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)
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
I can't easily visualize what it would be.
This is Turtle.
It would depend on the possion of the Moon as to on what side of the earth it
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.
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.
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,
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
With "drag" accounted for, it would conform to damped harmonic
oscillation and eventually settle at the equilibrium position, i.e. at
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.
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