A hypothetical question about gravity

[Subductionzon]

I see D.H. did a more exact analysis that showed that the air has to be a supercritical fluid and no longer a gas. The reason that the pressure gets so high in the calculations is that the Ideal Gas Law assumes that air is infinitely compressible. As we go down the hole not only would the air be increasing in pressure due to the weight of all of the gas above us, the density of that gas would be increasing with that increasing pressure resulting in exponential growth to the pressure and density. Sooner or later the IGL will fail.

 

[D H]

I did something dyslexic with my calculation, using R=3678 km rather than 6378 km.

In any case, an adiabatic assumption is arguably better than the isothermal assumption. This yields a temperature of about 48000 kelvin at the center of the Earth and a pressure of about 56 million atmospheres.

Details:
Gravity is constant at 10 m/s^2 down to the halfway point. This yields a lapse rate of 9.953 K/km (c.f. the adiabatic lapse rate of 9.8 K/km for g=9.80665 m/s^2). This makes the temperature at the halfway point be about 32000 kelvin. Below the halfway point gravity drops linearly toward zero at the center. This makes temperature quadratic below the halfway point,
$$T(z)=T(R/2) + \alpha*(z-R/2)\left(1-\frac{z-R/2} R\right)$$
where $\alpha$ is 9.953 K/km. This yields a temperature of 48000 kelvin at the center of the Earth. The assumption of adiabatic conditions means
$$P=P_0 \left(\frac{T}{T_0}\right) ^{\frac{\gamma}{\gamma - 1}}$$
Using $\gamma=1.4$ yields a pressure of 56 million atmospheres at the center of the Earth.

Bottom line: This air-filled tunnel is very hot and under extremely high pressure.

 

[poco9964]

But a lot of it. 1 bar is from the 100km column above surface at approx. constant g. In the tunnel center the column would be 63 times higher, however at decreasing g. My question was if that is enough pressure to make air liquid.


If you don't make the tunnel to wide, I doubt it. The polar bears could reach the Antarctic and eat all the penguins.


What do you mean? What should stop the air from falling to the center?

The same thing that causes a lack of Air in caves or mine shafts. I understand this hole will be opened both ends, but air will not flow from one end to the other. It would be like getting a pipe and attaching fans either end blowing inwards. They would reach a point where they would cancel each other out. In the centre, I would assume as it is equal forces being excreted. A better analogy would be to block both ends of a pipe so no air can escape, and then use a plunger mechanism on both ends to push them together. You will notice it takes tremendous force to do so. I can not see why the same effect should not occur in this hole.

The lack of gravity will also make air lighter than at the surface, so even though it got denser it does not mean it would get heavier. Gravity would begin to have less of an effect on its mass, the same as it did our own. And at zero point it will have no effect on its mass.
It's increase in density will increase air resistance the further you go down the hole. Yet another factor that I feel would prevent you reaching the core. Also the hotter it got, the more updraught would be created like thermals. giving even more reason as to why the air, like us would not make it to the core. Even if you could control the heat at the core, you would struggle to control the heat generated under the air pressure.

 

[mfb]

The same thing that causes a lack of Air in caves or mine shafts.

Can you explain this, please?
I understand a lack of oxygen (as it is consumed by humans, and finite without ventilation). But I never heard of underpressure without sealing and pumping.

It would be like getting a pipe and attaching fans either end blowing inwards.

This pipe will have a higher pressure inside.

A better analogy would be to block both ends of a pipe so no air can escape, and then use a plunger mechanism on both ends to push them together. You will notice it takes tremendous force to do so.

This is a good sign of a high pressure.

The lack of gravity will also make air lighter than at the surface, so even though it got denser it does not mean it would get heavier.

Weight gets lower, density increases. This means that a lot of air is in the center.

 

[poco9964]

Can you explain this, please?

