# Ellipse Application Problem

• May 9th 2009, 09:59 AM
chrozer
Ellipse Application Problem
The space shuttle travels in an elliptical orbit about the earth. The shuttle orbits such that it reaches a maximum height above the surface of the earth of 200 miles and a minimum height above the surface of 100 miles. The center of the earth is one focus of the elipse. Assume that the center of the ellipse is at the origin and the major axis is horizontal. Also assume the earth is circular. the approximate diameter of the earth is 8000 miles. Write and equation for the shuttle's orbit.

So far I know that the equation for an ellipse is $\frac {(x-h)^2}{a^2} + \frac {(y-k)^2}{b^2} = 1$.

I'm kinda confused on information given in the problem. I don't know how it is applied to the equation of an ellipse.
• May 9th 2009, 11:59 AM
earboth
Quote:

Originally Posted by chrozer
The space shuttle travels in an elliptical orbit about the earth. The shuttle orbits such that it reaches a maximum height above the surface of the earth of 200 miles and a minimum height above the surface of 100 miles. The center of the earth is one focus of the elipse. Assume that the center of the ellipse is at the origin and the major axis is horizontal. Also assume the earth is circular. the approximate diameter of the earth is 8000 miles. Write and equation for the shuttle's orbit.

So far I know that the equation for an ellipse is $\frac {(x-h)^2}{a^2} + \frac {(y-k)^2}{b^2} = 1$.

I'm kinda confused on information given in the problem. I don't know how it is applied to the equation of an ellipse.

Make rough sketch. You can pick all necessary values from the drawing.

Since the center of the ellipse is at the origin the equation which you have to use is:

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$
• May 9th 2009, 02:24 PM
chrozer
Quote:

Originally Posted by earboth
Make rough sketch. You can pick all necessary values from the drawing.

Since the center of the ellipse is at the origin the equation which you have to use is:

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

So would the equation be $\dfrac{x^2}{4250^2}+\dfrac{y^2}{4196.7^2}=1$?

I'm still kinda confused on the drawing:

• How did you figure the distance of the blue line?
• If the diameter of the earth is 8000, then the radius should be 4000, why is there a 4100?
• And what is with the 50, 8300, and 4150?
• May 10th 2009, 01:22 AM
earboth
Quote:

Originally Posted by chrozer
So would the equation be $\dfrac{x^2}{\bold{\color{red}4150}^2}+\dfrac{y^2}{ 4196.7^2}=1$?

I'm still kinda confused on the drawing:

[*]How did you figure the distance of the blue line?
.......
[*]If the diameter of the earth is 8000, then the radius should be 4000, why is there a 4100?.......That's the distance from the center of the Earth to the satellite at it's minimum height.
.......
[*]And what is with the 50, 8300, and 4150?.......The diameter of Earth + maximum height + minimum height ist the complete major axis of the ellipse. 8000+200+100 = 2a.
a = 4150
Since the center of the Earth is a focus the excentricity of the ellipse is 4150 - 4100 = 50
.......

As you may know the ellipse has the property: $\bold{b^2 + e^2 = a^2}$

I have drawn the minor semi-axis in blue. The slanted line has the length a.

...
• May 10th 2009, 08:16 AM
chrozer
Quote:

Originally Posted by earboth
...

Ok...I see now. Thanks alot.