## ieatpi 2 years ago 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.

1. hal_stirrup

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

2. ieatpi

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

3. ieatpi

$\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?

4. 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.

5. ieatpi

typo, should be dy

6. 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

7. ieatpi

ok I'll give that a shot, thanks!