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ieatpi Group Title

A space probe is shot upward from the earth. If air resistance is disregarded, a differential equation for the velocity after burnout is v dv/dt = -ky^-2, where y is the distance from the center of the earth and k is a positive constant. If y0 is the distance from the center of the earth at burnout and v0 is the corresponding velocity, express v as a function of y.

  • one year ago
  • one year ago

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  1. hal_stirrup Group Title
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    write the equation again am not sure what am reading is right.

    • one year ago
  2. ieatpi Group Title
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    \[v \frac{ dv }{ dt } = -ky ^{-2} \]

    • one year ago
  3. ieatpi Group Title
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    \[\int\limits_{}^{}Vdv = \int\limits_{}^{} -ky ^{-2}dy\]\[\frac{ 1 }{ 2 }V ^{2}=ky ^{-1} + c\]\[V=\sqrt{2k \frac{ 1 }{ y } + c}\] So having done that, does c turn in to the initial velocity?

    • one year ago
  4. phi Group Title
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    except that it is dt not dy I was thinking V= dy/dt so you could say dt= dy/V and sub in for dt in the equation.

    • one year ago
  5. ieatpi Group Title
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    typo, should be dy

    • one year ago
  6. phi Group Title
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    that makes it easier. to find c, plug in V0 for V and Y0 for y in your 2nd equation and solve for c

    • one year ago
  7. ieatpi Group Title
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    ok I'll give that a shot, thanks!

    • one year ago
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