Chapter 6.2, Problem 42E

a.

To determine

To find: the angular velocity needed to maintain a LEO satellite.

Answer to Problem 42E

  $4.1\text{ radians per hour}$

Explanation of Solution

Given information:

Linear velocity $\left(\nu \right)=\text{17000 miles per hour}$

Radius of earth $=3960\text{ miles}$

Distance of LEO satellite’s lowest orbit above Earth $=200\text{ miles}$

Calculation:

The distance between from the satellite to the Earth $\left(r\right)$ is evaluated as follows:

  $\begin{array}{c}r=200+3960\\ =4160\text{ miles per hour}\end{array}$

The angular velocity can be evaluated as follows:

  $\begin{array}{c}v=r\omega \\ \text{17000 miles per hour}=\left(\text{4160 miles}\right)\omega \left[\text{Put the value of }\nu \text{ and }r\right]\\ \omega =\frac{\text{17000 miles per hour}}{\text{4160 miles}}\\ =4.1\text{ radians per hour}\end{array}$

Thus the angular velocity of satellite in the lower orbit is $4.1\text{ radians per hour}$ .

b.

To determine

To find: distance between LEO and Earth.

Answer to Problem 42E

290 miles

Explanation of Solution

Given information:

Linear velocity $\left(\nu \right)=\text{17000 miles per hour}$

Radius of earth $=3960\text{ miles}$

Angular velocity $\left(\omega \right)=4\text{ radians}$

Calculation:

The distance between from the satellite to the Earth $\left(r\right)$ is evaluated as follows:

  $r=3960+r_1$

The angular velocity can be evaluated as follows:

  $\begin{array}{c}v=r\omega \\ \text{17000}=\left(3960+r_1\right)\left(4\right)\left[\text{Put the value of }\nu \text{ and }r\right]\\ 3960+r_1=\frac{\text{17000}}{\text{4}}\\ 3960+r_1=4250\\ r_1=4250-3960\\ =290\text{ miles}\end{array}$

Thus the distance is 290 miles.

c.

To determine

To find: the angular velocity of any LEO satellite.

Answer to Problem 42E

  $3.8\text{ radians per hour to }4.1\text{ radians per hour}$

Explanation of Solution

Given information:

Linear velocity $\left(\nu \right)=\text{17000 miles per hour}$

Radius of earth $\left(r\right)=3960\text{ miles}$

Distance of LEO satellite’s lowest orbit above Earth $=200\text{ miles}$

Distance of LEO satellite’s highest orbit above Earth $=500\text{ miles}$

Calculation:

The distance between from the satellite of highest orbit to the Earth $\left(r\right)$ is evaluated as follows:

  $\begin{array}{c}r=500+3960\\ =4460\text{ miles per hour}\end{array}$

The angular velocity can be evaluated as follows:

  $\begin{array}{c}v=r\omega \\ \text{17000 miles per hour}=\left(\text{4460 miles}\right)\omega \left[\text{Put the value of }\nu \text{ and }r\right]\\ \omega =\frac{\text{17000 miles per hour}}{\text{4460 miles}}\\ =3.8\text{ radians per hour}\end{array}$

Thus the angular velocity of highest orbit is $3.8\text{ radians per hour}$ .

The angular velocity of lowest orbit evaluated in part a. is $4.1\text{ radians per hour}$ . So, the angular velocities of any satellites between 200 and 500 miles above earth ranges between $3.8\text{ radians per hour}$ and $4.1\text{ radians per hour}$ .