Velocity at a place for fluid with a given pressure gradient? In a spherical coordinate system, the origin is a sink with fluid extending to infinite distance.
The pressure gradient generated in the fluid is:
$$\nabla P \alpha \frac{1}{r^2}$$
where $r$ is the distance measured from origin.
I need to calculate square of velocity of a fluid element at any given point. I am getting $v^2 \alpha \frac{1}{r}$ but the answer is $v^2 \alpha \frac{1}{r^2}$.
I know I need to integrate the relation:
$$v \frac{dv}{dr} = \frac{1}{\rho} \nabla P$$
But integrating will yield $v^2 \alpha \frac{1}{r}$.
What am I doing wrong?
 A: I think the answer should be
$$v \propto \frac{1}{\sqrt r}.$$
Indeed, if we assume a purely radial, steady, incompressible, inviscid flow, then the Navier-Stokes equation along the radial coordinate reduces to the form you used:
$$v\frac{dv}{dr} = - \frac{1}{\rho}\frac{dP}{dr} \propto -\frac{1}{r^2}.$$
But
$$v\frac{dv}{dr} = \frac{1}{2}\frac{dv^2}{dr},$$
Then we have
$$\frac{dv^2}{dr} \propto -\frac{1}{r^2} \quad \Rightarrow \quad v^2 \propto \frac{1}{r} \quad \Rightarrow \quad v \propto \frac{1}{\sqrt r}.$$
One can also arrive at the same solution without using the radial and steady flow assumptions, and that's by exploiting the analogy to gravitational force. The pressure gradient
$$\frac{dP}{dr} \propto -\frac{1}{r^2}$$
is equivalent to a gravitational field as given by Newton's law. Thus, if pressure is the only force acting on the fluid element, one can imagine an associated potential energy which is proportional to $-\frac{1}{r}$. Conservation of energy then implies that kinetic energy is proportional to $\frac{1}{r}$, and thus
$$v^2 \propto \frac{1}{r} \quad \Rightarrow \quad v \propto \frac{1}{\sqrt r}.$$
