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# Throwing Stuff Down A Mineshaft

Author Topic:   Throwing Stuff Down A Mineshaft
onifre
Member (Idle past 1457 days)
Posts: 4854
From: Dark Side of the Moon
Joined: 02-20-2008

 Message 46 of 69 (493431) 01-08-2009 7:48 PM Reply to: Message 10 by cavediver01-08-2009 9:21 AM

Re: Thought experiment
 The acceleration decreases all the way to the centre, where it is zero. As you overshoot, the acceleration grows again (linearly with depth) but in the opposite direction.

So this means you will land at the surface on the other end...?

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subbie
Member (Idle past 250 days)
Posts: 3509
Joined: 02-26-2006

 Message 47 of 69 (493462) 01-08-2009 11:35 PM

I have a couple of questions, generally, about a few things generally discussed in this thread.

1. What effect would the rotation of the Earth have on the falling goat? Wouldn't the poor thing keep banging on the sides of the hole all the way down, then all the way back up, then all the way back down again, then all the way back up again again, then . . . . (making myself dizzy so stopping now).

2. Re Dyson spheres: I'm confused about whether there's no gravity at every point inside the sphere. Gravity operates according to the inverse square law, no? Therefore, it seems to me that if you are closer to one side than the other that the gravitational pull from the closer side would be stronger and, thus, draw you to that side. I'm quite sure that people who know lots more about physics than I do have worked this out and I'm just wrong, but if someone could explain it to me, I'd appreciate it.

Those who would sacrifice an essential liberty for a temporary security will lose both, and deserve neither. -- Benjamin Franklin

We see monsters where science shows us windmills. -- Phat

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lyx2no
Member (Idle past 3222 days)
Posts: 1277
From: A vast, undifferentiated plane.
Joined: 02-28-2008

 Message 48 of 69 (493468) 01-09-2009 12:17 AM

Terminal velocity
Would not our terminal velocity drop to near nothing as the gravitational force decreases and the air density increases? We'd float gently down to meet the goat in the center, our oscillations damped to insignificance.

I couldn't begin to figure out what the density of 6.4 [6] meters depth of air would be in a constant G field yet alone a declining one.

([n]=10n)

Genesis 2
17 But of the ponderosa pine, thou shalt not eat of it; for in the day that thou eatest thereof thou shalt sorely learn of thy nakedness.
18 And we all live happily ever after.

Rrhain
Member (Idle past 378 days)
Posts: 6349
From: San Diego, CA, USA
Joined: 05-03-2003

 Message 49 of 69 (493479) 01-09-2009 2:56 AM Reply to: Message 47 by subbie01-08-2009 11:35 PM

subbie writes:

quote:
I'm confused about whether there's no gravity at every point inside the sphere.

Indeed, assuming that there is no other gravitational field anywhere else in the universe, there is no gravitation pull inside a perfectly uniform, spherical shell.

quote:
Gravity operates according to the inverse square law, no? Therefore, it seems to me that if you are closer to one side than the other that the gravitational pull from the closer side would be stronger and, thus, draw you to that side.

Yes, but you need to take into account that as you get closer to one side of the sphere, you have a lot more mass behind you. Because the shell is spherical, it exactly balances out: The gravitational pull you feel in one direction is perfectly balanced by the gravitational pull in the opposite direction.

That is, suppose you have a shell with an internal diameter of 8000 km and a thickness of 1 km.

Suppose you are standing on a point on the inside of the shell. How much mass is "beneath" you? You can visualize a plane spreading out tangent to the point of contact that lops off a thin shaving of the sphere...kinda like what happens when they do LASIK on you.

But that means the entire mass of the rest of the sphere is "above" you and pulling in the opposite direction. Due to the geometry of the sphere, it just so happens that the gravitational pull of the small-but-close mass is exactly equal to the gravitational pull of the large-but-far mass.

This is true for all points inside the sphere: The pull on one side exactly counters the pull on the other.

Now, if you have a different shape such as a cube, then there will be gravitational field gradient inside.

Now, you could fake a gravitational pull by rotating the sphere. Along the equator in relation to the axis of rotation, you'll have the largest acceleration, diminishing to zero at the poles.

Rrhain

Thank you for your submission to Science. Your paper was reviewed by a jury of seventh graders so that they could look for balance and to allow them to make up their own minds. We are sorry to say that they found your paper "bogus," specifically describing the section on the laboratory work "boring." We regret that we will be unable to publish your work at this time.

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Huntard
Member (Idle past 801 days)
Posts: 2870
From: Limburg, The Netherlands
Joined: 09-02-2008

 Message 50 of 69 (493498) 01-09-2009 7:20 AM Reply to: Message 46 by onifre01-08-2009 7:48 PM

Re: Thought experiment
 Onifre writes:So this means you will land at the surface on the other end...?

No. At the other end you velocity is 0, you will be pulled back into the hole and return to where you started, then back again, and again, and again.... You get the picture. Of course, only if you are in a vacuum. When there is air ressistnace and friction, you will eventually stop in the center.

