A particle moves in a circular orbit about the earth and directly over the equat

Question

A particle moves in a circular orbit about the earth and directly over the equator.
1) Calculate the work done by gravity during one orbit of the particle about the earth.
2) The particle has a period of 24 hours. Calculate the radius of the orbit.
3) The particle has a period of 24 hours. Calculate the acceleration due to gravity at this point without reference to Newton’s Universal law of gravitation.
4) The particle has a period of 24 hours. Calculate the acceleration due to gravity at this point using Newton’s Universal law of gravitation?
5) What is the angular velocity across the sky as observed by a person in the Rochester city?

Explanation / Answer

1) No work is done as potential energy unchanged 2) m v^2 / R = G M m / R^2 R v^2 = G M v = 2 pi R / (24 * 3600) R^3 = G M * 7.46 * 10E9/ (4 pi^2) solve for R 3) a = G M / R^2 (since F = m a = G M m / R^2) solve for a 4) a = v^2 / R centripetal acceleration solve for a 5) angular speed is zero since particle rotates at same speed as the earth

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