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blue up
22nd Jul 2011, 09:47
Strange how the mind wanders during long night flights.

We got talking about physics and this question came up.


If you could drill a 1cm diameter hole right through the earth to the point on the other side of the world (and I guess we can assume spherical planet rather than oblate spheroid) and filled it with water, what would the pressure be in the centre, assuming no compression and no heating?
The water at the exact centre would be 'weightless' since there would be no (or negative?) gravity, wouldn't it?

I must admit that I wouldn't know where to start.:uhoh:

forget
22nd Jul 2011, 09:52
I must admit that I wouldn't know where to start.

You'll need a drill.

Davidsoffice
22nd Jul 2011, 09:52
That's interesting.... When you initially poured water into the hole, velocity would build up as a result of gravity. At what point would the gravity, or lack of, defeat the velocity?





Does it really matter?

Would the molten matter in the middle come flooding out?

Fitter2
22nd Jul 2011, 10:17
Integrating the gravitational force of all the individual elements of any spherical body, the total effect is that of a point mass at the centre of the sphere. There would be no acceleration force in the centre, but the pressure would be simply density x radius x gravitational constant.

Neglecting compressibility (and the various engineering impracticalities), 6.4e6 Pa. Don't go there...........

Integral calculus is a wonderfully powerful tool - clever chap, that Leibniz.

beaufort1
22nd Jul 2011, 10:20
Water would soon boil off as it's about 6000C in the centre. :8

Slasher
22nd Jul 2011, 10:49
Good FOs used to bring along stacks of p0rn for the crew on
long trips. This would entail soft and hard core depending on
who you were flying with that night. Physics, maths, and any
deep-and-meaningfuls were unheard of, but occasionally the
Co manual was rewritten once the p0rn ran out.

tony draper
22nd Jul 2011, 11:04
Would be better to fill it with mercury, prevent the hole from collapsing in on itself.
:)

Akrotiri71
22nd Jul 2011, 11:31
If you could drill a 1cm diameter hole right through the earth to the point on the other side of the world (and I guess we can assume spherical planet rather than oblate spheroid) and filled it with water, what would the pressure be in the centre, assuming no compression and no heating?
Being involved on a daily basis with hydraulics and hydrostatic pressure, this is an easy calculation for me.
Here goes:
Data:
Distance to the centre of the earth = 3960 miles-ish, which =20,908,800 feet. TVD. (Total Vertical Depth).
Fluid Density,(Weight of fresh water) = 8.33 ppg.
Constant factor = 0.052, which is derived by:
http://upload.wikimedia.org/math/8/7/f/87fb1cdc83074c5288b0dcbd133094d5.png
Formula:
Fluid Density(ppg) x 0.052 x TVD(ft).
8.33 x 0.052 x 20,908,800.
The pressure at the centre of the earth would be:9,056,855.80 psi.
*Note. The diameter of the hole has no bearing on the pressure that is exerted on the bottom of said hole.

Takan Inchovit
22nd Jul 2011, 11:31
If you could drill a 1cm diameter hole right through the earth to the point on the other side of the world

Lets hope there wasn't a Chinaman sitting on the ground at the point where the drill bit came out! But if the drill came out under the sea, then where would you be?

G-CPTN
22nd Jul 2011, 12:03
There are relatively few locations where terra firma exists on both ends of the proposed hole:-
Map Tunneling Tool (http://www.freemaptools.com/tunnel-to-other-side-of-the-earth.htm)

RegDep
22nd Jul 2011, 12:11
There are relatively few locations where terra firma exists on both ends of the proposed hole:-

Yes but why would you drill from Greenland to Antarctic in the first place :ooh:

Funny tool, however :ok:

blue up
22nd Jul 2011, 12:30
So, if you dropped a single drip of water down your hole....it would go backwards and forwards past the centre of the earth until air resistance stopped it at the centre. What would it weigh? Surely there would be no gravity at the centre? If so, would the TVD need to be factored to allow for a reduced weight for the innermost part of the column of water or would this only apply in the event of compressibility?

