The International Space Station is orbiting the Earth at an altitude of about 34

Question

The International Space Station is orbiting the Earth at an altitude of about 340 Km above the Earth's surface, or a distance of 6400 Km from the center of the Earth. If the Space Station makes one revolution around the Earth in 90 minutes,

1)what is the size of the tangential velocity of the Space Station?

2)with these conditions, what is the size of the centripetal acceleration experienced by the space station?

3)compare the value of the centripetal acceleration to the acceleration of gravity at the earth's surface(wht percentage of g is this value?)

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Explanation / Answer

radius = 6400+340 km = 6730 Km

angular speed = [2*pi]/90min = [2*pi]/1.5hr = 4.19 rad/hr

1) tangential speed = angular speed * radius = 4.19 * 6730 = 28198.7 Km/h

2) centrepital acc. = [v^2]/R = (28198^2)/6730 = 118146.7 Km/h^2

3) g = 9.8 m/s^2 = 9.8*3.6 km/h^2 = 35.28 Km/h^2

centrepetal acc./g = 118146.7/35.28 = 334883 %

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