Question

WorkEarth’s attraction The force of attraction on an object below Earth’s surface is directly proportional to its distance from Earth’s center. Find the work done in moving a weight of w lb located α mi below Earth’s surface up to the surface itself. Assume Earth’s radius is a constant r mi.

Accepted Answer

Identify the variable distance from the center of the Earth. Let $$ x $$ represent the distance from the Earth's center in miles. Since the object is $$ \alpha $$ miles below the surface, its initial position is $$ x = r - \alpha $$, and the final position at the surface is $$ x = r . $$Express the force of attraction $$ F(x)  as a $$function of $$ x . $$Since the force is directly proportional to the distance from the Earth's center, we write $$ F(x) = kx $$, where $$ k  is $$the constant of proportionality. Given the weight $$ w  at $$distance $$ r $$, we can find $$ k  by $$using $$ F(r) = w $$, so $$ k = \frac{w}{r} . $$Rewrite the force function using the constant $$ k $$: $$ F(x) = \frac{w}{r} x . $$This force acts upward as the object moves from $$ x = r - \alpha  to$$ $$ x = r . $$Set up the integral for work done, which is the integral of force over the distance moved. Since the force varies with $$ x $$, the work $$ W  is $$given by $$ W = \int_{r - \alpha}^{r} F(x) \, dx = \int_{r - \alpha}^{r} \frac{w}{r} x \, dx . $$Evaluate the integral to find the work done: integrate $$ \frac{w}{r} x $$ with respect to $$ x $$ over the limits $$ r - \alpha  to$$ $$ r . $$This will give the total work done in moving the weight from below the surface to the surface.

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Key Concepts

Variable Force and Work

Proportional Force Inside Earth

Setting up the Integral with Limits