anonymous
  • anonymous
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.
Mathematics
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anonymous
  • anonymous
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.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
write the equation again am not sure what am reading is right.
anonymous
  • anonymous
\[v \frac{ dv }{ dt } = -ky ^{-2} \]
anonymous
  • anonymous
\[\int\limits_{}^{}Vdv = \int\limits_{}^{} -ky ^{-2}dy\]\[\frac{ 1 }{ 2 }V ^{2}=ky ^{-1} + c\]\[V=\sqrt{2k \frac{ 1 }{ y } + c}\] So having done that, does c turn in to the initial velocity?

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phi
  • phi
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.
anonymous
  • anonymous
typo, should be dy
phi
  • phi
that makes it easier. to find c, plug in V0 for V and Y0 for y in your 2nd equation and solve for c
anonymous
  • anonymous
ok I'll give that a shot, thanks!

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