Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: thedoc on 02/11/2016 17:23:01
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Sreejith AJ asked the Naked Scientists:
If I were to drill a hole through the earth all the way to the other side and jumped into it, assuming I had enough protection against the heat and the pressure, would i make it to the other side? Assuming I do, how would my exit/entry look like?-Sreejith
What do you think?
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Neglecting air resistance (and other impracticalities), you would oscillate back-and-forth between the two sides of the planet with simple harmonic motion. You would accelerate downwards from one surface, reach a velocity maximum in the region of the Earth's centre and then decelerate to a standstill at the opposite surface.
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Sreejith AJ asked the Naked Scientists:
If I were to drill a hole through the earth all the way to the other side and jumped into it, assuming I had enough protection against the heat and the pressure, would i make it to the other side?
If there is air in the hole you'll barely pass the center of the Earth.
If the hole is a vacuum and you ricochet off the walls of the tunnel, (which you inevitably would because of Earth's rotation), then you won't get close to the exit either.
Come to think on it, if your capsule/craft is metal, even non-ferrous, the Earth's magnetic field will act as a brake ...
https://www.youtube.com/watch?v=JN-A3RryOC8 (https://www.youtube.com/watch?v=JN-A3RryOC8)
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Neglecting air resistance (and other impracticalities), you would oscillate back-and-forth between the two sides of the planet with simple harmonic motion. You would accelerate downwards from one surface, reach a velocity maximum in the region of the Earth's centre and then decelerate to a standstill at the opposite surface.
That's theoretically true - if a perfectly symmetrical spherical object, such a ball-bearing of finest precision, were dropped into a hole drilled through the centre of a perfectly symmetrical planet. The ball-bearing would oscillate endlessly back and forth.
However, the OP posits that "you", ie presumably a human being, jumps into hole drilled through "Earth".
Earth and humans aren't perfectly symmetrical. Humans have more mass in their head than in their feet, and Earth has more land-mass in the Northern Hemisphere than in the Southern.
Wouldn't these asymmetries affect the oscillations, until the OP's human eventually stepped out of the hole, probably at the South Pole end?
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Neglecting air resistance (and other impracticalities), you would oscillate back-and-forth between the two sides of the planet with simple harmonic motion. You would accelerate downwards from one surface, reach a velocity maximum in the region of the Earth's centre and then decelerate to a standstill at the opposite surface.
That's theoretically true - if a perfectly symmetrical spherical object, such a ball-bearing of finest precision, were dropped into a hole drilled through the centre of a perfectly symmetrical planet. The ball-bearing would oscillate endlessly back and forth.
However, the OP posits that "you", ie presumably a human being, jumps into hole drilled through "Earth".
Earth and humans aren't perfectly symmetrical. Humans have more mass in their head than in their feet, and Earth has more land-mass in the Northern Hemisphere than in the Southern.
Wouldn't these asymmetries affect the oscillations, until the OP's human eventually stepped out of the hole, probably at the South Pole end?
The crust is pretty thin compared to the Earth as a whole. If you average out the oceanic and continental crust thickness, it averages out to ~13.3 km. It is also less dense (2.2 g/cc) than the Earth as a whole 5.52 g/cc. In the end it works out that the entire crust amounts to about 1/400 the mass of the entire Earth. And while the continental crust is quite a bit thicker than the oceanic crust, you also have to remember that it is floating on the mantel, so it extends deeper into the mantel than the oceanic crust does, displacing some of the denser mantel. If fact, it displaces exactly the same mass of mantel as the mass of the continent crust floating there. The continental landmass does not add to the total mass in that part of the Earth. This isn't to say that the mass distribution is totally homogeneous, but the effect of these variations will be extremely small, and likely swamped out by other factors ( the oblateness of the Earth for one.).
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The point might be, we only guess on what's 'inside'
And you will bungy jump, beware of spin
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If we really want to find out "what's inside" like yor-on says, why don't we drill a really deep hole into the Earth.
We weren't able to do such a thing in the past, because we didn't have the technology. But now we've got powerful modern machines which could bore through the Earth's crust and find out what's actually down there. Perhaps 20 or 30 kilometers down, there might be vast reserves of planetary methane gas, as some theorists have speculated. This would supply our energy needs for centuries.
Couldn't we at least drill an exploratory hole, to find out. Or isn't that allowed these days?
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bore through the Earth's crust and find out what's actually down there
There was an attempt to do this, in the 1960s: The Mohole Project.
They made the task easier by drilling in deep ocean, where the crust is thinner.
