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Homework Statement
The force of attraction on an object below Earth's surface is directly proportional to it's distance from Earth's center.Find the work done in moving a weight of ##w## lb located ##a## miles below Earth's surface up to the surface itself. Assume Earth's radius is a constant r miles.
Homework Equations
##F = k*x##
The Attempt at a Solution
I would like someone to check this over. The work I have done so far (no pun intended). We are given that the force in magnitude is proportional to the distance, or ##F = k*x## where ##x## is the distance from the center of Earth to the object. Now the question is a little confusing, how to find k. I think we can assume they mean at the surface of Earth it weighs ##w ## lbs, therefore ##w = k* r## and ##k = w/r##, where ##r## is the radius of earth. We want the work for ##a## miles below the surface of earth.We have the integral:
##Work=\int\limits_{r-a}^{r}\frac{w}{r}xdx= \frac{w(2ar-a^2)}{2r}##
This does not agree with book answer
##\frac{w}{2}(2ar-a^2)##
What did I do wrong?
Also as a check if this makes sense, I found the total work to move a mass of weight w lbs from the center of the Earth to the surface of the Earth is :
##\int\limits_{0}^{r}\frac{w}{r}xdx= \frac {1}{2}wr##
This implies if you drilled a tunnel through the surface of the Earth all the way to the opposite surface and dropped an object of weight w lbs, what would be the work done on the mass? ##2*1/2wr= wr##, where r is the radius of earth.
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