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Suppose the space shuttle is in orbit 470 km from the Earth’s surface, and circles the Earth about once every 94. 0 minutes. Find the centripetal acceleration of the space shuttle in its orbit. Express your answer in terms of g, the gravitational acceleration at the Earth’s surface. Related posts:Another terrestrial planet with twice the mass of Earth and its surface?What is the magnitude and direction of the centripetal acceleration?How long does it take a space shuttle to clear the atmosphere after liftoff?
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centrip accel = v^2/rwe need to find the speed and the radiusthe speed is the distance traveled/timethe distance is the radius of the orbit, or 2 pi r; the time is the period Pso v=2 pi r/P and v^2=4 pi^2 r^2/P^2then a=v^2/r = 4 pi^2 r/P^2r=radius of earth + 470 km = 6400 km + 470 km = 6870 km = 6. 87×10^6mP=94 mins = 94 minsx60 sec/min = 5640sa=4 pi^2 (6. 87×10^6)/(5640)^2 = 8. 5 m/s/s => 8. 5/9. 8 = 0. 87 g