'(s)\,\mathrm{d}s}_{\mathrm{d}\vec{\ell}}$$ In this definition, $s$ is a parameter of the path of integration, which ranges from $0$ at the beginning of the path to $1$ at the end. So in your case, $s = 0$ at infinity and $s = 1$ at $P$. One possible realization of this, if you assume that $P$ is a point on the $x$ axis, is $$\vec{r}(s) = \biggl(\frac{r_P}{s}, 0, 0\biggr)$$ Using the above realization, we have $$\begin{align} \vec{r}'(s) &= \biggl(\underbrace{-\frac{r_P}{s^2}}_{\mathrm{d}\rho/\mathrm{d}s}, 0, 0\biggr) = -\frac{r_P}{s^2}\hat{x} & \vec{F}\bigl(\vec{r}(s)\bigr) &= -G\frac{Mm}{(r_P/s)^2}\hat{x} \end{align}$$ Note that this makes the direction of $\vec{F}$ radially inward, as it should be, at all points along the line of integration.

Plugging these into the formula for work, we get $$W = \int_{0}^{1} \biggl(-G\frac{Mm}{(r_P/s)^2}\hat{x}\biggr)\cdot\biggl(-\frac{r_P}{s^2}\hat{x}\biggr)\,\mathrm{d}s = G\frac{Mm}{r_P}\int_0^1\mathrm{d}s = G\frac{Mm}{r_P}$$

"Guide questions" for self-study

In case you'd like to ponder this more deeply, here are some "guide questions" to think about:

  • Convince yourself that, as $s$ ranges from $0$ to $1$, $\vec{r}(s)$ traces out the path on which you want to integrate.
  • How does each component of $\vec{r}$ vary as $s$ changes from $0$ to $1$? Larger, smaller, or the same? Try making a drawing that shows it.
  • What are some other possible formulas for $\vec{r}(s)$ that would describe the same path?
  • What are some other paths, and the formulas that describe them, which would also be suitable for this physical situation? (E.g. if you don't assume $P$ is on the $x$ axis)
  • What is the sign of the x-component of $\vec{r}'(s)$?
  • Is that consistent with how your position changes as you move along the integration path?

A comparison

It's been pointed out that an answer to another, similar question says that $\mathrm{d}\vec{r}$ is not a vector directed along the path. After looking into it, I believe that's not correct, for a similar reason to what you did here: that answer uses the definition $$\mathrm{d}\vec{r} = \mathrm{d}r\,\hat{r} + r\,\mathrm{d}\theta\,\hat{\theta} + r\sin\theta\,\mathrm{d}\phi\,\hat{\phi}$$ but that expression is for the differential of the coordinate vector (of which my $\mathrm{d}\rho\,\hat{\rho}$ is the radial part). $\mathrm{d}\vec{\ell}$ is a different entity, which is directed along the integration path.

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  • $\begingroup$ Thank you, I think this is the reason, but I still find it difficult to understand it. In this way, I am integrating with respect to a negative increment, is that possible? And then, if I write the negative $dr$ as $-dr$, where I assume the symbol $dr$ to represent a positive quantity, I am in the same trouble as before. The dot product is a number, right? It should be positive here (when going from larger to smaller r) and negative in the reverse case. Ok, but then why writing a positive number with an explicit negative sign and an hidden negative sign (embedded in $dr$)? $\endgroup$
    – Emmet
    Jun 19 '20 at 12:59
  • $\begingroup$ This question has the same problem physics.stackexchange.com/questions/130579/… and a user answered that $d\vec{r}$ is not a vector directed along the path, which seems strange to me and different from what you wrote here. Is he wrong too? $\endgroup$
    – Emmet
    Jun 19 '20 at 15:06
  • $\begingroup$ Oh, you know I might have made a mistake about the direction of $\mathrm{d}\vec{r}$. I'll have to review this later and possibly edit it. $\endgroup$
    – David Z
    Jun 19 '20 at 20:16
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    $\begingroup$ I need to wrap my head around it and think about it too. Thank you for your help. Before accepting an answer I'll give the question some other time to see if you, or somebody else, comes up with something enlightening. $\endgroup$
    – Emmet
    Jun 20 '20 at 5:31
  • $\begingroup$ @Emmet I reviewed this and I believe my original answer was right, or close to right, although it probably wasn't explained very well. I've edited it to be more clear. BTW good idea to give it some time for further responses. $\endgroup$
    – David Z
    Jun 20 '20 at 9:58
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$\begin{equation*} W = \int_{\infty}^{P} \vec{F}(r) \cdot d\vec{r} \end{equation*}$

$\vec r = r \,\hat r$ is the displacement from the centre of the mass $M$ and $d\vec r=dr\,\hat r$ is incremental displacement.

The gravitational force is attractive and so in the opposite direction to $\hat r$ thus $\vec F(r) = -G \dfrac {Mm}{r^2} \hat r$

The integral now becomes $\displaystyle \int _\infty^r \left (-G \dfrac {Mm}{r^2} \hat r\right)\cdot (dr\,\hat r)\int _\infty^r -G \dfrac {Mm}{r^2}\,dr$ which will give you your required positive value.

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  • $\begingroup$ Thank you. But why $d\vec{r}=dr \hat{r}$ and not $-dr \hat{r}$? $\endgroup$
    – Emmet
    Jun 19 '20 at 13:25
  • $\begingroup$ Because the sign of $dr$ s entirely determined by the limits of integration. If you go from $r=5$ to $r=3$ then will be negative as you might have expected, $dr =\text {final - initial} = 3-5 =-2$. $\endgroup$
    – Farcher
    Jun 19 '20 at 14:23
  • $\begingroup$ And is it ok to integrate with respect to a negative increment? $\endgroup$
    – Emmet
    Jun 19 '20 at 14:29
  • $\begingroup$ If you integrate from a large number to a small number the increments are going to be negative. Try the following $\int_1^2x\,dx$ and $\int_2^1x\,dx$. $\endgroup$
    – Farcher
    Jun 19 '20 at 15:20

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