Asked by Nora
What is the ratio of the wight of a man when he is in a tunnel 1000 km below the surface of the Earth to his weight at the surface?
...I'm not sure how to proceed, not because I don't know how to do it, but because I have no idea how to come up with a(g) 1000 km inside the earth. I tried applying the law of shells (or whatever it's called) and:
a(g) = [GM(ins)]/(R^2)
*M(ins) = p*(4/3)*pi*R^3
so:
a(g) = [G * p*(4/3)*pi*R^3]/(R^2)
= 3 G M/(4 * pi * R^2)
...but when I end up plugging that back in, I get a value of a(g) that is like 3.0, which doesn't make any sense since we're closer to the center of the earth, so shouldn't a(g) be larger??
...I'm not sure how to proceed, not because I don't know how to do it, but because I have no idea how to come up with a(g) 1000 km inside the earth. I tried applying the law of shells (or whatever it's called) and:
a(g) = [GM(ins)]/(R^2)
*M(ins) = p*(4/3)*pi*R^3
so:
a(g) = [G * p*(4/3)*pi*R^3]/(R^2)
= 3 G M/(4 * pi * R^2)
...but when I end up plugging that back in, I get a value of a(g) that is like 3.0, which doesn't make any sense since we're closer to the center of the earth, so shouldn't a(g) be larger??
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