Sorry, I have not explained myself very well. I was trying to explain the convection effect, that heat makes molecules move faster. And that the force of gravity is temporarily over come all the time these molecules are excited enough. When they cool down gravity grabs them again and brings them back down. This effect will actually prevent the air density becoming any denser. The air will reach a certain point where the heat generated from the pressure above it prevents it going any lower due to the convectional heat (Pressure) being generated beneath it. As gravity gets weaker and the air gets hotter, there becomes less chance they could ever meet. I can not do the maths, but I would expect that to occur long before gravity started to drop off. I hope I have made myself a little clearer.

 

[Buckleymanor]

The hole doesn't need to be kilometers wide, so there's no need to worry about that.

I should imagine that the hole would need to be quite a bit wider than a couple of kilometers.
The Earth is tilted and rotates at a tilt so if you made a hole from the north to south poles the hole would have to be wide enougth to acount for the tilt.
Else you would hit the side.

 

[Ken G]

Get rid of the tilt by tilting your head. The "tilt" doesn't mean anything unless you are worrying about the relation to the Sun, and seasons and so on.

 

[Buckleymanor]

Get rid of the tilt by tilting your head. The "tilt" doesn't mean anything unless you are worrying about the relation to the Sun, and seasons and so on.

Well I was worrying that you would hit the side of the tunnel if the hole was not wide enough.Ok. I can tilt my head but that won't stop the Earth traveling around the Sun at a tilt. You and I are traveling with it and if we were to throw ourselfs down a hole dug at an angle or tilt would we not hit the side at some point on our journey.

 

[A.T.]

I can tilt my head but that won't stop the Earth traveling around the Sun at a tilt.

What does the Sun have to do with it? The relative motion of you and Earth is a result of Earth’s gravity.

 

[Buckleymanor]

What does the Sun have to do with it? The relative motion of you and Earth is a result of Earth’s gravity.

Is it not also a result of the Earth's motion around the Sun or can this be ignored.
If you were at a distance from Earth watching the person throw himself down the hole you would see he was moveing with the Earth in it's orbit around the Sun.

 

[Subductionzon]

Perhaps there would be a very small movement due to tidal forces as you fell. I will have to think about the fall and the motion of the barycenter of the Earth and the Sun and that of the Earth and the Moon. Even so they should not have a significant effect. In the evacuated hole you would pass through the Earth in about 40 minutes. Not long enough to worry about a slight change in pull from the Sun or Moon. In the case of the air filled tube aerodynamic effects would keep you from hitting the sides.

 

[Aero51]

Where are you getting this 10 m/s acceleration from? That doesn't make any sense. The acceleration of gravity has to decrease to 0 as you go to the center. In fact, you can check wikipedia where it clearly shows a linear decrease to 0 starting at the Earth's surface.

 

[D H]

Where are you getting this 10 m/s acceleration from? That doesn't make any sense. The acceleration of gravity has to decrease to 0 as you go to the center. In fact, you can check wikipedia where it clearly shows a linear decrease to 0 starting at the Earth's surface.

It clearly does not show that. Look at the solid curve, the one labeled "PREM", on the graph (depicted below). Starting from the left (the center of the Earth), it shows gravitational acceleration as rising more or less linearly, reaching a maximum at the core/mantle boundary. From this maximum, gravitational acceleration initially drops with increasing distance from the center but then flattens and even starts rising again. It drops again in the upper mantle. The value at the surface of the Earth is less than anywhere in the upper or lower mantle.

 

[Aero51]

Opps, I was assuming constant density, which is clearly not true. In either case there is only a small fraction of the core which has a predict acceleration greater than 10 m/s2.

 

[D H]

Opps, I was assuming constant density, which is clearly not true. In either case there is only a small fraction of the core which has a predict acceleration greater than 10 m/s2.

However, almost all of the mantle has acceleration greater than 10 m/s². Using a constant acceleration of 10 m/s² from the surface down to halfway to the center and a linear dropoff to zero from that point inward provides a nice, simple model (with a couple of nice, simple numbers) that happens to closely match the much more detailed PREM with regard to the period of an object falling in an airless tunnel through the Earth.

