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Math Help - eccentricity application

  1. #1
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    eccentricity application

    can someone help me with this problem ( I'm quite new with some terms)

    A satellite is to be placed in an elliptical orbit, with the center of the earth as one focus. The satellite's maximum distance from the surface of the earth is to be 22380 km , and its minimum distance is to be 6540 km . Assume that the radius of the earth is 6400 km , and find the eccentricity of the satellite's orbit.

    Thanks in advance!
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  2. #2
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    I'll take a crack at this...

    If you have an ellipse with equation
    \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1
    or
    \frac{(y - k)^2}{b^2} + \frac{(x - h)^2}{a^2}= 1,
    then the eccentricity e is
    e = \frac{c}{a},
    where
    c = \sqrt{a^2 - b^2}.
    The eccentricity relates to the shape of the ellipse, and it tells us how far off-center the foci are. In ellipses, 0 \leq e < 1. If e is close to 0 then the foci are close to the center, and if e is close to 1 then the foci are close to the vertices.

    In an elliptical orbit, the perihelion (minimum distance) is a - c, and the aphelion (maximum distance) is a + c. Since you mention distances "from the surface of the earth," we'll have to add on the radius of the earth.

    Minimum distance:
    a - c = 6540 + 6400
    a - c = 12940

    Maximum distance:
    a + c = 22380 + 6400
    a + c = 28780

    You have two equations and two unknowns. Solve for a and c.
    a - c = 12940
    a + c = 28780
    2a = 41720
    a = 20860
    20860 + c = 28780
    c = 7920

    The eccentricity e, therefore, is
    e = \frac{c}{a} = \frac{7920}{20860} \approx 0.380.

    Edit: Fixed typo.


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    Last edited by yeongil; May 24th 2009 at 05:34 AM.
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  3. #3
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    thanks , but according with my book , the answer should be 0.38
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  4. #4
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    Oops, I plugged in the wrong value for a at the end. I fixed it, and you're right, e = 0.380.


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