A hypothetical question about gravity

[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.

 

[Buckleymanor]

No, it won't.

Maybe I have not explained well enough or I am wrong.If you took a cross section of the man falling on his decent he would be surrounded by an equal amount of mass which to all extents would make him weightless, similar to being in a spherical shell.The difference being that he would have gravitational pull of mass pointing up towards him from below the hole and vice versa from above.Which would keep him in the middle.Because gravity is inversly proportional would any outside effects be mostly marginal.

 

[Buckleymanor]

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?



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.

Realized a plumb bob is a good example.No matter where it's placed on Earth it points in a straight line towards the centre of the Earth and is unperturbed by coriolis, cut the string and if placed over your hole it will travell in a straight line towards the centre, local gravity is a lot stronger.

 

[Ken G]

But there's also the centrifugal force, so a plumb bob does not point directly at Earth's center and an object wouldn't fall to the center, just because of the spin of the Earth, unless the hole was cut through the rotation axis. The role of the orbit around the Sun is even trickier-- there the centrifugal effect is of order the Sun's tidal effect, so I think that would be small (the scale of tides is a meter or less, on average), but the coriolis effect could be large.

I said above that the coriolis deflection scales like 1/v, where v is the falling speed in the rotating frame, which comes from the idea that the acceleration is proportional to v, and the deflection scales like the acceleration times the time squared, while the time scales like 1/v. For some inexplicable reason, I concluded that the coriolis effect wouldn't matter if air resistance slowed the fall down, because then the air would carry the object along with it, but if air resistance makes v small, then 1/v gets large. So air resistance could make the coriolis deflection even worse, it depends on some additional details.

In truth it had been bothering me to discount the coriolis effect if the falling was slow, because usually the coriolis effect is most prominent for motions that occur on a timescale that is comparable to the rotation period of the frame. Motions that happen much more quickly (like draining a bathtub) cannot be affected by the coriolis effect, and motions that happen much more slowly (like the settling of metals to the Earth core) must average out the coriolis influences. So I think we should be expect maximum coriolis deflection for processes that require about a year for the man to fall, and the deflection would be of order the Earth radus. If we have no air resistance, the free fall time is of order the radius over the escape speed, so that's roughly 500 seconds, way shorter than a year. Since the deflection scales like acceleration times the time squared, and the coriolis acceleration scales like the inverse of the falling time, the deflection scales like the falling time. The ratio of the falling time, with no air resistance, to a year is about 10^-5, so to order of magnitude, the coriolis deflection would be about 10^-5 times the Earth radius, or say 10-100 meters.

However, if the air resistance was cranked up until the fall time was about a year, the deflection would be much more. But, if the air resistance was so great that the fall time was way more than a year, the effects of the Earth's orbit would cancel out, and we'd be back to a small effect. If the fall speed was the terminal speed in air now, say 100 m/s, then the fall time is on order a million seconds, which is like a few weeks or so, so that would produce a pretty large coriolis deflection indeed, maybe several to ten percent of the Earth radius, though a full calculation is needed to be more accurate. If the air gets even denser, it would eventually become too thick to allow any coriolis deflection, and the effect magnitude would drop again.

So, as usual, when you dig deeper into a seemingly simple problem, and start layering on all the complexities that a real-world calculation would need, you get surprised! Of course, the problem is never going to be a "real world" problem in the first place, but you were certainly right to bring up the possibility of not falling straight through the hole, even if it is drilled through the rotation axis.