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Since we know that Mount Everest is 8848 meter high, we just have to add that to the radius of the Earth and find the radius of the centripetal force. We have that the radius is 6408848 meters.
Now, we need the formulas:
F=(mv^2)/r and
F=(GMm)/r^2
We combine the tow formulas and get
(mv^2)/r=(GMm)/r^2
Solve for v and get
v=sqrt(MG/r),
G being 6.67 x 10^-11
M being the mass of the Earth, or 6.4 x 10^24
and r we already solved
So, plug in, and the answer we get is 7902.2 m/s

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1mo ago

The horizontal velocity required for a satellite to be placed in a circular orbit around Earth from the top of Mt Everest, which is approximately 8,848 meters above sea level, would be about 7,900 meters per second. This velocity is necessary to counteract the Earth's gravitational pull and maintain a stable circular orbit at that altitude.

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Q: At what horizontal velocity would a satellite have to be launched from the top of Mt Everest to be placed in a circular orbit around the Earth?
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