Derivation of a Polar EQ. for a Satellite

[M98Ranger]

The problem is fairly straight forward. I was wondering if a couple of you could look over my work and let me know if it is correct. Here is the problem:

A satellite is in an orbit around the earth. The highest point of its orbit is 1000 miles above the earth surface. The lowest is 300 miles above the earths surface.

a.) Assuming an elliptical orbit with focus at the center of a radius 4000 mile earth find it's eccentricity.

b.)Setting the center of the earth on the origin and setting the orbit along the x-axis find a polar equation for the orbit.

....Bear with me while I get down what I have in the next post.

 

[M98Ranger]

\(\displaystyle \[
\begin{array}{l}
a = \frac{1}{2}*major\,axis\,of\,elipse \\
V_1 = vertex\,on\;earth's\,side \\
V_2 = vertex\,on\;earth's\,opposite\,side \\
e = eccentricity \to e = \frac{c}{a}\,where\,c = \;center\;to\;focus\;d \\
\end{array}
\]\)
I chose the equation, \(\displaystyle \[
\frac{{a(1 - e^2 )}}{{1 - e\cos \theta }}
\]\) to represent my ellipse on the x-axis.

What I have:
\(\displaystyle \[
\begin{array}{l}
a = 4650 \to 2a = 9300 \to c = \frac{{2a - 2(a - ea)}}{2} \\
ea = c \to e4650 = \frac{{2a - 2(a - ea)}}{2} \to e = \frac{7}{{93}} \\
\end{array}
\]\) My equation therefore would be:\(\displaystyle \[
\frac{{a(1 - e^2 )}}{{1 - e\cos \theta }} \to \frac{{4650(1 - \left( {\frac{7}{{93}}} \right)^2 )}}{{1 - \left( {\frac{7}{{93}}} \right)\cos \theta }} = r
\]\)

Assuming all is correct how would I simplify it properly. I assume I would just multiply top and bottom by \(\displaystyle \[
\frac{{93}}{7}
\]\) Am I correct? I am especially wary about the fact that the question wants me to use the center of the earth on the origin as the focus and to make my equation based on that parameter. Does my equation fit the requirements of the focus being on the origin? How could I find that out myself? Thanks for all thoughts in advance!!

Oh also, I have obtained \(\displaystyle \[
V_1 = \left( {4300miles,\;180^ \circ } \right) \to V_2 = \left( {5000miles,\;0^ \circ } \right)
\]\) from the information. But, it seems to me that the focus MUST be at the origin based on the equation I chose and the fact that the directrix is x= something on the negative side of the x-axes for that equation. I may be wrong, and my worries illegitamate. But I would like to hear what someone else thinks about it.

here is my final equation in what I believe is appropriate polar form, please correct me if you think otherwise...sorry for the sporadic nature of my post....: \(\displaystyle \[
\frac{{430000}}{{93 - 7Cos\theta }} = R
\]\)

 

[soroban]

Hello, M98Ranger!

I got the same answer
. . . with a slightly different approach (simpler formulas).

A satellite is in an orbit around the earth.
The highest point of its orbit is 1000 miles above the earth's surface.
The lowest is 300 miles above the earths surface.

a.) Assuming an elliptical orbit with focus at the center of a radius 4000 mile earth.
find it's eccentricity.

b.) Setting the center of the earth on the origin and setting the orbit along the x-axis.
find a polar equation for the orbit.

                  |     :
                  |   * * *
                * |     :       *
              *   |     :         *
             *    |     :          *
                  |     :
          V2*     |     C           * V1
        - - o - - + - - o - - - - - o - -
   (-4300,0)*     |  (350,0)        * (5000,0)
                  |     :
             *    |     :          *
              *   |     :         *
                * |     :       *
                  |   * * *
                  |     :

For simplicity, I placed the ellipse on a rectangular system.

A focus is at the origin.
Aphelion \(\displaystyle V_1\) is at (5000,0); perihelion \(\displaystyle V_2\) is at (-4300,0).
The center of the ellipse \(\displaystyle C\) is at (350,0).

Hence: \(\displaystyle \,c\,=\,350,\;a\,=\,4650\;\;\Rightarrow\;\;e\:=\:\frac{c}{a}\:=\:\frac{350}{4650}\:=\:\frac{7}{93}\)

The general form of an ellipse is: \(\displaystyle \L\,r\;=\;\frac{a}{1\,\pm\,e\cos\theta}\)

\(\displaystyle \text{So far, we have: }\L\,r\;=\;\frac{a}{1\,-\,\frac{7}{93}\cos\theta}\)

When \(\displaystyle \theta\,=\,0,\;r\,=\,5000\)
\(\displaystyle \;\;\text{Hence: }\L\:5000\:=\:\frac{a}{1\,-\,\frac{7}{93}\cos 0}\;\;\Rightarrow\;\;a\,=\,\frac{430,000}{93}\)


\(\displaystyle \text{Therefore: }\L\;r\;=\;\frac{\frac{430,000}{93}}{1\,-\,\frac{7}{93}\cos\theta} \;\;\Rightarrow\;\;r \;= \;\frac{430,000}{93\,-\,7\cos\theta}\)

 

[M98Ranger]

Thankyou very much Soroban. Especially thanks for the different approach. It's always nice to see different approaches to the same answer. Especially if you want to make sure that it is the correct answer.