Let's calculate the difference in force, $\Delta F$, experienced by the rocks. Because $\Delta r$ is very small compared to $r$,

$$\Delta F = F_{\text{out}} - F_{\text{in}} \approx\frac{dF}{dr}\Delta r = -\frac{2GMm}{r^3} \Delta r.$$

What's the significance of that $dF/dr \times\Delta r$? Or in a general case $y = dy/dx \times \Delta x$? What's with this $\Delta x/dx$?

Please tell me which topic it is so I can just study it . I'm searching book after books. Sorry for uploading image. Downvotes are welcomed but just tell me from where to study it and then you can close my question.

  • 1
    $\begingroup$ If we write force as $F$ as a function of distance $r$, we can write $F(r+\Delta r) - F(r) \approx F'(r) \Delta r$ as a first order approximation. You can see this by writing the Taylor series for $F$ around $r$ and just keeping up to the linear term. $\endgroup$
    – chris97ong
    Commented Oct 24, 2021 at 8:14
  • $\begingroup$ By definition, $$\frac{dy}{dx} = \lim_{\Delta x\to0} \frac{\Delta y}{\Delta x}$$ So if $\Delta x$ is small, $\Delta y \approx \frac{dy}{dx}\cdot\Delta x$ $\endgroup$
    – PM 2Ring
    Commented Oct 24, 2021 at 9:39
  • $\begingroup$ @ Roaming Electron there are some notes here at: sfu.ca/~boal/390lecs/390lec8.pdf The $1/r^3$ term is interesting! $\endgroup$
    – user315366
    Commented Oct 24, 2021 at 9:54

1 Answer 1


you start with:



$$F(r+dr)= -\frac{G\,M\,m}{(r+dr)^2}= -\frac{G\,M\,m}{r^2}\,\frac{1}{(1+\frac{dr}{r})^2}$$

take the Taylor expansion for

$$\frac{1}{(1+\frac{dr}{r})^2}\overset{\text{Taylor}}{=}1-2\,\frac{dr}{r}\quad \Rightarrow\\ F(r+dr)\mapsto -\frac{G\,M\,m}{r^2}+\frac{2\,G\,M\,m\,dr}{r^3}\\ F(r+dr)-F(r)=\frac{2\,G\,M\,m\,dr}{r^3}\\ $$


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