Sixth Power Law

Suppose you have a spherical particle being pushed up a slope of angle $\theta$ by the current:

Assume that the system is dominated by inertial forces not viscous forces, in which case the force on the particle is equal to the momentum change per second of the fluid striking it. If the flow velocity is $v$ then the amount of water hitting the particle per second is a cylinder of area $\pi r^2$ and length $v$, where $r$ is the particle radius. If the liquid density is $\rho_f$ the mass of water per second is $\rho_f\pi r^2 v$ and the momentum change is:

$$ mv = A\rho\pi r^2 v^2 $$

where $A$ is a fudge factor that gives the percentage of the liquid's momentum that gets transferred to the particle. So the force F pushing the particle up the slope is:

$$ F_{up} = A\rho_f\pi r^2 v^2 \cos\theta $$

The force down the slope is $mg\sin\theta$, and if $\rho_p$ is the particle density the force in terms of the radius is:

$$ F_{down} = \tfrac{4}{3}\pi r^3 (\rho_p - \rho_f) g \sin\theta $$

Remember to subtract off the liquid density to allow for the upthrust due to the fluid displaced. Now just equate the two forces to get the velocity at which the particle is in balance and we get:

$$ \tfrac{4}{3}\pi r^3 (\rho_p - \rho_f) g \sin\theta = A\rho_f\pi r^2 v^2 \cos\theta $$

or:

$$ r = \tfrac{3}{4}A\frac{\rho_f}{\rho_p - \rho_f} \frac{1}{g\tan\theta} v^2 $$

So for the particle that can just be moved by the flow we find:

$$ r \propto v^2 $$

and therefore:

$$ m \propto r^3 \propto v^6 $$

Your question is actually slightly wrong as the size is proportional to the velocity squared. It's the mass of the particle that is proportional to the sixth power of the velocity.