Akrotiri71
22nd Jul 2011, 12:45
Would be better to fill it with mercury, prevent the hole from collapsing in on itself.:)
Good point Mr D.
However, Mercury would be too heavy.
To offset the pressure at the centre of the earth, (experts calculate it's 360 gigapascals :rolleyes:, in real money that's 52,213,585 psi). You would then require a fluid with a density of only 48.024 lbs/gallon. (Mercury being 112.455 lbs/gallon).

forget
22nd Jul 2011, 12:59
Fluid Density,(Weight of fresh water) = 8.33 ppg.

The pressure at the centre of the earth would be:9,056,855.80 psi.

Eh? Surely the force of gravity decreases as you move nearer to the centre of the Earth with stuff at the centre weighing nothing. So these numbers are what we used to call ------ wrong.

Akrotiri71
22nd Jul 2011, 13:09
It's generally accepted that a normal single drip/drop of water weighs +/-0.003 oz, at normal atmospheric pressure.
I have absolutely no idea what it's weight would be at the centre of the earth, or how the characteristics of the earth's core would effect said drip/drop.
I did the above hydrostatic calculation based purely on your request of what the hydrostatic pressure would be at the centre of the earth using water, and without any outside physical effects factored in.
There is no difference in pressure, at any one point, in a column of fluid. The pressure in the centre of the column is the same as the outside of the column.

Fitter2
22nd Jul 2011, 13:26
Yes but why would you drill from Greenland to Antarctic in the first place?

Spain to New Zealand, however...........

Nemrytter
22nd Jul 2011, 14:08
So, if you dropped a single drip of water down your hole....it would go backwards and forwards past the centre of the earth until air resistance stopped it at the centre.
Yes.
What would it weigh? Surely there would be no gravity at the centre?
There's lots of gravity at the center, but it's pulling you in all directions equally so the net effect is no acceleration, and hence no weight.
That only works at the exact center, though. Away from the center the weight will slowly increase until it reaches the normal surface value.

Fareastdriver
22nd Jul 2011, 14:26
If you dropped a 9mm. ball down the hole how far up the other side of the Earth, or how close to the surface it would rise to, until it falls back down the hole again.

vulcanised
22nd Jul 2011, 15:21
Bet if you encountered oil at some point you'd lose interest in drilling any further.

Desert Dingo
22nd Jul 2011, 15:41
Here are the calculations that show that it will take 42 and a bit minutes for you to drop through the hole and get to the other side of the earth.
Hole Through the Earth Example (http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/earthole.html)

What is not shown in that link, but I have seen somewhere else, is the the hole does not have to pass through the centre of the earth. It can be between any two points, and the journey time remains the same. An oblique hole has lesser acceleration deceleration forces and I guess you would have to be on a frictionless trolley so rubbing against the walls has no effect, but it will take the same time no matter which direction you take.

G-CPTN
22nd Jul 2011, 16:02
Away from the center the weight will slowly increase until it reaches the normal surface value.

Does it keep increasing above the Earth's surface?

GroundedSLF
22nd Jul 2011, 16:09
All very interesting, but we seem to be overlooking the main question - how would you hit the right starting point - given that you would need to be 6,000 odd miles above the earth with the worlds best hand drill....

cdtaylor_nats
22nd Jul 2011, 18:21
It's not just the Earth's gravity. The moon would exert a considerable influence.

tony draper
22nd Jul 2011, 19:04
The large super dense ball of molten nickle iron sitting balefully at the center of the Earth might also present a problem.:uhoh:

balsa model
22nd Jul 2011, 20:25
Fitter2: 6.4e6 Pa
This converts to about 63 atm or a column of water of about 630m.
Slipped a decimal, perhaps?

Fitter2
22nd Jul 2011, 21:19
This converts to about 63 atm or a column of water of about 630m.
Slipped a decimal, perhaps?

Several. Missed off a K. Well spotted. too used to working in psi - these new-fangled SI units are all very well..............