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course Phy 231
12/08/2011 @ 5:05 p.m.
Gold Ball
How can we tell that there’s not a gold ball of diameter 1000 meters, just under the ground below the physics lab?
Look up the density of gold, then figure out how much gravitational force that ball would exert on a 1 kg mass in the lab.
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Fg = [(6.67*10^-11 m^3/kg*sec^2) * (1.01 * 10^13kg) * 1kg] / (500m)^2 = 2.69*10^-3 N. This is the force applied to the object by the gold.
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Assuming that the density of the 1000-meter diameter sphere just below the ground is 2500 kg / m^3 (which is what it would be if the ball is typical earth-crust material), how much force does it exert on that 1-kg mass in the lab?
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Fg = [(6.67*10^-11 m^3/kg*sec^2) * (1.31 * 10^12kg) * 1kg] / (500m)^2 = 3.5*10^-4 N. This is the force applied to the object by the rocks.
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What’s the difference in these forces?
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The gold applies .00234 more Newtons than the rocks.
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Could we detect the difference in the lab? If so, how?
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I believe that using F=m*a, we could find that the gold will accelerate a 1kg object by .00234 m/s^2 faster than an equal mass of rocks.
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That's a .02% difference in acceleration, which would be difficult to measure by, say, dropping a ball.
'What effect would this have on the period of a pendulum of convenient length, and why would this be easier to check out?
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