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.
Answer
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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