Problem: An object below the surface of the earth is attracted by a gravitational force that is directly proportional to the square of the distance from the object to the center of the earth. Set up and analyze a model to find the work done in lifting an object weighing $P$ lb from a depth of $s$ feet (below the surface). Assume that the earth's radius is $4000$ mi and that $P$ is constant between $s$ and the surface.
My Attempt: The gravitational force $F=kx^2$, where $k$ is the constant of proportionality and $x$ is the distance from the center of the earth. The object weighs $P$ lbs and therefore $$P=k(4000-s)^2$$ $$\implies k=\frac{P}{(4000-s)^2}.$$ Thus the work done is $$W=\int_{4000-s}^{4000}Fdx=\int_{4000-s}^{4000}\frac{Px^2}{(4000-s)^2}dx$$ $$=\frac{P}{(4000-s)^2}\left[\frac{x^3}{3}\right]_{4000}^{4000-s}=\frac{P}{3}\left(4000-s\right)-\frac{4000^3P}{3(4000-s)^2}.$$
However, the answer is $$4000^2P\left(\frac{1}{4000}-\frac{1}{s}\right)\text{mi-lb.}$$
Where am I going wrong?
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$\ds{x}$ is the distance from the body, of mass $\ds{m}$, to the Earth Center. $\ds{G}$ is the Gravitational Constant. It's assumed the trajectory ia a straight line from $\ds{R_{\oplus} - s}$ to $\ds{R_{\oplus}}$ ( the Earth Surface ) with the origin at the Earth Center.
The Earth Mass Density is given by $\ds{{M_{\oplus} \over 4\pi R_{\oplus}^{3}/3}}$ which assumes the Earth as a homogeneous sphere of radius $\ds{R_{\oplus}}$. \begin{align} \mbox{Force} & = G\,\bracks{\pars{M_{\oplus} \over 4\pi R_{\oplus}^{3}/3}\pars{{4 \over 3}\,\pi x^{3}}} {m \over x^{2}} = m\,\ \overbrace{{GM_{\oplus} \over R_{\oplus}^{2}}}^{\ds{=\ g}}\,\ {x \over R_{\oplus}} = mg\,{x \over R_{\oplus}} \\[5mm] \mbox{Work} & = \int_{R_{\oplus} - s}^{R_{\oplus}}mg\,{x \over R_{\oplus}}\,\dd x = mg\,{R_{\oplus}^{2} - \pars{R_{\oplus} - s}^{2} \over 2R_{\oplus}} = \bbx{\ds{{s \over R_{\oplus}}\,\pars{1 - {1 \over 2}\,{s \over R_{\oplus}}}\,mgR_{\oplus}}} \end{align}