56 miles/ hr * .447 m/s / mile/hr = 25 m/s
Ac = v^2/R = 625 / 35.9 = 17.4 /s^2 (wow, g is only about 9.81)
F = m Ac
again F = m Ac
this time F = G Mearth m/R^2
and m Ac = m v^2/R
so
G Mearth/R^2 = v^2/R
so
v^2 = G Mearth /R
6.67*10^-11 is G
Cloverleaf exits are approximately circular. A 1220 kg automobile is traveling 56 mph while taking a cloverleaf exit that has a radius of 35.9 m. Calculate the centripetal acceleration of the car and the centripetal force on the car. Remember that 1 mph = 0.447 m/s.
2.A satellite is in orbit about Earth. Its orbital radius is 7.29×107 m. The mass of the satellite is 4362 kg and the mass of Earth is 5.974×1024 kg. Determine the orbital speed of the satellite in mi/s. 1 mi/s = 1609 m/s.
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by the way, the mass of a satellite in orbit does not matter. If you set your coffee cup outside the space station it stays alongside in theory. Notice that m cancels.
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