With which speed should we throw something, that it goes out of the gravity of the earth and start moving around the earth or goes off? With which speed does earth around its own axis?

1. The escape velocity can be obtained by equating the initial kinetic energy of the object with the magnitude of the potential energy of the object it has initially.

2. I'm assuming the actual question is

"With which angular speed does the earth rotate around its own axis?"

Find the time taken for the earth to make a complete round, then divide the angle (360 degrees or 2pi radians) by the time that you got.

Can you post what you get? :)

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Originally Posted by Unknown008

1. The escape velocity can be obtained by equating the initial kinetic energy of the object with the magnitude of the potential energy of the object it has initially.

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When computing the potential energy in this particular case, however, bear in mind that the "h" in E=mgh is the distance from the center of the planet, not the distance to the ground. So if you're trying to compute escape velocity at ground level (like how fast you'd have to throw a baseball), the "h" term is the radius of the earth: ~6371km.

What Unk and jc described gets you the "escape velocity," which is the velocity you need to throw an object straight up for it to completely escape earth's gravity and theoretically rise to an infinite height. It's also equivalent to the theoretical speed a falling object would attain if dropped from an infinite height. However the velocity needed to make orbit is significantly less. To calculate it: set the force due to gravity equal to the object's mass times centripetal acceleration for a circular orbit. For a low altitude orbit you get an answer that is equal to the escape veocity divided by sqrt(2).