Chapter 8.2, Problem 59E

Orbit of a Satellite Scientists and engineers often use polar equations to model the motion of satellites in earth orbit. Let’s consider a satellite whose orbit is modeled by the equation r = 22500/(4 − cos θ), where r is the distance in miles between the satellite and the center of the earth and θ is the angle shown in the following figure.

1. (a) On the same viewing screen, graph the circle r = 3960 (to represent the earth, which we will assume to be a sphere of radius 3960 mi) and the polar equation of the satellite’s orbit. Describe the motion of the satellite as θ increases from 0 to 2π.
2. (b) For what angle θ is the satellite closest to the earth? Find the height of the satellite above the earth’s surface for this value of θ.

To determine

To Evaluate: The graph of satellite for $r=3960\text{mi}$ and the motion of satellite.

Answer to Problem 59E

The orbit of satellite of $r=3960\text{mi}$ is shown in figure (2) and the motion of satellite is elliptic.

Explanation of Solution

Given

The orbit of the satellite of the given radius is,

$r=\frac{22500}{\left(4-\cos \theta \right)}$

Where, $r$ is the distance in miles between the satellite and the earth and $\theta$ is the angle in the frame of reference

Figure (1)

Calculation:

Graph of the circle of the radius 3690 is,

Figure (2)

Figure (2) shows the graph of circle $r=3960$.

From the graph, it can be noticed that if the angle increases from $0$ to $2\pi$ , the satellite travels counter clockwise around the earth.

Thus, the motion of the satellite is elliptic.

To determine

To find: The angle $\theta$ for which the satellite is closest to the earth. The height of satellite relative to the earth.

Answer to Problem 59E

The angle $\theta$ is to $\pi$ of which the satellite is closest to the earth. The height of satellite relative to the earth is $540\text{miles}$.

Explanation of Solution

Section1:

For the satellite to be closest to the earth, the value of $r$ be minimum. This is possible only when the constant part of the equation of radius $\left(4-\cos \theta \right)$ is maximum.

Range of the cosine value lies between $-1\text{and}\text{1}$ .

For angle $\theta$ to which the satellite is closest to earth,

$\begin{array}{c}\cos \theta =-1\\ \theta ={\cos }^{-1}\left(-1\right)\\ =\pi \end{array}$

Thus, the angle $\theta$ is to $\pi$.

Section2:

The satellite to be closest to the earth at angle $\pi$.

Substitute $\pi$ for $\theta$ in the given radius equation,

$\begin{array}{c}r=\frac{22500}{\left(4-\left(-1\right)\right)}\\ =\frac{22500}{5}\\ =4500\end{array}$

The distance of the satellite from the centre of the earth is $4500$ miles.

The height of the satellite above the earth's surface is $4500-3960=540$ miles.

Thus, the height of satellite above the earth surface is $540\text{miles}$.