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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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