If you bored a hole through the centre axis of the Earth and let the air in?

Question

how wide should be the hole to suck in the entire
atmosphere?

Escuerdo:

The density of Earth atmosphere at distances as large as distance to the Moon no longer obeys Boltzman distribution. The reason is that free path of molecules is much larger than gradient of gravitatiion of the Earth. Strictly speaking, if atmosphere was in equilbribrium everywhere, than it's mass would be infinite because energy of gravitation field is constant and finite (-zero).

I would rather prefer to think, that at depth about 50km the atmospheric pressure will be about 1000 atm, and the air will be no longer comressible and have nearly constant density ~1g/cm³ below that level. That is the volume of the hole should be roughly 4πR² x 10m, and its diameter about d=√(8R x 10m) ~ 20 km.

Answer 1

This is an interesting problem, and a basic solution requires a little bit of physics and differential equations, but the final answer appears at the end.

First, the assumptions: We will assume a constant temperature independent of depth. This isn't as far from the truth as one might think since just a little way down Earth's temperature goes up and stays pretty high through most of the mantle, and at a similar temperature for the core. We will also assume that Earth is a perfect sphere rather than an oblate spheroid, that it is totally solid, has constant density, and that its radius is equal to its polar radius.

That said, Gauss's theorem tells us that on the interior of a spherical shell of mass, this shell's gravity is balanced at all points and produces no net force, and outside this shell, it acts as though it's all concentrated at the center. Hence, we will treat the gravity acting on the air at a distance from the core r as equal to that produced by the matter lying below it.

Now the basic equations:

Hydrostatic equilibirum tells us:

dP/dr=ρ*g

where P is pressure, r is distance from the core, ρ is the density of the air, and g is the gravitational acceleration, which Gauss's law and Newton's law of gravity tell us is given by

g=-4*π*M*G*r/(3*R^3)

where π~3.14159, M is the mass of Earth, G is the universal gravitational constant and R is the radius of Earth. So the equation for hydrostatic equilibrium is

dP/dr=-4*π*ρ*M*G*r/(3*R^3).

The pressure equation of state also relates pressure and density:

P=ρ*k*T/Z

where k is Boltzmann's constant, T is the temperature in Kelvins, and Z is the average molecular mass of the air. We are assuming that radiation pressure is negligible.

Taking the derivative with respect to r, we find:

dP/dr=(k*T/Z)*dρ/dr

where we assume that the degree to which temperature and average molecular mass vary with height is negligible (a more careful analysis would account for these as well, but would be more difficult).

So, substituting the right hand side of this into the left hand side of the equation for hydrostatic equilibrium gives us:

(k*T/Z)*dρ/dr=-4*π*ρ*M*G*r/(3*R^3)

which can be re-arranged to give:

(1/ρ)*dρ=[-(4*π*Z*M*G)/(3*R^3*k*T)]*r*dr

Taking the integral of both sides yields:

ln(ρ)=[-(2*π*Z*M*G)/(3*R^3*k*T)]*r^2 + C

for some constant C, or

ρ=A*Exp[[-(2*π*Z*M*G)/(3*R^3*k*T)]*r^2]

for some constant A.
This equation predicts that technically, the atmosphere only gets infinitely thin the farther away you get. But it never disappears entirely.

To find A we have to set a boundary condition, and the easiest way to do this is to come up for a requirement for how thin the atmosphere must be to be considered "no longer present" and set this as the boundary condition at r=R. Then we plug in the rest of our constants to find A.

π=3.14159...
Z=28.97 amu = 4.651x10^-26 kg
M=5.9736×10^24 kg
G=6.6743×10^-11 m^3*kg^-1*s^-2
R=6,356,752 m
k=1.380 650×10^−23 kg*m^2*s^-2K^-1
T=6000 K

Then we have for r=R

ρ=A*Exp[-147.5]
ρ=A*8.6976*10^-65

Now, if we want the density on the surface to be about equal to what it is on the surface of, say, the moon: 1 trillion times lower than it normally is, then we set
ρ=1.01,325*10^-7 kg*m^-3
so that we find
A=1.165*10^57 kg*m^-3.

So twice the integral of ρ from 0 to R will reflect the total mass in the hole over its cross-sectional area.

So mass/area=2*integral (0,R) [(-147.5*R^-2)*r^2)]
Integrating this numerically,
mass/area=927,719 kg*m^-2

Now, the total mass of Earth's atmosphere is just about 5.1480×10^18 kg.

So the area of the hole must be

5.549 x 10^12 square meters

If the hole were circular, then, its diameter would be
about 2,658,000 meters across.


That's about 2658 kilometers, or 1,652 miles.

That's about halfway across the continental United States in the East-West direction.

Answer 2

This looked like a stupid question, but the more you think about it the more you realise it is not. I assume a hole right through pole to pole and assuming the hole is lined where it passes any liquid the answer could be calculated. You would have to take into account, not just the simple volume of the atmosphere, but also the density gradient of the air "inside" the hole: I think it would be very dense towards the centre and temperature would also play a part.

I will star it but am not interested enough (or able) to work it out myself, sorry.

Answer 3

you can only dig so deep until the earth gets soft and the hole will close off, like poking a hole in pudding.
The atmosphere is a gas, which is compressible. You could fit the atmosphere into any size "hole" you want, depending on how much you compress it.
The hole in the ground would not suck in the atmosphere

Answer 4

i'm going to pass forward and make some mandatory assumptions. This ball is of countless density, meaning that the large volume of rigidity that the earth's gravity would have on it does not in basic terms overwhelm it whilst it have been given on the component of the middle. i'm going to additionally anticipate that there is not any air in this fictitious hollow, so there would be no resistance. additionally, i'm going to in elementary terms forget that maximum individuals of earth is roofed in oceans and water... that'll in basic terms foul issues up. I'lll ultimately anticipate that the ball is dropped from a undeniable top above the floor (in addition to, a undeniable top above the earth's middle) with out being "thrown" right down to grant it an preliminary speed. If all of those are real, meaning that the only rigidity performing on the ball is that of gravity. No resistance or something, then the ball would pass careening down the hollow, on no account achieving a terminal speed till it have been given to the middle of the earth, the place it would continuously lose speed till it reached the suitable top above the middle that it became initially dropped, yet on the choice area of the earth. case in point, if this occurred, and you dropped the ball down the hollow in dying valley and it got here up in mountains (an particularly not likely experience), then the ball would on no account come above the floor. this would pass on till yet another rigidity acted on the ball. i'm hoping I happy your interest.

Answer 5

Not as big as you would imagine, considering how relatively thin the depth of the atmosphere really is, in comparison to the thickness of the earth, but it's too late for me to research the actual numbers.

Answer 6

it depends on how wide the hole is because with that info. We already have knowledge of how deep the hole might be so, volume of the hole=(radius of hole squared)*(pi)*(and depth of hole) will be how much of the atmosphere will be sucked in.

Answer 7

My guess is a hole as big as the USA.

That will be pretty close to armageddon, if this happens.

Answer 8

Where would you put all the earth you dug from that hole?

Answer 9

Yeah what the other guy said ^^^^^^^^