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Math Help - Ellipse Application Problem

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by chrozer View Post
    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
    Attached Thumbnails Attached Thumbnails Ellipse Application Problem-satellit_erde.png  
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  3. #3
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    Quote Originally Posted by earboth View Post
    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?
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  4. #4
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    Quote Originally Posted by chrozer View Post
    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.
    ...
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  5. #5
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    Quote Originally Posted by earboth View Post
    ...
    Ok...I see now. Thanks alot.
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