How long will it take a steel ball dropped at the surface of the
mariana trench to reach the ocean bottom?  I am prepared to ignore the
effect of temperature on viscosity.  Assume that the ball has a diameter of 2cm.

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Request for Question Clarification by tutuzdad-ga on 16 Feb 2006 08:54 PST

Using the mathematical formula shown in this source, though the actual
depth of the ocean and the dimensions of the ball are not clearly
defined, we see that it would take a "steel ball" dropped at the
surface roughly 64 minutes to reach the bottom of the Mariana's
Trench.

http://faculty.smu.edu/lshampin/ma2343/assign.pdf
(page 4)

Does this answer your question sufficiently?

tutuzdad-ga

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Clarification of Question by ayespi-ga on 21 Feb 2006 16:40 PST

Not really.  See the more accurate analysis in the response below
yours.  The 64 min is off by 50%!!

Let's say your sphere will fall at somewhere in the neighborhood of 2
m/s.  The Reynold's number (Re = Ud/v, where U is velocity, d is
diameter, and v is kinematic velocity) is then somewhere around
40,000.  So the drag coefficient (C) is 0.43.
(http://www.fluidmech.net/jscalc/cdre01.htm).  The drag coefficent
curve is quite flat in this Reynold's number range, so it doesn't
matter if our 2 m/s guess is off.  1 m/s gives you a C of 0.44, and 3
m/s gives you a C of .43, so .43 is plenty close enough.  Drag
coefficient is

C = D/(0.5 rho_w U^2 A), where D is drag, rho_w is density of
seawater, U is velocity, and A is cross-sectional area.

To get terminal velocity, set D equal to the weight of the steel ball
in seawater:

D = g*V*(rho_s - rho_w), where g is the acceleration of gravity, V is
the volume of the ball, and rho_s is the density of steel.

The density of steel is about 7850 kg/m^3, and that of seawater is
about 1050 kg/m^3 (less at the surface, but more at the bottom of the
Mariana Trench).

So we have   C*(0.5 rho_w U^2 A) = g*V*(rho_s - rho_w)

or,

U = sqrt[2g*V*(rho_s - rho_w) / (C*rho_w*A)]

with 
C = 0.43
rho_w = 1050 kg/m^3
rho_s = 7850 kg/m^3
A = 0.000314 m^2
g = 9.8 m/s^2
V = 4.19E-6 m^3
U = ?????

Solving, U = 1.98 m/s.  (Did I get really lucky with my guess or did I cheat?)

As you know, the viscosity will increase with depth because the
temperature decreases.  However, the viscosity only changes the
Reynold's number, and we already noted that in this range, C is
insensitive to changes in Re.  So the answer should be fairly
accurate.  The depth of the Marianas trench is about 11,000 meters, so
it would take 93 minutes for the ball to reach the bottom.

Thanks. Your calculation seems entirely plausible. Two questions: 

1) Does the compression of the steel ball reduce the coefficient of
drag sufficiently to change its velocity?

2)Does the reduced gravity at that depth have any effect?

I think that these are probably miniscule and would only affect the
time insignificantly.

I definitely agree that those are both insignificant perturbations. 
The compressibility of steel is less than that of the water, so while
the ball hardly changes in size at all, the density of the water
increases from 1020 kg/m^3 at the surface to 1070 at the bottom.  The
gravity changes by less than 1 part in 1000.

Thanks for the answer.  This was the subject of a much heated debate
while diving not too far from the trench! The answers suggested by the
locals (actually stated categorically!) were in the range of over 9
hours.  I was unsure of the terminal velocity of a steel ball in
seawater.  This will silence them f'sure!