Single Variable Calculus Questions and Answers

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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 $\displaystyle \frac{16N}{\text{km}}$