Question 1137460: What will be the time (in seconds) taken by a small stone to reach the surface of the Earth if it is dropped from a height R/4 where R = 6 400 km, Mass of Earth =6 x 10^24 kg.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! The small stone is at rest when it is dropped from a height of 6400/4 = 1600 km
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Note The acceleration of gravity will be less for the stone at height 1600 km than if the stone is near the surface of the earth
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We know that
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F = ma = GmM/r^2, where M is the mass of the earth, G is the gravitational constant, m is the mass of the stone, M is the mass of the Earth and r is distance from the center of the earth to the stone
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The distance from the center of the earth to its surface is 6400 km, therefore r is 6400 + 1600 = 8000 km
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a = ( 6.677 x 10^-11 m^3 kg^-1 s^-2) * (5.972 x 10^24 kg) / (8000000)^2
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a = (39.875 x 10^13) / (6.4 x 10^13) = 6.23 m/sec^2
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distance = v(0)t +(1/2)at^2, where v(0) = 0, distance = 8000000 m
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Here, problems arise, notably the acceleration of gravity will increase as the object approaches the surface of the earth, air friction will increase as well
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What assumptions has your teacher given you about this problem?
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