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Author Topic: How long will a falling object take to reach the bottom of the deepest ocean?  (Read 3277 times)

chris

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How long will an object take to fall from the surface to the bottom of the deepest part of the deepest ocean? And how will the timing differ between objects of different sizes or masses?

RD

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If solid spheres made of the same material, (which was denser than water), were dropped into the ocean, larger ones will reach the bottom before smaller spheres because the mass of a solid sphere is proportional to the cube of it's radius whereas the (hydrodynamic) drag is proportional to the cross-section which is proportional to the square of the radius.

i.e. as the radii of the spheres increases their mass/weight/downward-force, (∝r3), increases faster than the corresponding (hydrodynamic) drag force opposing their fall (∝r2).
« Last Edit: 09/01/2014 07:18:43 by RD »

CliffordK

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Hmmm,
The way I would approach the problem would be to calculate the terminal velocity in water, assuming you reach the terminal velocity fairly rapidly.

(terminal velocity equation without buoyancy)


Vt is the terminal velocity
m is the mass of the falling object
g is the acceleration due to gravity, 9.8 m/s2
Cd is the drag coefficient
ρ is the density of the fluid (1.027 gm/cm3 for seawater)
A is the projected area of the object.

Ok, let's start with a shot put.
m = 7.260 kg
Cd = 0.47  (sphere)
diameter: 120mm (0.12m)  (area: 0.0113 m2)

So,
Vt = sqrt((2*7.260 kg*9.8 m/s2)/(1027 kg/m3 * 0.0113 m2 * 0.47 ))

Vt = 5.1 m/s

The maximum depth of the Mariana Trench is: 10.911 km  (10,911m)
(10911/5.1)/60 = 35.7

So, it takes your shotput about 36 minutes to fall to the bottom of the Mariana Trench.

Let me try the equation for terminal velocity with buoyancy (for a sphere) at the bottom of the Wikipedia page above.
'
(terminal velocity equation for spherical object with buoyancy)


d  is the diameter of the object (0.12m)
Vt is the terminal velocity
g is the acceleration due to gravity, 9.8 m/s2
Cd is the drag coefficient 0.47
ρ is the density of the fluid (1.027 gm/cm3 for seawater)
ρs is the density of the object
A is the projected area of the object (sphere).

Calculating the density of the shotput, mass/volume using the measurements above.
(7.260 kg) / (4/3 π (0.06 m)3)
8024 kg / m3

Vt = sqrt ((4 * 9.8 m/s2 * 0.12 m * (8024 kg / m3 - 1027 kg/m3)) / (3 * 0.47 * 1027 kg/m3))

Vt = 4.8 m/s

And I come up with about 38 minutes to fall to the bottom of the  Mariana Trench when including buoyancy

You can calculate the terminal velocity for different objects.

The calculation is dependent on the mass (based on the cube of the radius), and the frontal surface area (based on the square of the radius).  So, a more massive object would tend to reach a higher terminal velocity, and fall to the bottom quicker.

syhprum

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Is there any significant increase in the density of the water when you get 10K down or is to all effects not compressible ?

CliffordK

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Hmmm,

I was thinking about that.

According to Wikipedia, the density of water at the bottom of the Mariana Trench increased by 4.96%. 

So, if the density was about 1.027 g/cc at the surface (salt water), it should be about 1.078 g/cc at the bottom.

If one assumes the density change is linear, then the velocity change would be based on the square root of the inverse of the density, and would be non-linear. 

So...  An integral?
Just plugging in 1.078 into the first equation (no buoyancy) above, one gets:
Vt(surface) = 5.11 m/s (35.6 minutes)
Vt(bottom) = 4.98 m/s (36.5 minutes)
Average: 5.04 m/s  (36.0 minutes based on average terminal velocity).

For the second equation (with buoyancy)
Vt(surface) = 4.77 m/s (38.1 minutes)
Vt(bottom) = 4.64 m/s  (39.2 minutes)
Average: 4.70 m/s  (38.7 min based on average terminal velocity).

Anyway, so the change (for the shotput) is down to within a minute or so.  As mentioned, it should be a non-linear effect, but it is probably within the accuracy of my calculations.  The density portion of the calculation would be bounded by the surface/depth calculations.

The temperature and currents might also affect the calculations.  Even Plankton might effect it somewhat.
« Last Edit: 10/01/2014 22:56:38 by CliffordK »

CliffordK

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Is the terminal velocity equation a function of density or viscosity?

One may get quite a difference calculating the velocity in tar or molasses, and for those substances it may be highly temperature dependent.  Hopefully the equations used are ok for seawater.

syhprum

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Also the depth to which the falling body goes depends on whether or not it is compressible a relatively low density object could begin by falling slowly but as it became compressed would speed up but a non compressible low density object may never reach the bottom.

CliffordK

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So you should attach a Styrofoam float to your cannon ball.


http://www.mbari.org/expeditions/Northern09/L5/sept6.htm
See full sized cup, and shrunken cups from pressure.

I assume there are some objects that may just float, say 1km deep.  However, I believe that submarines use both density and dive planes to determine the depth.  Many objects may compress somewhat with the extreme pressures of the deep.

syhprum

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Just a thought I wonder how deep a cannon ball would go on Jupiter

CliffordK

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Just a thought I wonder how deep a cannon ball would go on Jupiter

You have two issues on Jupiter.
Temperature & Density.

A Tungesten Alloy would give you a MP of around  3,410 C (6,170 F). 
According to this chart, from here, you'd barely get 10,000 km depth (just over 10% of the depth of Jupiter) before your tungsten ball would melt.  It may not ever truly vaporize due to the extreme pressures.

According to this model, posted here, you'd get much deeper, about 70% of the depth before it would become buoyant. 

If your blob remained coherent, at the extreme pressures, it may have a greater density than surrounding lighter elements, and may continue to descend, although undoubtedly Jupiter's core not only has high pressures, but has collected heavy elements.

 

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