To use some reference let's take the ISS.

The ISS has an orbital speed of 7600m/s. At the altitude of the ISS ~400km the force of the gravity it's about 8m/s2. It will take about 16 minutes to the ISS to achieve a vertical speed of 8000m/s.

My question is, why it doesn't fall.

I know about orbit is falling but going to fast the you "miss" the earth. But you will be falling faster each time due to the gravity being strong at just 400km.

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    $\begingroup$ As it moves round the orbit, the direction of the gravity vector changes, so over one orbit, the velocity vectors all cancel out giving a net velocity in the 'vertical' direction of zero. $\endgroup$ – MrPhooky Mar 2 '17 at 16:21
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    $\begingroup$ Yeah, I was so dumb, it's circular motion, the gravity itself though it produces an acceleration in an object, it's a force which is applied to the speed vector and make it "point down" so it doesn't escape earth. $\endgroup$ – jisuskraist Mar 2 '17 at 16:49
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    $\begingroup$ The gravity is the centripetal force keeping it in the circular motion. $\endgroup$ – MrPhooky Mar 3 '17 at 15:48
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    $\begingroup$ ISS flies 8000m/s sideways, above surface of Earth. Simultaneously, it accelerates, falling downwards. After 16 minutes, ISS will just about clear the Earth below and continue falling at 8000m/s right past Earth's edge (passing it by about 400km margin). Now rotate the whole image to the new orientation of "down" at that position... $\endgroup$ – SF. Mar 4 '17 at 23:36

The ISS is moving in a circle. Certainly, the Earth is accelerating it "downwards", but on the other side of the orbit downwards is in the opposite direction. After one orbital revolution, the net acceleration of the space station is zero. Simple animation.

The direction of "vertical velocity" is constantly changing, so to use this as a quantity we would want to use a rotating frame of reference. In such a reference frame, we have an inertial force perfectly balancing the force of gravity.


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