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Math Help - Elliptical motion, related rates

  1. #1
    Junior Member Tclack's Avatar
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    Elliptical motion, related rates

    1. The problem statement, all variables and given/known data

    A satellite is in an orbit around earth. The distance from the center of the earth is described by

    r= 4995/(1+.12cos@) R earth= 3960 mi

    find the rate at which the altitude is changing at the instant where @=120 degrees. d@/dt= 2.7 degrees/min

    2. Notes
    altitude equals r - (R earth)

    "@" describes the angle the satellite forms with the Perigee of earth (the closest point)

    3. The attempt at a solution

    a = r - (R earth) = 4995/(1+cos@) - 3960

    da/dt= [-4995(-sin@)d@/dt]/(1+cos@)^2

    by plugging in the values I get: ~ 46,700 mi/min

    The answer from back of book is 27.7 mi/min


    I do find it a mystery that the R earth is not used, that may be a key to solving it. Help!
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  2. #2
    Eater of Worlds
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    r=\frac{4995}{1+\frac{3}{25}cos{\theta}}

    \frac{dr}{dt}=\frac{374625sin{\theta}}{(3cos{\thet  a}+25)^{2}}\cdot\frac{d{\theta}}{dt}

    I am using radians, so be careful with the 2.7. That is \frac{3\pi}{200} rad.

    So, we get:

    \frac{dr}{dt}=\frac{374625sin{\frac{2\pi}{3}}}{(3c  os{\frac{2\pi}{3}}+25)^{2}}\cdot\frac{3\pi}{200}=\  frac{44955\sqrt{3}{\pi}}{8836}\approx 27.6843 \;\ \frac{mi}{min}

    The reason the R is not used is because the given equation has it incorporated and already gives the distance from the CENTER of the Earth.
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  3. #3
    MHF Contributor
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    Quote Originally Posted by Tclack View Post
    1. The problem statement, all variables and given/known data

    A satellite is in an orbit around earth. The distance from the center of the earth is described by

    r= 4995/(1+.12cos@) R earth= 3960 mi

    find the rate at which the altitude is changing at the instant where @=120 degrees. d@/dt= 2.7 degrees/min

    2. Notes
    altitude equals r - (R earth)

    "@" describes the angle the satellite forms with the Perigee of earth (the closest point)

    3. The attempt at a solution

    a = r - (R earth) = 4995/(1+cos@) - 3960

    da/dt= [-4995(-sin@)d@/dt]/(1+cos@)^2

    by plugging in the values I get: ~ 46,700 mi/min

    The answer from back of book is 27.7 mi/min


    I do find it a mystery that the R earth is not used, that may be a key to solving it. Help!
    1. you need to use radians ...

    120^{\circ} = \frac{2\pi}{3} radians

    2.7^{\circ}/min = \frac{3\pi}{200} \, rad/min

    2. your derivative w/r to time has an error ...

    r = 4995(1+0.12\cos{\theta})^{-1}

    \frac{dr}{dt} = -4995(1+.12\cos{\theta})^{-2} (-.12\sin{\theta}) \cdot \frac{d\theta}{dt}<br />

    \frac{dr}{dt} = \frac{599.4\sin{\theta}}{(1+.12\cos{\theta})^2} \cdot \frac{d\theta}{dt}<br />

    recalculate.
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  4. #4
    Junior Member Tclack's Avatar
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    Why Radians?

    OK, I see I took the derivative wrong. But why does it only work in radians?
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  5. #5
    Flow Master
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    Quote Originally Posted by Tclack View Post
    OK, I see I took the derivative wrong. But why does it only work in radians?
    The standard derivatives of the trig functions are obtained using radian measure. eg. \frac{d}{d \theta} \sin \theta = \cos \theta is only true when \theta is measured in radians.
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