Chapter 4.4, Problem 74AYU

To determine

To findThe distance between the object and the planet

Answer to Problem 74AYU

The distance between the object and the planet must be between 192,400 and 384,400 kilometres.

Explanation of Solution

Given:

  $F=G\frac{m_1{m}_2}{r^2}$

  $m_1{m}_2=$ The masses of the two bodies

  $r=$ Distance between the two bodies.

  $G=$ Gravitational constant $=6.6742\times {10}^{-11}{\text{ newtons meter}}^2{\text{ kilogram}}^{-2}$

Calculation:

Let the object be at a distance r from the planet. Then, the object will be 384,400 -r kilometres from the satellite.

Substitute $5.9742\times {10}^{24}$ for $m_1$ , $7.349\times {10}^{22}$ for $m_2$ , and $6.6742\times {10}^{-11}$ for $G$ in the given formula to find the force on the object due to planet.

  $F=6.6742\times {10}^{-11}\frac{\left(5.9742\times {10}^{24}\right)\left(7.349\times {10}^{22}\right)}{r^2}$

Simplify

  $F\approx \frac{2.9\times {10}^{37}}{r^2}$

Substitute $5.9742\times {10}^{24}$ for $m_1$ , $7.349\times {10}^{22}$ for $m_2$ , $6.6742\times {10}^{-11}$ for $G$ , and $384,400-r$ in the given formula to find the force on the object due to satellite.

  $F=6.6742\times {10}^{-11}\frac{\left(5.9742\times {10}^{24}\right)\left(7.349\times {10}^{22}\right)}{{\left(384,400-r\right)}^2}$

Simplify.

  $F\approx \frac{2.9\times {10}^{37}}{{\left(384,400-r\right)}^2}$

The force on the object due to satellite is greater than that due to planet.

  $\frac{2.9\times {10}^{37}}{{\left(384,400-r\right)}^2}>\frac{2.9\times {10}^{37}}{r^2}$

Subtract $\frac{2.9\times {10}^{37}}{r^2}$ form each side.

  $\frac{2.9\times {10}^{37}}{{\left(384,400-r\right)}^2}-\frac{2.9\times {10}^{37}}{r^2}>0$

Multiply both the numerator and the denominator of both the terms with the least common multiple.

  $\begin{array}{l}\frac{2.9\times {10}^{37}}{{\left(384,400-r\right)}^2}\cdot \frac{r^2{\left(384,400-r\right)}^2}{r^2{\left(384,400-r\right)}^2}-\frac{2.9\times {10}^{37}}{r^2}\cdot \frac{r^2{\left(384,400-r\right)}^2}{r^2{\left(384,400-r\right)}^2}>0\\ \frac{2.9\times {10}^{37}{r}^2-2.9\times {10}^{37}{\left(384,400-r\right)}^2}{r^2{\left(384,400-r\right)}^2}>0\end{array}$

Simplify.

  $\begin{array}{l}\frac{2.9\times {10}^{37}{r}^2-2.9\times {10}^{37}\left(1.477\times {10}^{11}-768,800r+r^2\right)}{r^2{\left(384,400-r\right)}^2}>0\\ \frac{2.9\times {10}^{37}{r}^2-4.33\times {10}^{48}+2.25\times {10}^{43}r-2.9\times {10}^{37}{r}^2}{r^2{\left(384,400-r\right)}^2}>0\\ \frac{2.25\times {10}^{43}\left(r-192,444\right)}{r^2{\left(384,400-r\right)}^2}>0\end{array}$

The zero of the numerator is 192,444 and that of the denominator are 0 and 394,400.

Separate the real number line into four intervals using the zeros.

  $\left(-\infty ,0\right)\left(0,192,400\right)\left(192,400,\text{ 384,400}\right)\left(384,400,\infty \right)$

Select a number in each interval and evaluate $f\left(x\right)=\frac{2.25\times {10}^{43}\left(r-192,444\right)}{r^2{\left(384,400-r\right)}^2}$ to determine if $f\left(x\right)$ is positive or negative.

Interval	$\left(-\infty ,0\right)$	$\left(0,192,400\right)$	$\left(192,400,\text{ 384,400}\right)$	$\left(384,400,\infty \right)$
Number chosen	-100,000	100,000	200,000	400,000
Value of f	$f\left(0\right)=-4.8\times {10}^{15}$	$f\left(100,000\right)=-7.3\times {10}^{32}$	$f\left(200,000\right)=2.3\times {10}^{31}$	$f\left(400,000\right)=-1.87\times {10}^{33}$
Conclusion	Negative	Negative	Positive	Negative

For all x in the intervals $\left(192,400,\text{ 384,400}\right),f\left(x\right)>0$

Thus, the solution set of the given inequality is $\left\{x|192,400<x<384,400\right\}$

The distance between the object and the planet must be between 192,400 and 384,400 kilometres.

Conclusion:

Therefore, the distance between the object and the planet must be between 192,400 and 384,400 kilometres.