 

[Ken G]

Well I was worrying that you would hit the side of the tunnel if the hole was not wide enough.Ok. I can tilt my head but that won't stop the Earth traveling around the Sun at a tilt. You and I are traveling with it and if we were to throw ourselfs down a hole dug at an angle or tilt would we not hit the side at some point on our journey.

OK, I see what you are saying. You are worried about the tidal effects of the Sun and Moon gravity on the falling object. We know the scale of that effect by virtue of how far it shifts the equipotentials of the oceans, which I believe is typically perhaps a meter or so. We can estimate a similar scale for the deflection of the falling object-- I think we should expect that a hole much larger than a few meters should not need to worry about hitting the side of the shaft, but your point is well taken that this kind of effect needs to be considered to be sure.

 

[the_emi_guy]

I think the "hitting the sides of the tunnel" would be an issue if the tunnel went straight from equator to equator. During the 45 minute journey, the hole on the opposite side of the Earth has moved over 1000Km. I suppose you could dig the tunnel curved to compensate, but this means we now need a second tunnel, curved the opposite way, for the return trip. This project is getting expensive.

 

[mfb]

Install a maglev rail system inside. The involved forces are small, and it keeps you on track with a precision of a millimeter.

 

[sophiecentaur]

I don't think the following have been injected into this thread yet but (and this is essentially theoretical, of course).
1. For a planet of uniform density, no atmosphere, cold core etc. the oscillation would be simple harmonic motion and its period would be exactly the same as that of a satellite in (grazingly) low orbit. You would emerge from each end of the tunnel just as a satellite passed overhead each time.
2.An experiment that could easily be done (well. . . fairly easily) if you could get hold of a nice spherical asteroid which happened to be made of all granite (ish) rock. You could drill a hole through the centre and then drop a small pebble down the hole. Its period of oscillation would also be about 90 minutes. This is because the period of oscillation is independent of the amplitude of oscillation. (It would be relatively easy to stop it spinning to avoid bumping into the sides). I imagine the same thing with a rock and a grain of sand would fail because of electric forces becoming significant.

 

[poco9964]

Sorry I must be missing something here, let me give a similar scenario.
In a vacuum.
If I had a scaled ball (Earth), with a hole in it, in space, and I dropped a scaled ball (Man) through it. Would it not come out the other side?
Would it inevitable end up in the middle stuck?
If it does not fall out and there is no friction to slow it down, how exactly does it lose the energy put into it each trip by gravity?

My problem is this, I do not think solid doughnut shapes in space collect matter in their centre and begin to fill up, from the inside out.
Space is effectively the only vacuum we could be talking about that has so little resistance to even begin to suggest we could free fall at 17,700 miles/hr. Space contains some gases, but I won't go there. So let's say we accelerate at this speed and neither the decrease in gravity or the decrease to our mass slows our decent to the centre. How does gravity slow us down again given we are traveling at this speed and there is no air resistance at the other end? Does our mass increase slow us down? or does gravity some how apply the breaks to us given enough time?

Basically my biggest issue with this is if there is no resistance, a true vacuum then gravity would oscillate them back and forth for as long as it is present. There would be absolutely nothing to slow this up and down motion. And effectively creating perpetual motion machine. Now I know nobody is going that far, or at least I hope not?

 

[sophiecentaur]

Remember that any spin of the rock and the hole would rapidly clog up with captive stuff. The KE would be largely lost. Also, nothing would ever 'just turn up' in the ideal spot for this experiment to happen on its own. There would be some angular momentum to upset things. You would definitely need to be there to make it work.

 

[mfb]

In a vacuum.
If I had a scaled ball (Earth), with a hole in it, in space, and I dropped a scaled ball (Man) through it. Would it not come out the other side?

Assuming a perfect sphere without rotation and without other experimental issues, it would reach the other side.

Basically my biggest issue with this is if there is no resistance, a true vacuum then gravity would oscillate them back and forth for as long as it is present. There would be absolutely nothing to slow this up and down motion. And effectively creating perpetual motion machine. Now I know nobody is going that far, or at least I hope not?

