Physics hypothetical - ball falling in tube through earth - what happens?

Question

OK. So the earth is a perfect sphere and there are no issues with molten heat and such - this is all taken care of with very good insulation.

There is an open tube extending through the earth from the North Pole to the South Pole. Since the tube is through the axis of rotation - we can ignore any corriolis/centripital effects.

You drop an apple - or a billiard ball - into the tube - straight down so that it does not touch the sides - so we can dispense with any friction issues. And the tube is maintained at a vacuum - so we can dispense with any air resistance issues.

So, now that we have that down - we drop the ball into the tube - what happens? How does the ball behave as it falls toward the center of the earth, past the center of the earth, and beyond the center of the earth? How would it behave if we took air resistance into effect?

Enjoy!!!

And, let's ignore the movement of the earth around the sun for the moment.

And the earth has more or less a homogeneous density - or at least a density distribution that is spherically symmetrical about the center of the earth.

In response to British Soldier - this is offered as a fun diversion for people interested in physics - it is not a homework assignment I am looking to have answered. I think someone on here recently complained about the lack of interesting brain teasers in the physics section.

Answer 1

Well the force of gravity as determined by Newton is GMm/r^2 where M and m are the masses of the two objects and r is the distance of separation. So suppose the ball has mass m and is a distance r from the center of the Earth. Then the mass of Earth to be included is 4/3(pi)r^3*d where d is the density of the Earth. This is because we only use the mass in a ball of radius r. The gravitational effects from the other mass all cancel out.
Thus we can clearly see that the force at a given time is:

F = G * 4/3*pi*r^3*d*m / r^2 = 4*G*pi*d*m / 3 * r

And the acceleration of the ball is given by:

a = F/m = d^2/dt^2 (r)
Thus we have a second order differential equation:

d^2 r / dt^2 - 4*G*pi*d/3 * r = 0

And we can solve this quite simply to find the solution is:

r = Acos(wt + b)

Where A and b are constants depending on the intial conditions, and w is the frequency, given by:

w^2 = 4*G*pi*d/3

Thus the ball will oscillate back and forth, speeding up as it approaches the center of the earth, reaching a maximum exactly at the center, and then slowing down as it goes beyond, to come to a halt exactly at the surface on the other side. Assuming no friction, it will continue this way forever.

If we include air resistance, and assume the tube is much larger in radius than the ball, then there will be a drag force that can be calculated (although it produces some nasty differential equations) and the oscillation amplitude will be damped, so the ball goes less and less distance on each pass, eventually settling at the center of the earth (with imperceptible continuing oscillations of tiny amplitude).

Answer 2

Since the force of gravitational attraction increases nonlinearly as the distance between the two objects decreases, the ball would find itself accelerating MUCH quicker once it got very close to the center of the Earth.

However, at that point it would have so much acceleration and velocity it would overshoot the center and from the potential energy that it gained would go pretty much to the other end of the surface (other side of the earth, opposite of the starting end) and at this point the amount of potential energy gained is dispensed and must respond to the gravitational pull of the earth - possibly this oscillation of the ball going in and out of alternating sides of the earth would go indefinately when ignoring friction.

If we take Air resistence into account, then the ball would slowly lose some of the "potential energy" of the "fall" towards the center of the earth as heat and other forms of energy and so it would NOT make it quite to the end of the other surface and slowly, but surely it would oscillate in a "damped" motion where it would finally come to rest after some time in the center of the earth (center of the tube).

This is of course ignoring any and all circular/corriolis effects which could affect its path and contribute to other forms of loss.

Very interesting question - wonder where you came up with the idea. Thanks for asking - hope this suffices.

Answer 3

If we consider earth to be of uniform density then the ball will feel maximum force at the outer most part as it falls through earth the force will lessen. At the center there will be no force(rather the net force acting on all sides will be 0). But at that point net velocity will be maximum so it will continue to move. As soon as it leaves the center it will start decelerating. If we neglect air resistance it will continue upto other end where it will stop and reverse direction. In this way it will oscillate in SHM from one end to other. In presence of air resistance there will be damped oscillations about the center and ultimately the ball will halt at the center after a long period of time. If we consider actual density of earth which is non-uniform the ball will still oscillate between the same limits in absence of air but not in SHM because the maximum force will not be at outermost part but some where in between outermost part and center.

