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.

Question

anonymous · Answer

write the equation again am not sure what am reading is right.

anonymous · Answer

$$v \frac{ dv }{ dt } = -ky ^{-2} $$

anonymous · Answer

$$\int\limits_{}^{}Vdv = \int\limits_{}^{} -ky ^{-2}dy$$$$\frac{ 1 }{ 2 }V ^{2}=ky ^{-1} + c$$$$V=\sqrt{2k \frac{ 1 }{ y } + c}$$

So having done that, does c turn in to the initial velocity?

phi · Answer

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 · Answer

typo, should be dy

phi · Answer

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 · Answer

ok I'll give that a shot, thanks!