Chapter 1.8, Problem 115E

Orbit of a Satellite A satellite is in orbit around the moon. A coordinate plane containing the orbit is set up with the center of the moon at the origin, as shown in the graph, with distances measured in megameters (Mm). The equation of the satellite’s orbit is

$\frac{{(x-3)}^2}{25}+\frac{y^2}{16}=1$

1. (a) From the graph, determine the closest and the farthest that the satellite gets to the center of the moon.
2. (b) There are two points in the orbit with y-coordinates 2. Find the x-coordinates of these points, and determine their distances to the center of the moon.

To determine

The closest and the farthest points that the satellite gets to the center of the moon in the graph.

Answer to Problem 115E

The closest and the farthest points that the satellite gets to the center of the moon in the graph is $\left(-2,0\right)$ and $\left(8,0\right)$ respectively.

Explanation of Solution

Given:

The equation of satellite’s orbit is,

$\frac{{\left(x-3\right)}^2}{25}+\frac{y^2}{16}=1$

And the graph of orbit with the centre of the moon at the origin is,

Figure (1)

Calculation:

The closest and farthest points in satellite’s orbit can be found by the intersection points of equation $\frac{{\left(x-3\right)}^2}{25}+\frac{y^2}{16}=1$ and $y=0$(x-axis),

Substitute 0 for y in orbit equation $\frac{{\left(x-3\right)}^2}{25}+\frac{y^2}{16}=1$ for the x,

$\begin{array}{c}\frac{{\left(x-3\right)}^2}{25}+\frac{0^2}{16}=1\\ \frac{{\left(x-3\right)}^2}{25}=1\\ {\left(x-3\right)}^2=25\\ x-3=\sqrt{25}\end{array}$

Further solve,

$\begin{array}{c}x-3=\sqrt{25}\\ x=\left(\pm 5+3\right)\\ x=8,-2\end{array}$

So the closest and the farthest points that the satellite gets to the center of the moon in the graph is $\left(-2,0\right)$ and $\left(8,0\right)$ respectively.

To determine

The x-coordinates of points in the orbit with y-coordinates 2 and find their distances to the center of moon.

Answer to Problem 115E

The x-coordinates of points in the orbit with y-coordinates 2 is 7.3303 and $\left(-1.3303\right)$. and distance of the point $\left(7.3303,2\right)$ with the center of moon $\left(0,0\right)$ is 7.60megameters and distance of the point $\left(-1.3303,2\right)$ with the center of moon $\left(0,0\right)$ is 2.40megameters.

Explanation of Solution

Given:

The equation of satellite’s orbit is,

$\frac{{\left(x-3\right)}^2}{25}+\frac{y^2}{16}=1$

And the graph of orbit with the centre of the moon at the origin is,

Figure (1)

Calculation:

The x-coordinates of the points in the orbit with y-coordinates 2 is,

$\begin{array}{c}\frac{{\left(x-3\right)}^2}{25}+\frac{2^2}{16}=1\\ \frac{{\left(x-3\right)}^2}{25}+\frac{4}{16}=1\\ \frac{{\left(x-3\right)}^2}{25}=1-\frac{1}{4}\\ \frac{{\left(x-3\right)}^2}{25}=\frac{3}{4}\end{array}$

Further solve,

$\begin{array}{c}\frac{{\left(x-3\right)}^2}{25}=\frac{3}{4}\\ {\left(x-3\right)}^2=\frac{3}{4}\times 25\\ x-3=\sqrt{\frac{75}{4}}\\ x=\left(\pm \frac{5\sqrt{3}}{2}+3\right)\end{array}$

So the two value of x is $\left(\frac{5\sqrt{3}}{2}+3\right)=7.3303$ and $\left(-\frac{5\sqrt{3}}{2}+3\right)=\left(-1.3303\right)$.

And the distance of the point $\left(7.3303,2\right)$ with the center of moon $\left(0,0\right)$ is,

$\begin{array}{c}\sqrt{{\left(7.3303-0\right)}^2+{\left(2-0\right)}^2}=\sqrt{53.73+4}\\ =\sqrt{57.73}\\ =7.60\end{array}$

So distance of the point $\left(7.3303,2\right)$ with the center of moon $\left(0,0\right)$ is 7.60megameters.

And the distance of the point $\left(-1.3303,2\right)$ with the center of moon $\left(0,0\right)$ is,

$\begin{array}{c}\sqrt{{\left(-1.3303-0\right)}^2+{\left(2-0\right)}^2}=\sqrt{1.77+4}\\ =\sqrt{5.77}\\ =2.40\end{array}$

So distance of the point $\left(-1.3303,2\right)$ with the center of moon $\left(0,0\right)$ is 2.40megameters.