From: jafoust@cco.caltech.edu (Jeffrey Alan Foust)
Subject: Re: Deep-sea Diving on Europa
Organization: California Institute of Technology, Pasadena
Date: Sun, 16 Aug 1992 08:41:55 GMT

In article <1992Aug16.015343.28998@leland.Stanford.EDU> schmasey@leland.Stanford.EDU (Casey Carlton O'Hara) writes:
>In article <1992Aug15.073439.25744@news.iastate.edu> barrett@iastate.edu (Marc N Barrett) writes:
>>
>>   I have heard that Europa's oceans are quite a lot deeper than the oceans
>>on Earth, even though Europa is a much smaller world.  My question is: do we
>>have any deep-sea diving equipment that could survive the great pressures 
>>that exploration of the bottoms of the Europan oceans would entail?  I know
>>that the Mariana Trench on Earth has been visited before, but that trench is
>>only eight or so miles deep, which is nothing compared to the figures of 20
>>or more miles that I have heard bandied about for the possible depth of the
>>oceans on Europa.  
>
>The formula for hydrostatic pressure (the pressure exerted by a fluid
>due to the weight of the fluid above it) is something like:
>	p(h) = p(a) + d*g*h
>where p(h) is pressure at depth h, p(a) is atmospheric pressure, d is
>depth below surface, and g is acceleration due to gravity.  

It looks like you've got some terms confused: you have both d and h 
representing depth below the surface, but no term for the density of the
fluid.  Let's try a more rigorous derivation:

The equation of hydrostatic equilibirum states that:

	dp/dr = -rho*g					(1)

Where p=pressure, rho=density, g=acc. due to gravity, and r=radius here.
Actually, p, g and rho are functions of r, but for this simplification
we will assume that the density remains constant (the fluid is incompressible,
which is reasonable enough), and that g is also a constant (a reasonable
assumption if we limit ourselves to values of r that are near Ro, the 
planetary radius).  Now we can easily take the integral of (1):

	/r	 /r
	|	|
	| dp = -|rho*g dr				(2)
	|	|
	/Ro	/Ro

Which simplifies to:

	p(r) - p(Ro) = -rho*g*(r-Ro)			(3)

or:
	p(r) = p(Ro) + rho*g*d				(4)

where d=Ro-r => depth below the surface.  Given that p(Ro) will represent
only the pressure exerted by the atmosphere (which is negligable at great
depths on Earth and at any depth on Europa), we have the equation which the
previous poster wrote, with the needed correction.

>I have NO clue about the value of g on Europa, but I'll assume it's
>less than that on Earth's moon, which is about 1/6 that of Earth's
>gravity.

Another quick and easy derivation:

Using the expression for the force of attraction between two masses by
gravity:

	F = GMm/r^2					(5)

Where G=gravitational constant (6.67x10^-8 g^-1 cm^3 s^-2 in CGS units),
M and m represent the masses of the planet (or moon) and another arbitrary
object respectively, and r is the distance between them.  Combine this
with Newton's 2nd Law:

	F = ma						(6)

which represents the acceleration on the object by the force of attraction,
and we get:

	ma = GMm/r^2					(7)

The other object's mass drops out (as it should) and we get for the
acceleration due to gravity:

	a = GM/r^2 = g					(8)

Which we call g.  For Europa M=4.8x10^25 g and r=1.569x10^8 cm, so
g = 130 cm/s^2, which is just over 13% of the value of g for Earth 
(980 cm/s^2).

Now back to the equation for pressure.  For water, rho = 1 g/cm^3, so
for Earth:
	P = 980*d g cm^-1 s^-2				(9a)
	  = 980*d dyne/cm^2				(9b)
	  = 0.000967*d atm				(9c)
where d is in centimeters.  For Europa:
	P = 130*d dyne/cm^-2				(10a)
	  = 0.000128*d atm				(10b)

Now for applications to exploration of Europa.  Bathyspheres have gone down
to the bottom of the Marianas Trench, approx. 11km below the surface. So,
the pressures experienced down there are about (0.000967*1.1X10^6) atm, which
is just under 1100 atmospheres.  On Europa this would correspond to a 
depth of about 83 km.  It's impossible to say if this would take you to
the bottom of the Europan ocean, but it's no doubt quite possible to build
submersibles capable of withstanding much greater pressures if need be. 
It's a long way down the road, though...

Incidentally, Charles Sheffield's recent novel _Cold_as_Ice_ features
submersibles prowling around in the Europan ocean.  The figures he uses for
pressure compare very well to those I computed above.

>Wow, I'm actually applying stuff I learned in class.  Probably
>applying it wrong, though.  

No, you got it right, save for a few details.  As for the volume and detail
of what I wrote, all I can say is that the three years I've spent so far
at Caltech have done nasty things to my brain... :-)

-- 
Jeff Foust                        Senior, Geophysics/Planetary Science, Caltech
jafoust@cco.caltech.edu   			   	 jeff@scn1.jpl.nasa.gov
   "In other sports like tennis and basketball, the team that scores the most 
points sometimes loses." - an International Amateur Boxing Federation official