Right.
Maybe apart from gravitational waves ;).

 

[sophiecentaur]

But this is no more 'perpetual motion' than a simple orbit round the outside of the rock! And a much much more special case.

 

[D H]

An experiment that could easily be done (well. . . fairly easily) if you could get hold of a nice spherical asteroid which happened to be made of all granite (ish) rock. You could drill a hole through the centre and then drop a small pebble down the hole. Its period of oscillation would also be about 90 minutes.

About 120 minutes, not 90. Your 90 minute value is for an object whose average density is almost twice that of granite. Note: For an object with a uniform density equal to the average density of the Earth, the period would be about 84.3 minutes.

 

[sophiecentaur]

Fair enough. but 84 is pretty near 90, ain't it? Presumably granite is not 'average density'?

If the result of the test came out between one and two minutes, I'd be pretty pleased with myself. Well worth the cost of a launch - and the fees for Morgan Freeman, Clint and Tommy Lee Jones, too.

My point was that things scale. I also realized, during my thoughts on this, that there must be loads of small things out there (in Saturn's rings for instance) circling each other at these sort of rates.

 

[poco9964]

Assuming a perfect sphere without rotation and without other experimental issues, it would reach the other side.


Right.
Maybe apart from gravitational waves ;).

Rotation would not be an issue as the tunnel it is in would move relative to the trajectory of the falling object. My question was not weather it would reach the other side.

 

[D H]

Basically my biggest issue with this is if there is no resistance, a true vacuum then gravity would oscillate them back and forth for as long as it is present. There would be absolutely nothing to slow this up and down motion. And effectively creating perpetual motion machine.

That's correct. You've eliminated friction by postulating a pure vacuum. Note well: There is no such thing as a pure vacuum.

There are three classes of perpetual motion machines, classified per the law of thermodynamics that is being violated. Postulating a pure vacuum makes this is a perpetual motion machine of the third kind (ignoring gravity waves, of course). The solar system is very close to constituting such a perpetual motion machine. It's been around for 4.6 billion years, after all. It isn't exactly such a device. The interplanetary medium is not a pure vacuum, and even if it were, there are gravity waves.

 

[phinds]

That's correct. You've eliminated friction by postulating a pure vacuum. Note well: There is no such thing as a pure vacuum.

There are three classes of perpetual motion machines, classified per the law of thermodynamics that is being violated. Postulating a pure vacuum makes this is a perpetual motion machine of the third kind (ignoring gravity waves, of course). The solar system is very close to constituting such a perpetual motion machine. It's been around for 4.6 billion years, after all. It isn't exactly such a device. The interplanetary medium is not a pure vacuum, and even if it were, there are gravity waves.

But I don't see how this could be considered a PMM since no work is being done. I though "true" PMMs (which don't exist) had to do work. No?

 

[AlephZero]

Get rid of the tilt by tilting your head. The "tilt" doesn't mean anything unless you are worrying about the relation to the Sun, and seasons and so on.

Sorry to resurrect this subtopic, but that is wrong. Suppose the the Earth's rotation axis is perpendicular to its orbit (just to simplify visualising the situation) and the hole was drilled between two points on the equator. At some moment in time, the hole is pointing directly at the sun. 6 hours later, it is pointng 90 degrees away from the sun.

If you are oscillating inside the hole, there is nothing that will change your motion so that you won't hit the sides of the hole. Google "Coriolis effect" for the math.

If the hole is from pole to pole, you wpn't hit the sides. It doesn't matter that the Earth is orbiting around the sun, because you will follow the same orbit. Otherwise, satellites in orbit around the Earth would soon be "lost in space".

 

[D H]

If the hole is from pole to pole, you wpn't hit the sides. It doesn't matter that the Earth is orbiting around the sun, because you will follow the same orbit. Otherwise, satellites in orbit around the Earth would soon be "lost in space".

