Chapter 8.2, Problem 53E

Solution

To Find: The values of an ellipse.

  $a=237086.5,b=236571,c=15623,e=0.066$ .

Given:

The Moon’s apogee (farthest distance from Earth) is 2 $252,710mi$ , and its perigee (closest distance to Earth) is $221,463mi$ . Assuming the moon’s orbit of Earth at one focus.

Calculation:

Draw a diagram of the Moon’s orbit which is an ellipse and place the Earth at one of the foci(assumimg that its diameter is negligible). The semimajor axis is $a$ and the distance from the center to a focus is $c$ . So, the apogee is $a+c$ and the perigee is $a-c$ as shown:

  

Given the apogee is $252710$ , and the perigee is $221463$ , and can solve the system of an equations:

  $\begin{array}{c}a+c=252710\mathrm{......}(1)\\ a-c=221436\mathrm{.......}(2)\end{array}$

Add (1) and (2), to eliminate $c$ and solve for $a$ :

  $\begin{array}{c}2a=474173\\ a=237086.5\end{array}$

Solving for $c$ using equation (1):

  $\begin{array}{c}237086.5+c=252710\\ c=15623.5\end{array}$

To find $b$ , use the Pythagoras relationship of an ellipse :

  $\begin{array}{c}{a}^2=b^2+c^2\\ b=\sqrt{a^2-c^2}\\ b=\sqrt{{\left(237086.5\right)}^2-{\left(15623.5\right)}^2}\\ b=236571\end{array}$

The eccentricity of an ellipse:

  $\begin{array}{c}e=\frac{c}{a}\\ e=\frac{15623.5}{237086.5}\\ e=0.066\end{array}$

Hence, the values of an equation:

  $a=237086.5,b=236571,c=15623,e=0.066$