I hunt for the truth

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Stile
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 Message 51 of 69 (493500) 01-09-2009 7:26 AM Reply to: Message 46 by onifre01-08-2009 7:48 PM

Basically.... yes, precisely... no.
 onifre writes:So this means you will land at the surface on the other end...?

Disregarding all energy loses due to friction/air resistance... yes.

Well, not exactly, you wouldn't really "land". As in, you wouldn't shoot out the end of the hole, move to your side, and then come back down and "land" on the ground. With no lateral force, there would be nothing to move you away from being right over the hole.

If you dropped into a hole that went through the centre of the earth (disregarding air resistance), you would end up popping out the other end (feet first, so you'd feel "upsidedown") exactly the same height beyond the surface as you started... this "height" is measured from your centre of mass... not the top of your head or bottom of your feet or something like that. And then you'd start your drop all over again (head first this time) back to the other end and you'd end up hovering just above the hole, with your centre of mass at the exact same position you started in. This would repeat over and over again forever.

If the hole was sized just perfectly enough to fit you, and say you could spread your legs so that you didn't fall into it, and then brought your legs together to "jump in", you'd get through on the other side enough to spread your arms out on the surface and stop yourself (remember you'd be upsidedown at this point). Or you could not do anything, drop back in head-first, and you'd be able to regain your original position and spread your legs to stop yourself from doing another full oscillation.

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Stile
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 Message 52 of 69 (493502) 01-09-2009 7:40 AM Reply to: Message 47 by subbie01-08-2009 11:35 PM

 subbie writes:1. What effect would the rotation of the Earth have on the falling goat? Wouldn't the poor thing keep banging on the sides of the hole all the way down, then all the way back up, then all the way back down again, then all the way back up again again, then . . . .

I'm not positive about this, but I think that the rotation of the Earth wouldn't have any effect. That is, the goat is on the Earth already (that's where it jumps from to get into the hole). So that would mean the goat has all the rotational momentum and motion of the Earth. Now, the goat can't really run up to the hole and jump in... it would eventually hit the far side with it's own additional momentum. It would really have to straddle the hole (or be held over it somehow) and drop into it so that it can't reach the sides.

I do know that if the goat hits the sides on the way down, each hit will lose a bunch of energy... this will ensure that the goat won't make it all the way back up the other side. The poor banging goat will still oscillate a bit, but it would be similar to air resistance and the goat would eventually end up at the centre of the earth.

 2. Re Dyson spheres: I'm confused about whether there's no gravity at every point inside the sphere. Gravity operates according to the inverse square law, no? Therefore, it seems to me that if you are closer to one side than the other that the gravitational pull from the closer side would be stronger and, thus, draw you to that side. I'm quite sure that people who know lots more about physics than I do have worked this out and I'm just wrong, but if someone could explain it to me, I'd appreciate it.

Rrahin's explanation is much better and more in-depth. I'm just going to say that you're half-right. Stuff that is closer to you most certainly does pull you stronger. The part that you're half-wrong is that in a perfect sphere, if you're closer to one side, there's less "stuff" pulling stronger on you, and there's more "stuff" pulling weaker on you. The properties of a perfect sphere are such that these effects perfectly balance... lots of weak stuff = less strong stuff. Therefore you go nowhere.

Of course, this only works in an isolated, perfect sphere. In a non-perfect sphere, or a different shape, or a perfect sphere with something else on one side of it... you'll be pulled towards those imperfections depending on how significant they are.

Added by Edit: Oh... if anyone's wondering, this "Dyson" sphere stuff is a perfect sphere with another perfectly spherical cavity inside it, that is located at the exact centre.

Any deviation from that, like if you have a perfect sphere cavity in another perfect sphere, but the cavity is off-set to one side somewhat... would not produce the same effect. You would then be drawn towards the imperfection somewhat.

Edited by Stile, : Added Dyson Sphere explanation

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cavediver
Member (Idle past 2149 days)
Posts: 4129
From: UK
Joined: 06-16-2005

 Message 53 of 69 (493505) 01-09-2009 7:50 AM Reply to: Message 49 by Rrhain01-09-2009 2:56 AM

 Indeed, assuming that there is no other gravitational field anywhere else in the universe, there is no gravitation pull inside a perfectly uniform, spherical shell.

Relative to the shell, it doesn't matter what the external gravitational field is. The shell will respond to that field identically to the shell occupant, and again the occupant will be entirely weightless within the shell.

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elianna
Suspended Member (Idle past 4063 days)
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Joined: 01-06-2009

 Message 54 of 69 (493514) 01-09-2009 8:45 AM

The heart of the matter
A man who was having heart trouble went to the doctor to see what his options were. Naturally, the doctor recommended a heart transplant. The man reluctantly agreed, and asked if there were any hearts immediately available, considering that money was no object.

"I do have three hearts," said the doctor. "The first is from an 18-year old kid, non-smoker, athletic, swimmer, with a great diet. He hit his head on the swimming pool and died. It's \$100,000. The second is from a marathon runner, 25 years old, great condition, very strong. He got hit by a bus. It's \$150,000. The third is from a heavy drinker, cigar smoker, steak lover. It's \$500,000."

