Finding the change in radius that induces an $11\%$ drop in gravitational force between two bodies.

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How come differentials only estimate the answer and don't give an exact answer that you might get when you calculate the real figure using other methods?

Example: Let $F = Gm_1m_2/r^2$

Let $r=8$. What is the change in $r$ that would correspond to an 11% decrease in $F$?

My calculations is that $r$ changes by .44 but this isn't what I get when I calculate this by hand.

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Let $\rho$ be the new radius. Then you want to have $$0.89=\cfrac{\left(\cfrac{Gm_1m_2}{\rho^2}\right)}{\left(\cfrac{Gm_1m_2}{8^2}\right)}=\frac{64}{\rho^2},$$ so $$\rho^2=\frac{64}{0.89},$$ and so $\rho=\sqrt{\dfrac{64}{0.89}}.$ The difference, then, is $\sqrt{\dfrac{64}{0.89}}-8.$ An approximate answer is $0.48.$ I'm not sure what happened in your calculations, but you must have made a mistake somewhere.


Now I understand what you did. The principle behind differential approximation is that $$\Delta f\approx\frac{df}{dx}\cdot\Delta x,$$ and that this approximation can be made as close as we like, simply by making $\Delta x$ small enough. The problem is that "small enough" will vary from function to function. Here, you were able to get an approximation within about $0.04,$ which isn't bad, but to get a better approximation to $\Delta F$, we'd want a less substantial $\Delta r$