tehno said:
1.
Depending on temperature gradient profile taken I would expect increase of the liquid density by 4...6%.
2.
Near surface with t=20°C,viscosity \eta =10^{-3}Pas,if you apply the Stokes law to the sphere made of glass :
v=\frac{D^2g(\rho -\rho_{water})}{18\eta}>80\ m/s
This figure is unrealistically high which suggests Reynolds number is higher than the critical and the flow is certainly turbulent (all the way down but with gradually decreasing terminal velocity).
I agree in the part that the rigid glass sphere undergoes turbulent flow
.Reynolds number range ,with data of experimental velocity of sinking found as v~1m/s ,is:
Re=\frac{\rho D v}{\eta}=10000>>Rc
You are right that Stoke's law does not apply (unless D<1mm but that wasn't the case in the example) and the turbulent flow must be considered with quadratic velocity dependence:
F_{drag}::A*c_{0}*\rho_{w}*v^2
Problem is how the numerical coefficient of turbulent drag depends on the pressure and temperature?
That's very complicated with water.Water is unique among the substances
investigated in that ,at lower pressures and temperatures,its viscosity decreases with rising pressure instead of increasing!
At low low temperatures the viscosity passes through a pressure minima and than increases again.
I agree with you that in such a extreme depth as 10 km ,the viscosity is higher than in shallow water nevertheless.
How all that will affect Co I don't know.I guess it will increase so you can write:
\frac{v_{1}}{v}=\sqrt{\frac{c'\rho_{w}'(\rho-\rho_{w})}{c\rho_{w}(\rho-\rho_{w}')}}.
If I'm right ,the sinking rate is at at least 4.5% slower 10 km under the surface ( I took the density increase by 5%,guided by your estimation of 4...6%.)