See: https://en.wikipedia.org/wiki/Project_Mohole
Or you could just go to a place where the layer beneath Earth's crust comes up to the surface, as it does in Iceland.
Iceland is a part of the mid-Atlantic ridge that happens to be above sea-level. New crust is actively forming there.
Perhaps 20 or 30 kilometers down, there might be vast reserves of planetary methane gas, as some theorists have speculated.
Technically, drilling in deep water is quite difficult.
And natural gas behaves quite differently under extreme pressures.
In the movie "Deep Horizon", they illustrated, but didn't attempt explain that under extreme pressures, water and methane form a clathrate "snow" that gummed up all their pipes, and prevented their safety equipment from operating.
Indeed, it is thought that there are considerable deposits of clathrates around the world, which could potentially be mined for natural gas (provided that oceanic warming doesn't melt them first).
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"would oscillate endlessly back and forth"
Sorry, no perpetual motion devices allowed, but if in a perfect vacuum quite a while.
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"would oscillate endlessly back and forth"
Sorry, no perpetual motion devices allowed, but if in a perfect vacuum quite a while.
Assuming you could eliminate all other forms of friction, quite a long, long while. Then the only thing left to dampen the oscillations would energy loss through gravitational radiation, and this would take on the order of 3e39 years or 2e29 times the present age of the universe.
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"would oscillate endlessly back and forth"
Sorry, no perpetual motion devices allowed, but if in a perfect vacuum quite a while.
That's a interesting suggestion, but surely perpetual motion is fully allowed by Newton's First Law: "Every object remains at rest, or in uniform motion in a straight line, unless acted on by an exterior force."
And even suppose we discard Newtonian physics as obsolete. And replace it by Einsteinian physics, which says that an object naturally follows the geodesic path of curved space. Won't the object continue to follow it perpetually. Doesn't this show that no "device" is needed for perpetual motion. Such motion is natural and intrinsic. Isn't what you need, rather, an exterior "device" to stop the motion?
Janus, in the post above, gives insightful examples of such practical "stopping" influences. But is there any theoretical objection to "perpetual oscillation"?
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You would have to time your exit at the antipodes carefully or you might well start another trip down if you wanted to see what it is like in the middle you would have to look quick as you would be moving at Orbital speed
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You would have to time your exit at the antipodes carefully or you might well start another trip down if you wanted to see what it is like in the middle you would have to look quick as you would be moving at 11.6Km/h
You might want to check your maths and units there. I can run at 12 km/hr adn still watch the scenery go by.
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Yes, no wonder Mars Climate Orbiter crashed. The guys at NASA couldn't make out whether it was moving at 11.6Km/h, or was it 11.6Km/s, or what?
I entirely empathise with NASA. A speed such as 100 or 750 miles per hour, I can grasp immediately, but the metric stuff is not readily comprehensible.
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You would have to time your exit at the antipodes carefully or you might well start another trip down if you wanted to see what it is like in the middle you would have to look quick as you would be moving at 11.6Km/h
Assuming a uniform density for the Earth, the answer comes out to 7.9 km/sec at the center. Some of you might recognize that value as it turns out to be equal to the orbital velocity at the Surface of the Earth. It also turn out that it would take an object dropped into the hole ~84.48 mins to complete a round trip back to where you dropped it. This also equal to the period of a orbit at the Earth's surface. In other words, if you cleared all obstacles( including the air) from its path an object put in orbit just at the Earth surface, would meet up with your dropped object on the other side and then again back where they started. When the object reaches the center of the Earth, the orbiting object will have completed 1/4 of an orbit and for that moment they will have the same velocity.
In fact, if you imagine an imaginary line at a right angle to the hole that follows the dropped object, it will intersect with the orbiting object at all times. (Assuming they both start at one end of the hole at the same time.)
This give us a way to determine how far the dropped object has fallen at any time after release.
This will work out to be ~ (1-cos(0.00124t))R
Where t is the time in seconds, R the radius of the Earth and the cosine is in radians.
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You ever heard abouts clocks synchronizing their 'beats' Janus?
I find it fascinating :)
http://discovermagazine.com/2000/jul/featmath
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The point might be, that I've seen this before
Doesn't make it less
then again, I need more
So, what do you offer?
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Assuming a uniform density for the Earth, the answer comes out to 7.9 km/sec at the center. Some of you might recognize that value as it turns out to be equal to the orbital velocity at the Surface of the Earth. It also turn out that it would take an object dropped into the hole ~84.48 mins to complete a round trip back to where you dropped it. This also equal to the period of a orbit at the Earth's surface. In other words, if you cleared all obstacles (including the air) from its path an object put in orbit just at the Earth surface, would meet up with your dropped object on the other side and then again back where they started. When the object reaches the center of the Earth, the orbiting object will have completed 1/4 of an orbit and for that moment they will have the same velocity.