That's a non sequitur. Satellites in orbit around the Earth are perturbed by the Moon and the Sun. Modeling these perturbations is somewhat important even in low Earth orbit, and is very important for geostationary satellites. At geostationary altitudes and beyond, these "third body effects" are larger than are those that arise from the non-spherical nature of the Earth.

There would be no upper limit on the radius of an object orbiting the Earth if the Sun and Moon (and Venus, and everything else) were not present. The presence of the Sun limits the distance at which some object will stably orbit the Earth to about 500,000 to 750,000 kilometers or so (1/3 to 1/2 the Hill sphere radius). Objects orbiting further out than that will eventually be "lost in space".

 

[sophiecentaur]

Sorry to resurrect this subtopic, but that is wrong. Suppose the the Earth's rotation axis is perpendicular to its orbit (just to simplify visualising the situation) and the hole was drilled between two points on the equator. At some moment in time, the hole is pointing directly at the sun. 6 hours later, it is pointng 90 degrees away from the sun.

If you are oscillating inside the hole, there is nothing that will change your motion so that you won't hit the sides of the hole. Google "Coriolis effect" for the math.

I can't picture this in a way that agrees with your conclusion. If you are moving in a straight line (not touching the sides) and the axis of the hole is not parallel to your motion at all times then you will hit the sides. A trans-equatorial hole will not stay parallel to your motion if it follows the rule in the previous para that implies a 24hour rotation about the polar axis.
Which bit have I got wrong? Or did you mean 3 months and not 6 hours later?

 

[Ken G]

I think he was saying that you will indeed hit the sides. But the description is unclear, because we have all assumed the hole likes along the rotation axis of the Earth, so we are certainly not drilling a hole from equator to equator.

However, the point remains that since the rotation axis is tilted, there is a coriolis force due to Earth's revolution about the Sun. The best way to picture that is probably to picture Uranus, and imagine we have a 90 degree tilt and the hole lies along the rotation axis. That sounds like the situation AlephZero is really talking about. As the object falls, its distance to the Sun changes, so the object is, in effect, in a slightly elliptical orbit around the Sun. That could certainly cause it to hit the walls if they were not extremely broad, as Buckleymanor pointed out. I felt the scale of the effect would be small, based on the tidal forces, but that forgets about the coriolis force. So I think you're right-- we must not have a Sun-orbiting Earth in the no-air-resistance case. If there's air resistance, and we get a slow terminal speed, then the air should carry the object along and the coriolis effect will be much smaller (the deflection scales like 1/v over the time to fall, where v is the terminal speed).

 

[sophiecentaur]

In a few 90minute cycles, we could establish the principle, though, (a short period compared with 365 days) and just put up with the fact that practical details would put an end to the 'perpetuity' of our experiment.

 

[Buckleymanor]

I felt the scale of the effect would be small, based on the tidal forces, but that forgets about the coriolis force.

I am not sure that the coriolis force would have much effect having looked at it again.Fire a missile at a target and the Earth rotates benieth the moveing mass, the consiquence is it veares to one side of the target.
Gravity above a rotating sphere don't seem to have a sideways effect upon a moveing mass unless say a huge mountain protruded but most these experiments are done at sea.
Your man falling down a hole has an equal amount of mass more or less surrounding him in a close proximity, won't the proximity of the mass have a stabiliseing effect upon the falling man and keep him firmly in the middle of the hole as he falls.He is not moveing above the mass but through the middle of it.

 

[D H]

I am not sure that the coriolis force would have much effect having looked at it again.

Of course it will. The solution is simple: Make the tunnel walls frictionless. We have a magic tunnel already, what's the deal with adding yet another magical quality?

won't the proximity of the mass have a stabiliseing effect upon the falling man and keep him firmly in the middle of the hole as he falls.

No, it won't.

The coriolis force is an artifact from working in a rotating frame of reference. Transfer to an inertial frame and you'll get the same answer as you would doing the calculation in a rotating frame; it's just going to be a bit messier getting to that answer.