Chapter 8, Problem 17TP

Solution

a.

To Calculate the equation of Original and geo stationary orbits.

The original orbit $x^2+y^2={4200}^2$

The geostationary orbit $x^2+y^2={22240}^2$ .

Given:

Original circular orbit $\text{4200}\text{miles}$ .

geo stationary orbit $\text{22240}\text{miles}$ .

Calculation:

The original orbit is a circle centered at Earth's origin, that is, at $(0,0)$ and has radius $\text{4200miles}$ , so it's equation is

  $x^2+y^2={4200}^2$

The geostationary orbit is a circle centered at Earth's origin and has radius $22240\text{miles}$ , so its equation is

  $x^2+y^2={22240}^2$

Conclusion:

The original orbit $x^2+y^2={4200}^2$

The geostationary orbit $x^2+y^2={22240}^2$ .

b.

To Find the values of $a$ and $c$ .

Value $a=13220$

Value $9020=c$

Given:

Original circular orbit $\text{4200}\text{miles}$ .

geo stationary orbit $\text{22240}\text{miles}$ .

Calculation:

The distance between the point $(-4200,0)$ on the circle/ellipse and the focus $(0,0)$ is equal to $a-c$ , so

  $a-c=4200$

The distance between $(-4200,0)$ and $(0,22240)$ on the outer circle is equal to $2$ a, so obtain $a$ from it

substitute $a$ into the above equation and solve it for $c$ .

  $a-c=4200$

and

  $\begin{array}{c}2a=22240-(-4200)\\ =22240+4200\\ =26440\end{array}$

Dividing by $2$

Get $a=13220$ .

Substituting this into the above equation,

  $\begin{array}{c}13220-c=4200\\ 13220-4200=c\\ 9020=c\end{array}$

Conclusion:

Value $a=13220$

Value $9020=c$

c.

Equation module of transfer orbit.

Equation module of transfer orbit $\frac{{(x-9020)}^2}{13220}+\frac{y^2}{{9665}^2}=1$ .

Given:

Original circular orbit $\text{4200}\text{miles}$ .

geo stationary orbit $\text{22240}\text{miles}$ .

Calculation:

The distance between the point $(-4200,0)$ on the circle/ellipse and the focus $(0,0)$ is equal to $a-c$ ,

so

  $a-c=4200$

The distance between $(-4200,0)$ and $(0,22240)$ on the outer circle is equal to $2$ a, so obtain $a$ from it

substitute $a$ into the above equation and solve it for $c$ .

  $a-c=4200$

and

  $\begin{array}{c}2a=22240-(-4200)\\ =22240+4200\\ =26440\end{array}$

Dividing by $2$

Get $a=13220$ .

Substituting this into the above equation,

  $\begin{array}{c}13220-c=4200\\ 13220-4200=c\\ 9020=c\end{array}$

Substituting these values into the formula

  $c^2=a^2-b^2$

The following equation for $b$ :

  $\begin{array}{c}{9020}^2={13220}^2-b^2\\ b^2={13220}^2-{9020}^2\\ b^2=93408000\\ b\approx \pm 9665\end{array}$

Since it must be $b>0$ , Only consider the positive solution.

Also, must calculate the center of the ellipse. It's the midpoint of the major axis, which has endpoints $(0,-4200)$ and $(0,22240)$ . Hence, the coordinates of the center are

  $\begin{array}{c}{x}_0=\frac{-4200+22240}{2}\\ =9020,\\ y_0=0\end{array}$

Now the equation is

  $\frac{{(x-9020)}^2}{13220}+\frac{y^2}{{9665}^2}=1$

Conclusion:

Equation module of transfer orbit $\frac{{(x-9020)}^2}{13220}+\frac{y^2}{{9665}^2}=1$ .