In fact, if you imagine an imaginary line at a right angle to the hole that follows the dropped object, it will intersect with the orbiting object at all times. (Assuming they both start at one end of the hole at the same time.)
As we have been discussing orbits:
https://www.thenakedscientists.com/forum/index.php?topic=66442.msg526356#msg526356
and some years ago I learnt what quoted, I was planning to bring it up there. I even brought up the case of an object orbiting at zero altitude, and had said something "touching" the physical reasons of what quoted,
But minutes ago I saw this thread about that phenomenon.
I agree with you ... Only one "but": it must be just a lapsus, but your last paragraph is not clear. Surely you mean " ... an imaginary line, drawn at dropped object position, at a right angle to the hole that follows the dropped object ... "
I learnt that from a Physicist, a lady, through BBC radio. She did´t explained why that happens (rather complex for the program). I ruminated about it, and, as previously said, I managed to grasp its physical (and mathematical) reasons. Did you?
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In other words, if you cleared all obstacles( including the air) from its path an object put in orbit just at the Earth surface, would meet up with your dropped object on the other side and then again back where they started. When the object reaches the center of the Earth, the orbiting object will have completed 1/4 of an orbit and for that moment they will have the same velocity.
I learnt that from a Physicist, a lady, through BBC radio. She did´t explained why that happens (rather complex for the program). I ruminated about it, and, as previously said, I managed to grasp its physical (and mathematical) reasons. Have you?
I was waiting for a reply from Janus, or somebody else ... As this is an old thread, perhaps almost nobody have seen what I posted three days ago. So, I´ll send those reasons now. Hope you all will agree !
For the sake of an easier explanation, let us imagine Earth´s 90º/270º meridian section "(back) on a drawing board", and "x,y" ortogonal coordinates´s system axis, x on equator plan, and y from pole to pole, same line of drilled shaft.
Let us drop the object into N pole "hole". It will start an accelerated southward movement. For any position, Earth can be imaginary divided into two spherical segments: one in which all its mass is ahead of the object movement, and the other back. The ahead one will always pull to accelerate the object, and the other to decelerate it. In both cases proportionally (the force) to the mass of each segment, and inversely to the square of the distance to the segment C.G.
Not necessary to make any calculation to understand that, supposing no friction energy lost, the object will get maximum speed at Earth´s center, will reach S pole at zero speed, and then start a S —> N "mirror" movement ...
Let us now consider same object at N pole, and thrown eastward tangentially, perpendicular to N-S axis and with the necessary zero altitud orbiting speed. The object will always be pulled by the whole Earth, in the direction of its C.G.: the centripetal force. Velocity and position vectors will always be the result of applying basic laws to those magnitude vectors (initial position, initial speed, constant centripetal force), and object mass.
But as I previously commented to Janus:
"I hope now i´ve really found the error … But not in what you say: the error was mine !
Basic laws I was referring to are about vectors (force, acceleration, speed, linear momentum, impulse …). We can applied them separately to their components along perpendicular axis, typical x,y axis of coordinates (if within a plane).
But to apply them to components such as radial (circular movement for simplification) and tangential is erroneous, because their actual directions vary with time …"
(How do objects orbit in space?, post on 27/10/2017 in https://www.thenakedscientists.com/forum/index.php?topic=66442.msg526180#msg526180)
all those movement features can be analyzed independently for "x" and "y" components of each vector …
At any position of the orbiting object, "Y" COMPONENT of all those vectors will follow same basic laws, with exactly same initial and later figures as in the falling object case !!
We can say that in "y" sense orbiting object oscilates exactly as the linear movement of the falling object ...
Similarly, in "x" sense, ahead Earth segment will always pull to accelerate the object, and back one to decelerate it, also proportionally (the force) to the mass of each segment, and inversely to the square of "X" COMPONENT of the distance to the part C.G.: orbiting object will move IN "X" DIRECTION exactly as if we had ANOTHER SHAFT, along "x" axis, and we had dropped the object 90º before the N <---> S one (at 270º longitude).
If Earth were a perfect sphere, and isotropic, we could say orbiting object dynamic scenario (forces, accelerations, speeds, positions) would ALWAYS be the addition of the two scenarios constituted by the "x" and "y" components of all those vectors … because same fundamental laws apply both to all those two-dimension vectors, and independently to their "x" and "y" one-dimension components.