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Gravity at the bottom of a mineshaft
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[QUOTE="THarper, post: 4514243, member: 488975"] [h2]Homework Statement [/h2] The value of g measured at the bottom of a mineshaft is higher than that measured at the surface. Show that this implies the density of the Earth’s crust is less than 2/3 that of the mean density of the Earth. [h2]Homework Equations[/h2] g(r) = -GM(r)/r^2 G = 6.67384 × 10-11 m3 kg-1 s-2 [h2]The Attempt at a Solution[/h2] I started by showing that at a mineshaft depth of around 3500m, if the density of the Earth was constant from surface to center, then the formula for g would be g(r) = 4/3 Pi G ρ r where ρ is the density of the earth, and r is the distance from the center to the point of measurement. Using a measurement of the radius of the Earth of 6400km, g at the bottom of the shaft would be 0.055% smaller than on the surface. We know, however, that it is greater. I'm unsure as to how it would be shown that the density of the crust is less 2/3 than that of the mean. Assumptions may have to be made. Thanks in advance for any help! :) [/QUOTE]
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Gravity at the bottom of a mineshaft
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