How to find the minimum launch velocity $u$ given the following information. A space probe is fired vertically upwards from the surface of a planet with a radius $R$. If the atmospheric drag is ignored, the outward velocity ($v \space m/s$) of the probe at height $h$ metres above the surface of the planet is modelled by the differential equation:$\frac{dv}{dh}=-\frac{gR^2}{v\left(R+h\right)^2} $, where $v = u$ when $h=0$ and $g$ is the gravitational acceleration on the planet.
First the question ask me to verify that $v^2=u^2-\frac{2gR}{1\:+\:\frac{R}{h}}$ which I did successfully. Then the question ask me to determine the minimum launch velocity $u$ for the probe to escape the planet's gravity.
I am not sure how to mathematically represent this information. Since it is asking for the "minimum" could it be using derivative $=0$? But this doesn't make sense. Then Could it be $u^2 >\frac{2gR}{1\:+\:\frac{R}{h}}$?
FYI the answer is: $u\:=\:\sqrt{2gR}$
 A: I believe the approach for this type of "escape velocity" problem is to take the height, $h$, to be $\infty$ and the velocity, $v$ to be zero at this height.
If you take the expression for $v^{2}=u^{2}-\frac{2gR}{1+\frac{R}{h}}$ and solve for $u^{2}$ under these conditions, you should obtain the result in question.  In particular:
$$
0=u^{2}-2gR \rightarrow u^{2}=2gR \rightarrow u=\sqrt{2gR}
$$
because as $h\rightarrow \infty, 1+\frac{R}{h} \rightarrow 1$
I hope this helps.
A: I will address the question about $ v=0 $ at $ h= \infty$ because I think the other part got answered already. Consider this analogy, suppose you have a car and you fuel it for a journey of thirty kilometres, then after you reach the point 30 kilometers, the engine will slowly come to a hault.
Similarly, in this case, we can think of the kinetic energy as the 'fuel' of our satellite. As it moves more and more outwards from space, the kinetic energy depletes but the good thing is that even if it depletes kinetic energy, as it gets further and further away the pull of earth reduces. So, image it's like a large 'vacuum cleaner' trying to suck the earth in, if the car uses the fuel and gets farther and farther away, it's not easy for the giant vacuum cleaner to suck the car in.
So, here, the object would lose some energy as it goes to infinity. To find this energy, consider the minimum case, say that all the fuel is depleted in it's journey to infinity. We have enough equations to find relation between kinetic and potential energy if all of it's energy is depleted in journey to infinity, subtracting this from the actual kinetic energy (fuel of the body) from the starting amount ( fuel initially) we will get the fuel remaining after it crosses infinity. But we usually don't care much about the second part!
