Melt.

Try this link

http://www.straightdope.com/classics/a1_165.html

http://www.straightdope.com/classics/a1_165.html

Playintennis5274 wrote:

Eventually, it would hit the wall of the hole.

Same as if you dropped a loaf of bread.

Eventually, it would hit the wall of the hole.

Same as if you dropped a loaf of bread.

stop in the middle, no gravity

You'd go just a bit past the middle if you take into account non-vacuum:

"What we are dealing with here is essentially a cosmic pendulum. But let me explain a few basic concepts first. For one thing, we are going to disregard the problem of the earth's molten interior. There is no surer way to ruin a good discussion than to contaminate it with the facts.

Next, you must force yourself to accept the following notion: if you were somehow teleported to a cave in the center of the earth, you would find that you were weightless. This is because you would have approximately equal amounts of mass on all sides of you, which would cancel each other out.

Now then. If you jumped into a frictionless (and consequently airless) interpolar tube, you'd fall, obviously, picking up momentum as you went. As you approached the center of the earth the pull of gravity would decline and eventually (at the center) cease, but inertia would keep you going.

Once past center, though, the pull of the earth's mass behind you would begin to slow you down, at exactly the opposite rate that you'd accelerated. You'd come to a complete stop just at the brink of the Antarctic end of the tube, where you'd have an opportunity to wave gaily to the bunny rabbits or whatever they have out there before beginning to fall back in the opposite direciton. This process would continue forever.

Once we start figuring for the effects of atmospheric friction, of course, the situation changes. After a certain point in the course of falling you'd reach a top speed called "terminal velocity," where air resistance would counteract the accelerating effects of gravity. With less momentum, you'd only fall a relatively short distance past the center of the earth before you stopped and started heading in the other direction. Eventually you'd reach equilibrium at the earth's center.

I was going to calculate how long this would take, but twenty minutes of computation has produced no useful result and you didn't ask anyway. But watch where you're going."

--CECIL ADAMS (from website noted in prior answer)

Well Roger Prove it , Prove I am wrong, You cant So ????

Roger wrote:

<snipped>

Which prompts me to ask; Assuming the hole is open at both ends and there's a constant temperature all along the length of the hole, what would the plot of atmospheric pressure vs. distance from the surface look like?

I can't easily visualize what it would be.

Jeff

<snipped>

<snipped>

Which prompts me to ask; Assuming the hole is open at both ends and there's a constant temperature all along the length of the hole, what would the plot of atmospheric pressure vs. distance from the surface look like?

I can't easily visualize what it would be.

Jeff

<snipped>

--

Jeffry Wisnia

(W1BSV + Brass Rat '57 EE)

Jeffry Wisnia

(W1BSV + Brass Rat '57 EE)

Click to see the full signature.

This is Turtle.

It would depend on the possion of the Moon as to on what side of the earth it was on.

If the Moon was on the same side of the earth as where you dropped the ball in at. The Ball would stop before it got to the center of the earth.

If the moon was on the other side of the hole/ earth , when you dropped it in at. the ball would go just past the center of the earth before it stopped.

I might suggest you use a Trillium ball for the heat at the center of the earth is pretty warm.

TURTLE

TURTLE strikes again, typos and all!!!

down

it

ball in

in

earth

down

it

ball in

in

earth

At a certain depth which I forget how to calculate, it would go into synchronous orbit around the center of the earths mass.

--

Larry Wasserman Baltimore, Maryland

Larry Wasserman Baltimore, Maryland

Click to see the full signature.

down

Correct me if I'm wrong - ball would pick up a spin which would more than overcome air resistance and propel it at a higher speed out the hole on the opposite side of earth, at a greater velocity than the ball initially entered the hole. The balls mass would determine the height it would rise above earth's surface and as the ball no longer has a cylinder forcing it in a straight path the earth's rotation would cause the ball, once it falls, to land beside the hole. Plop.

Playintennis5274 wrote:

Assuming a spherically symmetric distribution of mass in the Earth, the gravitational force ("F") exerted on the ball is proportionate to and opposite in direction from its displacement ( "x") from the center of the earth, i.e., the phenomenon conforms to the following eqn:

F = -kx (aka, "Hook's Law")

Recall: F = ma where "a" is the second derivative of the displacement, "x"

So, Hook's Law is just a special case of the differential equation that discribes the Simple Harmonic Oscillator (SHO).

IDEALLY, the ball would simply oscillate from one end of the hole to the other.

With "drag" accounted for, it would conform to damped harmonic oscillation and eventually settle at the equilibrium position, i.e. at the center.

Assuming a spherically symmetric distribution of mass in the Earth, the gravitational force ("F") exerted on the ball is proportionate to and opposite in direction from its displacement ( "x") from the center of the earth, i.e., the phenomenon conforms to the following eqn:

F = -kx (aka, "Hook's Law")

Recall: F = ma where "a" is the second derivative of the displacement, "x"

So, Hook's Law is just a special case of the differential equation that discribes the Simple Harmonic Oscillator (SHO).

IDEALLY, the ball would simply oscillate from one end of the hole to the other.

With "drag" accounted for, it would conform to damped harmonic oscillation and eventually settle at the equilibrium position, i.e. at the center.

Quite correct but that is another factor: Unless the hole was between the poles, the ball would have a curved path. For the sake of continuing the argument, we can assume that hole would "curve with the pathway.

Some other posted said something about the ball going into "orbit." Conservation of energy says that the "orbit" would bring the ball back to the earth's surface.

I am far too lazy to "do the math" so I will check out the URL posted earlier in the thread.

Why don't you post back when you've drilled to within a few miles of the other side ... that should give everyone a good while to think this problem through.

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