Question

a ball thrown parallel to the surface of the earth from the top of Mount everest with a velocity more than escape velocity. it will​

Answer 1

Answer: Two results are possible depending on whether air resistance is negligible or present.

The velocity where the ball maintains a stable orbit (outside Earth’s atmosphere) is given  by:  v = square root (G * M(Earth)/r) where G is the Universal gravitational constant, M is the  mass of the Earth in KG, and r is the distance of the ball from the center of Earth. This is  known as the satellite orbit speed formula. All satellites at the same altitude move with the  same velocity. The higher the altitude, the lower the required velocity to maintain orbit.

G = 6.67 x 10^-11 N m^2 kg ^-2, Mass of the Earth is about 6.0 x 10^24 kg. At r = to  10,000 km (which is an altitude of about 3300 km or 3,300,000 m), the velocity to maintain  orbit works out to be 6326 m/s or 22,774 km/hr. At the top of Mt. Everest, r is 6380km, or  6,380,000 m. The orbital velocity at this altitude (ignoring air resistance) is 7,920 m/s, or  28,512 km/hr.

If v > the orbital velocity, we have escape velocity.

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