"Hey, why is that heart so expensive? He lived a terrible life!"

"Yes, but it's from a laywer. It's never been used."

http://reseller.name

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Huntard
Member (Idle past 801 days)
Posts: 2870
From: Limburg, The Netherlands
Joined: 09-02-2008

 Message 55 of 69 (493515) 01-09-2009 8:50 AM Reply to: Message 54 by elianna01-09-2009 8:45 AM

Re: The heart of the matter
Hello Elianna.

Though I appreciate the jokes, there is a thread for this here. That would be the place to post these. :)

I hunt for the truth

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Posts: 4754
Joined: 11-11-2003

 Message 56 of 69 (493532) 01-09-2009 9:45 AM Reply to: Message 54 by elianna01-09-2009 8:45 AM

Warning for Elianna
It almost seems that you are a spammer.

You link is broken so I going to give you the benefit of the doubt.

One more similar post though and you will be banned permanently.

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subbie
Member (Idle past 250 days)
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Joined: 02-26-2006

 Message 57 of 69 (493535) 01-09-2009 10:11 AM Reply to: Message 52 by Stile01-09-2009 7:40 AM

quote:
I'm not positive about this, but I think that the rotation of the Earth wouldn't have any effect. That is, the goat is on the Earth already (that's where it jumps from to get into the hole). So that would mean the goat has all the rotational momentum and motion of the Earth.

Well, here's where I'm running into trouble. Assuming the hole is at the equator, when the goat first goes in, he has a speed of appx. 1,000 miles an hour as a result of the Earth's rotation. However, as he falls, the relative speed of the Earth will drop but his won't. Thus, as I see it, he'll eventually bounce off the side. And, after the first bounce, there's nothing to keep him from bouncing off the other side, and so on.

Those who would sacrifice an essential liberty for a temporary security will lose both, and deserve neither. -- Benjamin Franklin

We see monsters where science shows us windmills. -- Phat

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Stile
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 Message 58 of 69 (493539) 01-09-2009 10:31 AM Reply to: Message 57 by subbie01-09-2009 10:11 AM

 subbie writes:Assuming the hole is at the equator, when the goat first goes in, he has a speed of appx. 1,000 miles an hour as a result of the Earth's rotation. However, as he falls, the relative speed of the Earth will drop but his won't. Thus, as I see it, he'll eventually bounce off the side. And, after the first bounce, there's nothing to keep him from bouncing off the other side, and so on.

Well, I have to admit I don't really know what's going on here anyway. And even when I did think I knew what was going on I got parts wrong (like the inside of the perfect sphere).

But... this is what I think (for whatever it's worth).

I agree with you that if the goat bounces of a side, he's likely to bounce of multiple sides and the shit goes to hell (..literally? :)).

However, I'm still not sure that the goat hits the side at all. Rotational speed is strange... it depends on how far you are from the centre. Yes, the relative speed of the Earth will drop (since we're getting close to the centre... each specific area of the Earth will be moving slower and slower). However, I'm not convinced that the relative rotational speed of the goat stays constant. As the goat moves closer to the centre of the Earth, I'm guessing that the goat's rotational velocity reduces equally with respect to how close the goat is to the centre of the earth.

Now, I may just be flat wrong about that and I don't have any hard facts to back myself up with this one, and I hope someone more knowledgeable chimes in to set us straight. But for now, I'm standing on the side that the goat's rotational velocity reduces itself as the poor fella falls so that the goat never touches the sides anyway. That is, neglecting air resistance and stuff like that. Long live the free falling goat!

I suppose my real answer is: *shrug* :(

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cavediver
Member (Idle past 2149 days)
Posts: 4129
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 Message 59 of 69 (493540) 01-09-2009 10:32 AM Reply to: Message 57 by subbie01-09-2009 10:11 AM

 Thus, as I see it, he'll eventually bounce off the side.

Probably not a bounce as it is a very smooth process. You will end up against one side of the tunnel which could introduce some nasty friction burns. Your contact will decelerate your rotational motion as you drop towards the centre and then re-accelerate you back to full rotational speed back on the other side.

Or just build the tunnel between the poles...

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lyx2no
Member (Idle past 3222 days)
Posts: 1277
From: A vast, undifferentiated plane.
Joined: 02-28-2008

 Message 60 of 69 (493564) 01-09-2009 1:32 PM Reply to: Message 59 by cavediver01-09-2009 10:32 AM

 Or just build the tunnel between the poles...

Two guys were hiking in Antarctica when they came across an old mine shaft going straight down into the ground.

Genesis 2
17 But of the ponderosa pine, thou shalt not eat of it; for in the day that thou eatest thereof thou shalt sorely learn of thy nakedness.
18 And we all live happily ever after.

 This message is a reply to: Message 59 by cavediver, posted 01-09-2009 10:32 AM cavediver has not yet responded

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