Chapter 1.5, Problem 118E

Gravity If an imaginary line segment is drawn between the centers of the earth and the moon, then the net gravitational force F acting on an object situated on this line segment is

$F=\frac{-K}{x^2}+\frac{0.012K}{{(239-x)}^2}$

where K > 0 is a constant and x is the distance of the object from the center of the earth, measured in thousands of miles. How far from the center of the earth is the “dead spot” where no net gravitational force acts upon the object? (Express your answer to the nearest thousand miles.)

To determine

The distance from the center of earth to the dead spot.

Answer to Problem 118E

The distance from the center of earth to the dead spot is $\underset{\_}{215000\text{mi}}$.

Explanation of Solution

Given:

The net gravitational force F acting on an object situated on imaginary line segment is drawn between the centers of the earth and the moon is $F=\frac{-K}{x^2}+\frac{0.012K}{{\left(239-x\right)}^2}$, where $K>0$ is a constant and x is the distance of the object from the center of the earth, measured in thousands of miles.

Formula used:

Quadratic formula:

The solution of a quadratic equation of the form $a{x}^2+bx+c=0,a\ne 0$ can be obtained by using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Calculation:

Note that, no net gravitational force act between centre of the earth and dead spot. Therefore, $F=0$.

Substitute $0$ for $F$ in the equation $F=\frac{-K}{x^2}+\frac{0.012K}{{\left(239-x\right)}^2}$.

$\begin{array}{l}0=\frac{-K}{x^2}+\frac{0.012K}{{\left(239-x\right)}^2}\\ \frac{K}{x^2}=\frac{0.012K}{{\left(239-x\right)}^2}\\ K{\left(239-x\right)}^2=0.012K{x}^2\\ K{\left(239-x\right)}^2-0.012K{x}^2=0\end{array}$

Simplify the above equation as follows.

$\begin{array}{l}K{\left(239-x\right)}^2-0.012K{x}^2=0\\ {\left(239-x\right)}^2-0.012x^2=0\\ 57121-478x+x^2-0.012x^2=0\\ 0.988x^2-478x+57121=0\end{array}$

Use the Quadratic formula to find the value of x.

Substitute $0.988$ for $a$, $-478$ for $b$ and $57121$ for $c$ in the quadratic formula.

$\begin{array}{c}x=\frac{-\left(-478\right)\pm \sqrt{{\left(-478\right)}^2-4\times 0.988\times \left(57121\right)}}{2\times 0.988}\\ =\frac{478\pm \sqrt{228484-225742.192}}{1.976}\\ =\frac{478\pm \sqrt{2741.808}}{1.976}\\ =\frac{478\pm 52.362}{1.976}\end{array}$

Simplify further as follows.

$\begin{array}{l}x=\frac{478+52.362}{1.976}\text{or }x=\frac{478-52.362}{1.976}\\ x=\frac{530.362}{1.976}\text{or }x=\frac{425.638}{1.976}\\ x\approx 268\text{or }x\approx 215\end{array}$

Note that, 268(000) miles is greater than the distance from the earth to the moon.

Thus, the distance from the center of earth to the dead spot is $\underset{\_}{215000\text{mi}}$.