Chapter 9.7, Problem 64E

(a)

To determine

To find: the polar equation of the orbit

Answer to Problem 64E

The polar equation of the orbit is $\frac{7978.8144}{1-0.9371\cos \theta }$ .

Explanation of Solution

Given:

Its low and high points above the surface of Earth were about 119 miles and 122,800 miles, respectively.

The radius of Earth is 4000 miles

  

Calculation:

Write the expression for the aphelion.

  $a\left(1-e\right)=119+4000$

Write the expression for the perihelion.

  $a\left(1+e\right)=122800+4000$

Divide the aphelion by the perihelion. Solve for e.

  $\begin{array}{l}\frac{a\left(1-e\right)}{a\left(1+e\right)}=\frac{119+4000}{122800+4000}\\ \frac{1-e}{1+e}=\frac{4119}{126800}\\ 126800-126800e=4119+4119e\\ 130919e=122681e\\ e=0.9370755964\end{array}$

Use to and the expression for the aphelion to find a.

  $\begin{array}{l}a\left(1-e\right)=119+4000\\ a=\frac{4119}{1-e}=\frac{4119}{1-0.9370755964}=65459.5\end{array}$

Use the equation given in exercise 49. Plug in the a and e.

  $\begin{array}{l}r=\frac{\left(1-e^2\right)a}{1-e\cos \theta }\\ r=\frac{\left(1-{0.9370755964}^2\right)\left(65459.5\right)}{1-0.9370755964\cos \theta }=\frac{7978.8144}{1-0.9371\cos \theta }\end{array}$

Conclusion:

Therefore, the polar equation of the orbit is $\frac{7978.8144}{1-0.9371\cos \theta }$ .

(b)

To determine

To find: the distance between the surface of Earth and the satellite

Answer to Problem 64E

The distance between the surface of Earth and the satellite is $11013$ .

Explanation of Solution

Given:

  $\theta ={60}^0$ .

Calculation:

Use the result of part a.

  $r=\frac{7978.8144}{1-0.9371\cos \theta }$

  $\theta ={60}^0$

  $r=\frac{7978.8144}{1-0.9371\cos {60}^0}=15013$

Subtract 4000 to find the distance to the surface of the Earth.

  $15013-4000=11013$

Conclusion:

Therefore, the distance between the surface of Earth and the satellite is $11013$ .

(c)

To determine

To find: the distance between the surface of Earth and the satellite

Answer to Problem 64E

The distance between the surface of Earth and the satellite is $38340$ .

Explanation of Solution

Given: $\theta ={30}^0$

Calculation:

Use the result of part a.

  $r=\frac{7978.8144}{1-0.9371\cos \theta }$

  $\theta ={30}^0$

  $r=\frac{7978.8144}{1-0.9371\cos {30}^0}=42340$

Subtract 4000 to find the distance to the surface of the Earth.

  $42340-4000=38340$

Conclusion:

Therefore, the distance between the surface of Earth and the satellite is $38340$ .