Imagine that there's a tunnel that crosses the planet from one  side to the opposite side, for example that starts in America and finishes in China, then throw a ball into it from America... would the ball go out in China?
119503
The ball would execute simple harmonic motion in addition to being hugely affected by Coriolis "force" due to the spinning Earth. \nIf the tunnel were not spinning and the ball was dropped straight down the center, the equation governing the ball's motion would be ma = GM(r)m/r^2 where M(r) refers only to the mass of the solid sphere under the ball, since uniform spherical shells do not exert any force on objects inside them. Assuming uniform density p, we have M(r) = (4/3)*Pi*r^3*p, so a(r) = G*(4/3)*Pi*r*p. If we let k = G*(4/3)*Pi*p, which is a constant, we get the ubiquitous equation a(r) = k*r, or d^2r/dt^2 = k*r, which we all know the solution to as the simple harmonic oscillator r(t)=A*cos(Sqrt[k]*t + B) where in our case A is the radius of the Earth and B=0.\nTo complicate matters, due to the presence of some pretty dense air, it will be necessary to add dampening conditions to the differential equation, giving a damped oscillator.\nAlso, the Earth is spinning and since the ball is not rigidly connected to it, it experiences a Coriolis "force" towards the tunnel walls in addition to its oscillation, which will surely cause it to at least stop prematurely.\nIt is more likely the ball will be destroyed by hitting the sides of the tunnel at high velocity. ;)
471759
NO!!  If people would stop (mis)quoting half-remembered precepts here, and do a spot of THINKING, instead, we might get somewhere.\n\n1.  If a moving spherical ball directly hits an identical free, stationary ball head on in an elastic collision, the moving ball STOPS, and the previously stationary one heads off with the original velocity of the other.  So ALL of the moving ball's K.E. has been transferred to the other ball.  \n\nHere are two ways to see this:\n\n1a.  The device known somewhat indelicately to British schoolboys as "Newton's balls" (but in squeamish America as "The Swinging Wonder") provides an excellent table-top demonstration of this.  Most undergraduate physics instructors must haul this out from time to time.\n\n1b.  You can easily solve it analytically, but a thought experiment will also demonstrate it.  In the originally stationary ball's frame, let the other ball approach with speed v.  Now shift (with an appropriately directed speed of v/2) to the frame in which the two equal balls are simply approaching each other, each with a speed in this frame of that same v/2 (oppositely directed, of course)!  Clearly, from the symmetry of the situation, they'll bounce off each other with the same (oppositely directed) speeds in this frame, still of v/2.  Now go back to the original frame: the previously moving ball is now completely stationary, while the originally stationary one is haring off with speed v.  (You may need to make little sketches of what is happening as viewed successively from each of these frames, to see this; I admit that that's what I did.)\n\n2a.  A similar approach can be used to solve the general case for an arbitrary ratio of masses, m (initially moving)/M (initially stationary).  Hint: have the balls approach one another in their center of mass frame with equal but opposite momenta of mv and  MV.  They'll clearly bounce off one another in such a frame, retaining the same values of their respective (but now oppositely directed) speeds.  Then you can do the other frame-jugglings from there. \n\nHowever, this analytical approach, done accurately, is a tedious mess.  So, to appreciate the general problem more physically, consider just the following:\n\n2b.  The case of m << M.\n\nConsider the "equal but opposite momentum" frame for these two balls, in which MV = mv.  This means that if ball M is VERY MASSIVE, the speed it will have in such a frame, V = (m/M)v; that is, it's VERY SMALL compared to little puny m's high speed.\n\nSo, transforming to the frame in which M itself (the big guy) is initially stationary takes only a speed shift of O(m/M)v.  (In fact identically so.)  This only affects the speed of puny m minutely, so m's initial K.E. is still of order O(m v^2).  (There's no need to do picky constant book-keeping for these purposes.)  \n\nNow after the collision: In the original frame, Big M is now moving (and in fact at 2V if you think about it.)  But, being loose about the constants, M's K.E is only O(MV^2).  \n\nSo, the ratio (m's K.E.)/(M's K.E.) ~ (mv^2)/(MV^2)\n\n= (m/M)(M/m)^2 (since mv = MV)\n\n= M/m.\n\nBINGO!  The little bugger still carries by far the larger amount of the kinetic energy.  The massive ball (i) moves sluggishly, and (ii) because  K.E. involves both mass but also the SQUARE of the speed, the TWO POWERS of (small) speed more than cancel out the effect of its (large) mass.\n\nThis is a well-known fact in particle collision theory, that little particles basically bounce off big ones.  \n\nYou could also say that it's well-known in "space billiards": when spacecraft head close to Jupiter to receive a "gravitational sling-shot effect" that sends them off to the outermost planets.  \n\nIt's more complicated because of 3-D effects there, but analogous frame-shifting methods can be used to show that the spacecraft can pick up considerable speed AND change
96977
Yes, but you notice the rubber ball is going slower now. Its because part of the momentum has been transfered from the ball to the sidewalk (eventually tranfered to the Earth itself).  This is called transfer of momentum.