Answer 4

If the Earth were perfectly spherical and of homogeneous density, local g(r) would be simply proportional to r. (It is wrong to think that g(r) for some reason increases as r ---> 0; in fact, by symmetry, g(r) = 0 at the centre provided its a point with a spherically symmetric mass distribution surrounding it.)

Under a LINEAR g(r) the ball would execute simple harmonic motion (SHM), with period the same as an "Earth-grazing satellite" (assuming negligible friction for both ball and satellite).

That means that the period would be about 84 1/2 minutes, or the time from one end to the other about 42 minutes.

[Douglas Adams, where are you when we need you to explain why 42 is indeed "the answer to almost everything"? I suspect that it was because he came across the following twist to this story as an English schoolboy.]

Here's the twist just referred to:

Incidentally, the most amusing thing about this hypothetically stated situation is that the hole could be bored --- in a non-rotating Earth --- in ANY direction whatsoever, and the resulting motion would STILL be SHM, with the SAME time properties.

[I proved that fact myself, as an English schoolboy.]

In reality, since the Earth has density variations throughout it, the motion becomes only quasi-SHM, and with temporal properties similar to, but not identical with, those of the uniform density case. But, assuming spherical (but not radial) symmetry, you could still send, or even ride, one of those suckers from any spot on Earth to any other, thereby achieving non-polluting world-wide and extremely swift transportation!

Live long and prosper.

Answer 5

at the surface the ball has max gravitational potential energy (GPE) and no kinetic energy (KE). it accelerates under the force of gravity as it falls, and gains KE while losing GPE. at the center it has max KE and min GPE, with no net gravitational force since the earth's mass is now evenly distributed around the ball. now it slows down under the influence of gravity, losing KE and gaining GPE until it reaches the opposite pole. this motion would repeat indefinitely.

that's without air resistance. with air resistance or friction, it would lose some energy with each 'bounce' just as real balls do on the surface, but it would come to rest in the center rather than on the surface.

acceleration due to gravity would vary continuously as the ball fell, but i don't think it's necessary to work that out to say what happens in general terms.

below, G M are you sure about that? i don't think the force due to the spherical shell above the ball cancels out. shouldn't it be a sphere around the ball, just touching the surface, that cancels out? basically i don't think you can assume the earth is a point mass any more if you're inside the earth.

Answer 6

It accelerates until it reaches the center of the earth, then decelerates, finally stopping at the opposite end of the tube, begins to fall back towards the center of the earth, etc. It goes back and forth like something bouncing on a very large spring. If their is air resistance, then it does not accelearate beyond terminal velocity, so it only goes a little beyond the center of the earth before it stops and reverses direction.

Answer 7

there is a lot of problems with your question. First, the earth isn't a perfect sphere. Nothing known to man is a perfect sphere. Second you state that we can ignore the centrifical (or centrifugal) forces, friction and air resistance, so what are you looking for? This is like asking me what i would be if I hadn't gone to college.

Answer 8

u have given us things we can ignore, things we can dispense with and things that are taken care of....i reckon ur smart enough to answer that urself...however, let me offer u this train of thought. If the ball was passing, resistence free, through the centre of the earth, I would like to think it would be wondering why someone smart enough to be asked this question would then offer it to yahoo answers, ie 2 million strangers that cant be bothered to punctuate correctly(thats me, for information)

Answer 9

Its a simple harmonic motion (SHM)

m d2r/dt2=mMG/r^2

d2r/dt2=MG/r^2

M=4/3pi*r^3, earth volume

thus d2r/dt2= 4/3pi*r

if you add in air resistance, you got damping motion.

Answer 10

it comes out the other end!