A hole is drilled form North pole through the center of Earth to South pole?

Question

Well, you know the drill

Two small balls separated horizontally by distance d = 1cm
dropped into the hole.

What is horizonatal separation between the balls when they
reemerge 85 minutes later after one full period of oscillation?

The answer is almost as cool as Euler's formula, please
keep it in closed form.

Ignore air resistance

I forgot to mention that the hole is
circular, which is important. My fault.

85 minutes is non-datum, I just tried to give better feeling and explanation of the setup.

Note to catarthur (unrelated to this problem):

Equations like F(r) = d²x/dt² = 0 are integrated using conservation of energy.
F(r) = mdv/dt
v(t)F(x) = mvdv/dt
v(t)F(x) = md(v²/2)/dt

dx/dt dP(x)/dx = -d(mv²/2)/dt
dP(x)/dt = -d(mv²/2)/dt
P(x) + mv²/2 + Eo = 0

See also
http://answers.yahoo.com/question/index?qid=20070312114334AAjk9Kj

Again, the balls are small 'probe' particles, they feel only pre-existing external field and do not interact due to their negligible masses.

Answer 1

Actually, that is incorrect. The initial angular velocity of the two balls is equal to the rotation of the earth so they will be moving together. Saying otherwise is like arguing that when you jump up and down, you will move really far to the side because the earth is rotating around the sun at about 30 km/s (not per hour, per SECOND). As far as the balls are concerned, the earth is a closed stationary system. Relativity would make the same argument.

You mention the actual time it takes for this oscillation. I wonder if this is suppose to mean something or is just a red herring. To be perfectly honest, I have to disagree that no change in their distance will occur. When the two balls are at the center of the earth, they will probably collide. Of course, I am using the term collide very very loosely here. It is more like they keep getting closer until they are just barely touching, and then they are touching, and then they can't get any closer. So they spend some time stuck together before they reach the center of the earth. at this point, they can be treated as just one object because there is no tangental components to their velocities. They negate each other. If this had occurred in a fast collision where they separated immediately, then, you could argue they will have the same distance apart on the other side. However, the balls are too close together for that to happen. When the balls pass the center, their velocities are purely radial away from the center. They experience no forces to pull them apart. on the contrary, they are experiencing a force (gravity from the center of the earth) that pulls them together as well as back down. So, they should probably exit together with no distance between them.

All of that was just theorizing and guessing though. Something like this is already way over simplified and impossible to test. I think my line of reasoning makes sense though.

EDIT: Wow, there are a lot of spam posts here. Firstly, drilling through the Earth does not create a volcano. Those are formed at the meeting of two tectonic plates not at the north pole. These answers are really ridiculous.

Secondly, it doesn't matter precisely how big the particles are. If they are they are the size atoms, peas, bowling balls, or even the moon, they will still collide. In an ideal system they must travel right next to each other the entire time pulling directly towards the center. They both will pass through the precise center at the exact same moment in time. However, assuming that their size tends towards zero, then their collision would be instantaneous and thus cause them separate out after they pass through the center. This way conservation of momentum will be conserved. Anyway, the falling down process becomes identical to flying up process. They will be the exact same distance apart by the time they come out. Actually, I want to try some calculations here.

The arc length of the two balls starts out as 1 cm. Let's see, this equation states: ArcLength=Radius*angle where angle is constant throughout the dropping because all forces are radial, not tangental. So, the two particles would collide when the ArcLength is the width of one atom. I will be using the polar radius because the actual position of the tunnel was stated.

angle = 0.01/Radius of Earth = 0.01/ 6,356,752 = 1/635,675,200

diameter of H is 50 pm = 50 x 10^-12

5x10^-11 = R / 635,675,200 or R = 5x10^-11 * 635,675,200 = 0.0317 m

So, if we drop 2 hydrogen atoms at 1 cm apart and assume no interaction with each other as they fall, they will collide 3 cm before reaching the center of the Earth (or 30 cm is the atom was as big as cesium). Alright, so the question is whether or not that is a significant amount of time. To judge that we need to know how fast they are moving at that time. I just attempted to integrate the potential energy (f will mean integral from now on). Also, Me is the total mass of earth while me is the partial mass of some radius, x, of the earth.

Me = k * Re^3
k = Me / Re^3

me = k * x^3
me = (Me / Re^3) * x^3

U= -f F dx = -f (G*me*m/x^2) dx
= -G*m f [ (Me/Re^3) * x^3 /x^2]
= -G*m * (Me / Re^3) * f x dx
= " " * (1/2) x^2 | x=Re -> x=0
= " " * (1/2) * [ 0^2 - Re^2]
= " " * (1/2) * (-Re^2)
= (1/2)*m*G*Me/Re

Conservation of energy: potential Energy, U, becomes Kinetic Energy, K.

