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There are 2 balls in a vacuum, next to each other but not touch. They are on the edge of a surface they will both leave the table at exactly the same time. One gets pushed harder than the other. The earth is completely flat. Apparently both balls will hit the floor at exactly the same time and even further be at the same vertical height from the floor at any point in time. How is this possible?

If you push the one ball much much much harder than the first so that it is going at a considerable speed 100m/s compared to 1m/s say. The balls can't possible hit the floor at the same time, right?

Now the earth is round. You hit the one ball so hard it goes into orbit the other ball just falls of the table straight down like before. As the one ball has gone into orbit it will never reach the ground. However the slow ball will reach the ground much faster. How can a rule be different? If it doesn't work at the extreme then it shouldnt work at all.

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In your first two paragraphs, where the Earth is assumed to be flat, why do you think the balls wouldn't hit the Earth at the same time? A very basic calculation, $z = z_0 - gt^2/2$, is all you need to show that they are at the same vertical height at all times, and it's really not the purpose of this site to walk you through that calculation, if that's what you're looking for. But if there is some more subtle point you're getting at, the question might be okay if you edit it to clarify that. – David Z Jan 14 '11 at 2:12
up vote 5 down vote accepted

The confusion arises because there are two different versions of what Earth's surface looks like, and two different models of how gravity works between the case where the ball goes into orbit and the case where both balls hit the ground at the same time.

We often approximate gravity near the Earth's surface by saying that it is constant everywhere. This approximation is pretty good if we are only interested in exploring distances much smaller than Earth's radius. In this approximation, the two balls will always hit the ground at the same time, no matter what speed they leave the surface with.

But, in reality, the Earth is round, and if you push the ball so fast that it travels a distance which is comparable to the radius before it lands, you must take a more sophisticated picture of gravity and the Earth into account. In this picture, it is possible that the fast moving ball will remain in orbit forever, or even fly away from the Earth never to return.

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The answer is that you are using a Discworld assumption (i.e. Earth is a plane) to derive conservation of momentum in the horizontal directions. Motion in this direction is then decoupled from the motion in vertical direction and so both balls have the same altitude at all times.

As you might imagine, Earth is round, so the above assumption fails at large distances. So you have to use full Newton's gravitational law and investigate possible trajectories. They are in general ellipses (which is the correct shape of the trajectory you thought was a parabola previously) and in particular circles. This is the trajectory that the orbiting ball will take.

Usual disclaimers about ignoring all effects of air, uneven Earth's surface, spinning of the ball and everything else that hasn't been explicitly mentioned.

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Your confusion arises because a lack of understanding general relativity. Einstein tells us that gravity is a fictitious force.

Furthermore, Einstein tells us that all objects will follow timelike geodesics.

The difference in behaviour that you note are merely due to a change in initial conditions. In the examples you give the variables are position, momentum and direction, these determine the particular geodesic each ball will follow. The difference in why one ball smacks into the earth and not the other is related to particles of the earth obstructing the path of the ball that was directed at its center. If you could imagine that there was a precise hole through the earth along the path of the downward moving ball, you would see that it would in fact oscillate for eternity from one side of the earth to the other; just as the orbiting ball would oscillate around the earth for eternity (barring the influence of non-gravitational forces, or effects of cosmological constants, or extreme gravitional fields that create singularities).

If you are looking for a reasonable introduction to the subject of relativity, I would recommend Relativity DeMystified by David McMahon as a reasonable primer to Gravitation by Misner Thorne and Wheeler.

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With respect to McMahon's books, please see the cooperative effort to make errata sheets here – Eduardo Guerras Valera Jan 31 '13 at 16:23

The balls' height above the Earth is the same for both balls at each time because of the principle of relativity - that is true even in Newton's physics, without all the special effects that occur near the speed of light.

The principle says that the phenomena obey the same laws of physics from a viewpoint of a uniformly moving observer. So if you have two systems that are moving, without respect to one another, in the $x$ and/or $y$ direction, it doesn't change the fact that the "depth" at time $t$ after any ball leaves the desk goes like $$-z = gt^2/2$$ In some sense, you may imagine that the forces and the laws of physics act on each component of the velocity separately. So the velocity of a ball in the horizontal direction (one of them) has no impact on its motion in the vertical direction.

Just to include an amusing correction: the Earth is not completely flat, after all haha.

Because it is not flat now, the very speedy ball may escape to outer space. The principle of relativity doesn't hold anymore: the presence of the round Earth distinguishes objects that are at rest with respect to the surface from those that move horizontally. It's because if you move the whole physical system in a linear direction, the background looks different because the Earth changes its location. This wasn't the case for the flat Earth which was invariant under horizontal translations.

More precisely, you may still introduce a principle of relativity, but with an extra correction. You may use two frames - one that is static with respect to the Earth and another one that is rotating around an axis through the Earth's center exactly in such a way that the speedy ball will look like a ball at rest. However, the laws in rotating frames are not quite identical to the laws in the inertial frames. Fictitious forces have to be added.

In the second, rotating frame, however, there is an additional force - the centrifugal force - that needs to be added to the laws of physics. Because of this centrifugal force, the speedy ball is able to escape from the gravitational field while the slow ball is not. More precisely, there's a centripetal force (into the center) that acts on the ball at rest - which is moving in the rotating frame - and this centripetal force keeps the (originally) slow ball in the Earth's gravitational field.

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Nowhere does it say that one ball will be faster than the other. It only says that one is pushed harder than the other. If the one pushed harder has more mass, then the amount of force on it might simply have been mecessary for the same acceleration.

In addition, I read this as the ball is pushed horizontally -- perpendicular to the force of gravity. That would have no affect on the fall to a hypothetical flat earth.

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