# Single Variable Calculus, Chapter 3, 3.7, Section 3.7, Problem 20

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The equation $\displaystyle F = \frac{GmM}{r^2}$ represents the Newton's Law of Gravitation. Where $F$ is the magnitude of the force exerted by a body of mass $m$ on a body of mass $M$, $G$ is the gravitational constant and $r$ is the distance between the bodies.

a.) Find $\displaystyle \frac{dF}{dr}$ and explain its meaning. What does the minus sign indicate.

b.) Suppose it is known that the Earth attracts an object with a force that decreases at a rate of $\displaystyle \frac{2N}{\text{km}}$ when $r = 20,000$km. How fast does this force change when $r = 10,000$km?

\begin{equation} \begin{aligned} \text{a.) } \frac{dF}{dr} &= GmM \cdot \frac{d}{dr}\left( \frac{1}{r^2}\right)\\ \\ \frac{dF}{dr} &= GmM \cdot \left( \frac{r^2 \cdot \frac{d}{dr} (1) - 1 \cdot \frac{d}{dr} (r^2) }{(r^2)^2}\right)\\ \\ \frac{dF}{dr} &= GmM \cdot \frac{\left( r^2 \cdot (0) - 1 (2r) \right)}{r^4}\\ \\ \frac{dF}{dr} &= GmM \cdot \left( \frac{- 2\cancel{r}}{r^{\cancel{4}}} \right)\\ \\ \frac{dF}{dr} &= \frac{-2GmM}{r^3} \end{aligned} \end{equation}

$\displaystyle \frac{dF}{dr}$ represents the rate of change of force with respect to the distance between the bodies. The minus sign indicated that the force decreases as the distance between the bodies increases.

\begin{equation} \begin{aligned} \text{b.) } \frac{dF}{dr} &= -2\frac{GmM}{r^3} && \text{where } r = 20000 \text{ and } \frac{dF}{dr} = -2\\ \\ \cancel{-2} & = \cancel{-2} \frac{GmM}{r^3}\\ \\ GmM &= r^3 = (20,000)^3 = 8\times10^{12} \end{aligned} \end{equation}

when $r = 10,000$km,

\begin{equation} \begin{aligned} \frac{dF}{dr} &= \frac{-2\left(8\times10^{12}\right)}{(10000)^3}\\ \\ \frac{dF}{dr} &= \frac{-16N}{\text{km}} \end{aligned} \end{equation}

$F$ decreases at a rate of $\displaystyle \frac{16N}{\text{km}}$