U = K = (1/2)m*v^2 = (1/2)*m*G*Me/Re

v^2 = G*Me/Re (which are all constants)
v = 7,918 m/s

t = d/v = 0.01/7918 = 1.26 * 10^-6 seconds

So we have to double the distance and the time because we are talking about falling to the center and then moving past.

The question is do we think 2.5 x 10^-5 seconds and/or 6 centimeters are a significant time for two particles to be colliding when they are on the size of a Hydrogen atom or d = 50x10^-12 m. Well, this is a judgement call that you will have to make. If we were to take in real physics, the two particles would probably try react together or repel each other but we are ignoring that fact. Collision cannot be ignored in this case. there is no way you can say they don't collide. However, we don't know if they stay stuck together or not.

IN CONCLUSION, my opinion is that no matter the size of the particle, they will collide at the center of the Earth and stay stuck together. However, there is ONE other answer. The two could collide and simply bounce back which means that their exit distance is going to be identical to the entering distance or 1 cm. These are the ONLY TWO CHOICES. I assume the correct answer would be the latter for any test or exam since they are likely to neglect how long the two collided particles are touching. I hope this helped.

Answer 2

They would both head for the center of the earth. But they would collide on the way. The separation when they emerge on the other side would depend on their diameter and elasticity. Larger balls would collide earlier and have more time to separate.
If the elasticity is zero, they would stick together and emerge with zero separation. If the collision is perfectly elastic, they would separate at the same velocity that they came together. the problem become more complex because they are no longer move directly toward the center of the earth.

If, however the person asking the question finds a way to eliminate the collision, the balls emerge with the same separation as when they entered. Each travels on a line to the center of the Earth. Its momentum brings it up to the surface on the exact opposite side. From there, it reverses direction, falls back through the center and continues to its starting point.

Answer 3

Absent air resistance, heat melting, inelastic collision with each other, Satan, and all the other complications of going through the earth (you know the drill).

Uh, 1 cm. They end up where they started. The earth rotated under then by 2 pi radians * (85 minutes / 1 sidereal day). But they're still 1 cm apart.

Is that closed enough for you?

Edit: Okay, so you actually care about the coriolis force from the horizontal component of motion. Sigh. Coriolis forces are caused by the devil, so I was including them in the things I was ignoring.

I have to go now, but will think about it later.

Answer 4

It's going to simply be 1 cm! Both objects oscillate with the same frequency--as you know the frequency is simply proportional to the density of the Earth. So the end of the cycle for one will occur at the same time as the end of the cycle for the other--they will both be back to their original starting places, 1 cm apart. BTW, trictly speaking, the two objects would collide in the tunnel, making the problem not trivial, but I think you want to ignore that.

Answer 5

The balls will fall, right. But (ideally) they will only fall half way through (up becomes down at that point) and will stay there, oscillating. Besides never emerging on the other side. they went in rotating around an axis midway between the 2. They maintain that distance, without friction or touching anything.
They do continue some oscillation due to degree of tilt of the earth. They don't re-emerge.

Answer 6

he is asking a question that doesn't care about the heat

anyway the probe will have a harder time in the middle of the earth because of the heat and pressure it has to endure. It probably takes a little longer. And if i'm right it won't even make it to the other side because it will stay in the middle of the pippe making it immovable. Gravity will pull it in all ways.

Answer 7

Your question of course is hypothetical . due to the fact that if you did drill through the center of the earth you wouldn't be around to see the results ., in fact none of would , IT would be like 2 giant volcano's at either pole spewing out molten rock and gases and eventually imploding .. melting ice , flooding , the sun would be blocked out by a black cloud , our oxygen level would be depleted, and the gases would choke us , the cloud would blind us ....so what was your question ?

Answer 8

Lets see, two balls are dropped in a hole drilled through the center of the....Oh NO, my eyes have gone crossed!

Answer 9

would the said ball if made it through the earths core role out of the hole in the south pole or would it shoot out like a rifle?????

Answer 10

We need to know the effect gravity will have - will it pull the ball toward the wall of the pipe, thus apart, or will it pull the balls together. Since each ball is closest to its respective wall than its partner, the net effect of gravity will be very, very slightly greater towards its respective wall. So the balls will be slightly farther apart when they emerge, guess by 1mm.