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.
Best Answer
you start with:
$$F(r)=-\frac{G\,M\,m}{r^2}$$
hence
$$F(r+dr)= -\frac{G\,M\,m}{(r+dr)^2}= -\frac{G\,M\,m}{r^2}\,\frac{1}{(1+\frac{dr}{r})^2}$$
take the Taylor expansion for
$$\frac{1}{(1+\frac{dr}{r})^2}\overset{\text{Taylor}}{=}1-2\,\frac{dr}{r}\quad \Rightarrow\\ F(r+dr)\mapsto -\frac{G\,M\,m}{r^2}+\frac{2\,G\,M\,m\,dr}{r^3}\\ F(r+dr)-F(r)=\frac{2\,G\,M\,m\,dr}{r^3}\\ $$