# Sink rate

1. Feb 17, 2007

### kasse

If I drop a stone on the surface of the Pacific Ocean, over a very deep pit, will it then keep a constant speed towards the bottom?

I've heard that some places it will take years before the stone gets to the bottom. Is this true? Which physical laws can I use to prove/disprove this?

2. Feb 17, 2007

### Hootenanny

Staff Emeritus
It will initally and then reach a terminal velocity at which it will remain until it reaches the bottom of the pit. Assuming of course the depth of the pit is less than the thickness of oceanic crust (~5-10km)
You should start by looking up Fluid Friction and in particular Viscous resistance (http://hyperphysics.phy-astr.gsu.edu/hbase/airfri.html#c3)

3. Feb 17, 2007

### arunma

Keep in mind that ideally, there are three forces acting on this stone. The first is obviously gravity. The second is a constant buoyant force acting directly opposite to gravity. Since both of these forces are constant, you can just consider the stone to have a smaller effective weight for the purposes of your calculations. Finally there is a retarding drag force, which is proportional to the stone's velocity. Thus you need to solve a first order differential equation in the velocity variable in order to find the speed of the stone at any given time. It turns out that the speed of the stone will initially increase, and then level off to a terminal velocity, which it will maintain for the rest of its trip down.

Of course, the situation is far from ideal. There are many water currents that can affect a sinking stone, and these currents are very difficult to predict. So I'm not sure how well the above model will approximate the actual situation. But if the stone sinks at its terminal velocity, I'm fairly certain that it won't take years to reach the ocean floor.

4. Feb 17, 2007

### tehno

Nope.Nonsense.

5. Feb 17, 2007

### AlephZero

A reasonable guess at the max depth would be 10000 m (no doubt Google will find a better number).

1 year = 365.25*24*3600 seconds

So if it took one year to fall, the average speed would be 0.3 mm/sec.

A "stone" the size of a dust particle might possibly fall that slowly (because viscous drag affects small objects more than large ones) but a "normal sized" stone would not.

6. Feb 18, 2007

### tehno

Hmm guys ,it seems You didn't notice how I answered kasse's two Qs above.
"Nope"is my first ,negative answer ,to his question of constant velocity of stone sinking.No,a typical stone of typical size will not sink at constant terminal velocity.Let's take that the bottom of deep Ocean pit is 10km below the surface.10 kilometers underneath the pressure is so enormous that increase in water density is no more negligible.
The differences aren't big but egzistent however:See for instance:
http://www.engineeringtoolbox.com/fluid-density-temperature-pressure-d_309.html
That means both :a "buoyant force" increase and "liquid drag force" increase.
Meaning? : slower velocity of sinking to be expected 10 km underneath than in shallow waters.
"Nonsense" is my comment about second question ( rumors kasse heard about year long trip of stone).Typical stones have specific densities $\ro>3000\ kg/m^3$.
The increase of liquid density is not sufficient,even in deepest Ocean's areas,to slow down sinking of a typical stone to such rate as kasse indicated.

Last edited: Feb 18, 2007
7. Feb 20, 2007

### zoki85

Interesting problem.
Liquid water is very complicated medium for theoretical treatmens.
And not very much compressible.What increase in water density do you expect in depth like 10,000 m (high pressure and I reckon it must be bloody cold there )?
Lets consider sinking of a smooth glass marble with:
D=1 cm
and
$$\rho =2500 kg/m^3$$

What kind of flows (turbulent or not) do you expect to take place?

Last edited: Feb 20, 2007
8. Feb 20, 2007

### tehno

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).

Last edited: Feb 20, 2007
9. Feb 21, 2007

### zoki85

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%.)

Last edited: Feb 21, 2007
10. Feb 21, 2007

### tehno

I don't know either.I just said "I expect".
Here is the link where you can find unusual physical characteristics of liquid water:
http://www.lsbu.ac.uk/water/anmlies.html
And this in particular:
http://www.lsbu.ac.uk/water/explan5.html#Pvisc

So,it looks like viscosity is lower in cold ocean depths than in shallow waters.
Will that effect C0 or how I don't know...

Last edited: Feb 21, 2007
11. Feb 21, 2007

### Mk

Wait wait wait, what if the force of gravity gets weaker the closer to the center of gravity you get to?

12. Feb 21, 2007

### tehno

Negligible.
The Ocean on Earth is not 1000 miles deep...

13. Feb 22, 2007

### zoki85

Thank you.Great link.I saw it before .
I looked up in my fluid mechanics books for data about change of C in regard to viscosity and pressure and found nothing.

14. Feb 22, 2007

### tehno

Well that's not big surprise having on mind that C0 should be factor which describes "geometry of object" and not physical state of a liquid in turbulent flow.
For rigid spherical object C0=1 per definition.
For blunt object that is different.

So the equation is: Fdrag~$A\rho v^2$.

It has to be emphasized this just an aproximate equation.
I don't know of any rigororous mathematical computations/simulation of turbulent flow of water around rigid sphere. Any links?
It is merely my expectation increased density is the factor that slows down the marble.Sorry I can help you more .Not my area of expertise.
.

Last edited: Feb 22, 2007
15. Feb 23, 2007

### zoki85

I'm getting your point.Usually they write the formula for turbulent drag force like:
$$Fd=0.5*Co*A*\rho *v^2$$
This is a VERY aproximative formula.Example is the calculation of terminal velocity of the glass marble (same dimension as before R=0.005 m),in a free fall through the air.
Standard parameters for air are $$\rho=1.29 kg/m^3,\eta=180 mPas$$

With Co=1 the result is $$v=16 m/s$$.
Only!

Undoubtly it's a turbulent flow,but real velocity is 5 times or more higher than that.

16. Feb 23, 2007

### tehno

Why do you think C0=1 in the formula?
I just took that value in per unit system for the refference point
valid for all smooth spheres.IOW,I defined it =1 becouse sphere is taken as standard shape with respect to other shapes.For calculation of terminal velocity of the marble the numerical coefficient of drag c0 is probably much smaller than 1 (as your result indicates).

17. Feb 26, 2007