419775
If there was a hole right through the center of the earth?\n\nAlright.  In order to understand the answer, you have to understand the difference between potential energy and kinetic energy.  Potential energy is stored up energy.  (Like a rubber band stretched and ready to snap.)  Kinetic energy is moving energy.  (Like the rubber band in the process of snapping.  \n\nThe moment that a person steps over the hole through the earth., that person has complete potential energy.  The earth's gravitation is pulling on them, but they are not yet falling.  As they fall, that potential energy is converting to kinetic energy.  They fall faster and faster.  Once they reach the center of the earth, they have no potential energy, only kinetic energy.  \n\nThis kinetic energy keeps them moving down the hole towards the other side of the earth.  As they are moving away from the earth's gravitational pull, they slow down.  The kinetic energy is converting back to potential.  \n\nOnce they reach the surface, they once again have full potential energy and zero kinetic.  \n\nSo, they yo-yo back and forth between the two mouths of the tunnel.\n\nWhat ends up happening depends on if you are taking air resistance into concideration.  \n\nIn a complete vacuum, the person would yo-yo between the two openings of the tunnel forever.   The person's potential and kinetic energies would have an inverse relationship.  Back and forth.\n\nIf you are taking air resistance into account, then some of the energy is lost from friction between the person falling and the air.  (It's lost through heat energy.)  During each yo-yo, the person's fall would be less and less.  Eventually, all of the energy in the person would be lost through the heat energy and the person would be stuck in the center of the earth, suspended there for all time.\n\nKinda cool, huh?
77661
this would all depend on the angle of the secant, the depth of it, and, of course, the velocity at which you would be traveling upon entering the opening.  In general, you would simply roll till you ran out of momentum.  gravity still pulls to the core.
161418
China has Started a manned Space programme and has said it's aim is to land on the Moon.\n\nSo the Yanks will Try to get get there first.\n\nBut their Space programme is still not up and running properly since the Columbia disaster.
323195
Same plane and not parallel as other answers have said, but the reference to moving balls complicates things. The balls are not guaranteed to meet, and since they are in unidirectional motion, their paths will not cross if they diverge initially. Such paths are not extended in both directions as the definition of a line requires.
11666
Hmm. I think air would not flow as well. I would guess the atmosphere would stay hovering relative to each plane (6 of them), and perhaps intersect at the corners.\n\nEach side of the earth would have varying surface temperatures, unevenly distributed, which would make the air dramatically different temperatures. This would cause the air to move, possibly faster so it could be quite windy.\n\nHee hee what a weird question, I like it :-)
77175
No way to know.  What is the shape of "everything"?  I think it is round and that if you go to one end, you'll end up back at the same point if you keep going the same direction.\n\nGood question.
490079
The atmosphere would freeze out on the dark side of the planet. \n\nEven rotating once a year, the greenhouse effect would not be enough to prevent this. Even in Antarctica it can get cold enough for CO2 ice to form during its long dark night. As the CO2 is sucked out of the atmosphere it gets much colder quite quickly. The oceans on the day side would join the frozen air air since as the air pressure dropped, they would evaporate too and the water vapor would meet the same fate. The first cycle would probably take many years to occur due to residual heat in the crust and the amout of time it will take to freeze out most of the CO2 to get the ball rolling quickly.\n\n\nBits of the frozen atmosphere would thaw as it came into sunlight but diffusion would eventually carry anything that thawed to the cold side where it would precipitate out and stick.  I suppose in equilibrium there would be some sort of VERY thin atmosphere, but not much.  I could not tell you how slowly the Earth has to rotate for this to occur. Without ozone, mcuh of the water that evaporates will be photolyzed quite quickly and the hydrogen will mostly escape to space. So the water ice will be lost over time.\n\nSo, you would have a big old mountain of frozen water, nitrogen and oxygen, with less and less water each year on the dark side, and nitrogen/oxygen snows not far from the dayside terminator. \n\nThings are much more stable if the earth is tidally locked to the sun. All the stuff falls on the night side and just stays there.
491444
if yo get half way you float! which means the earth is hollow! but if you make it through you end up in the Indian ocean if you live in north America because north America is in the north hemisphere and western hemisphere bu t the Indian ocean is in the eastern and southern
304671
The gavitational force decreases as you go towards the centre of the earth from its surface. At the center it is zero. but Why will you fall off if there is no force there? If proper calculations are done you will find that an object falling in such tunnel will oscillate in it. As soon as it reaches the surface at the other end it will turn and again fall inwards. Burning is another issue. :)
647727
No, and something from outside it would have to break Jupiter apart if it came apart, and if it did, the number and size of pieces it broke into would depend on what hit it, how big it was, and how fast it was traveling first.  And it would have to be much bigger than Earth to do any damage to Jupiter.  There are many other things to be concerned about - like bad movies.
50744
The object will follow a parabolic trajectory to the earth, as observed from outside the plane.  On the plane, it will appear to fall straight down in your (moving) frame of reference.  Air resistance will retard this motion, but it's magnitude depends on the drag coefficient of the object dropped and the medium in which it is falling.
257423
No, because if this pit when through the entire earth it would end up on the other side of the earth.
1285
No, the Earth will continue in its orbit around the sun. But as the sun runs out of hydrogen fuel and balloons into a red giant, it will become so huge that Earth will be completely incinerated. Sorry to break it to you.
307120
Yes.  There is a tunnel to get to the infield.