# Calculus Problem concerning Work

I came across a question in my calculus book that asked me to calculate the work done my gravity in pulling a mass of $$100$$ pounds from a height of $$500$$ miles above the earth to the surface of the earth. In the book, we had already derived that the work accomplished in moving an object from a point $$r_1$$ to the point $$r$$ is:

$$W = \frac{GmM}{r} - \frac{GmM}{r_1}$$

where $$W$$ is the work, $$G = 1.07 * 10^{-9}$$, the gravitational constant, and $$M = 13.1 * 10^{24}$$, the mass of the earth, and $$m$$ is the mass of the object be pulled down from distance $$r_1$$.

The surface of the earth is the point on the surface on the earth that is $$R$$ away from the centre where $$R = 4000 * 5280$$. $$r$$ and $$r_1$$ have to be in feet. Using this information, I managed to determine that the answer to the problem is $$\frac{22528 * 10^6}{3}$$ foot-poundals. However, the next question is the question that stumped me:

Suppose that we use $$32m$$ as the constant force of gravity and calculate the work done by gravity in pulling an object over the same path as in Exercise 2. Before calculating, determine whether the answer will be greater or smaller than the answer in Exercise $$2$$ and then carry out the calculation.

Here was my reasoning to solve this problem:

The constant force of gravity is given as $$G = 1.07 * 10^{-9}$$. The force of gravity has now been changed to $$32m$$ = $$32 * 100$$ (in the original problem, the mass of the object was 100 pounds) = $$3200$$. Obviously, this new constant of gravity is greater than the original, so the answer must be that the work will be less. Then, I simply changed the constant force of gravity in the above equation given to be $$G = 3200$$ instead of $$G = 1.07 * 10^{-9}$$. When I performed the calculation, my answer was incorrect. In fact, it was way off. The answer to this problem is supposedly $$8448 * 10^6$$ foot poundals.

This confuses me for $$2$$ reasons:

$$1.$$ Why is the work done not lesser? If the constant force of gravity is now $$32m$$, the amount of work done is less since earth's pull increased.

$$2.$$ Does this problem have more to it than just calculating using the derived equation? Considering this is a calculus book, I'm assuming the author wanted me to do something related to calculus rather than just order of operations.

The book I'm working out of is Calculus by Morris Kline.

Thanks.

• For #1, I'm just gonna go with basic physics. Work = Force x Distance $W = F\cdot D$. Because the force of gravity is greater over the same distance, the amount of work done increases. That's one way I might reason with myself. – Christopher Marley Sep 22 '18 at 3:00

You are confusing $$G$$, the gravitational constant, with $$g$$, the gravitational acceleration at the earth's surface. The standard value for $$g$$ in imperial units is $$32.174 ft/sec^2$$ The $$32$$ is a rounded version of this. We have $$g=\frac {GM}{r^2}$$ and your constants give $$g \approx 31.012 ft/s^2$$ which is close to the standard value. The force of gravity on an object is $$mg$$. This is very close to the force on the object at the surface of the earth, but at $$500$$ miles altitude the force has decreased by a factor $$\left(\frac {4000}{4500}\right)^2 \approx 0.79$$. Using the constant force of $$32m$$ overstates the force as the altitude rises, so it should increase the work done. I would expect the increase to be around $$10\%$$ as the force was dropping from $$1$$ to $$0.8$$ so the average is around $$0.9$$.
• As a quick clarification, how did you calculate the decrease by a factor of $0.79$? Thanks. – S. Sharma Sep 22 '18 at 3:25
• The distance from the center goes from $4000$ miles to $4500$ miles, a ratio of $\frac 98$. I inverted and squared that. – Ross Millikan Sep 22 '18 at 3:29
I wanted to add this here as clarification for anyone who is overthinking it like me. Basically, $$W = F * d$$. Thanks to the explanation by @Ross Millikan, the "constant force of gravity" it mentions is basically the gravitational acceleration at Earth's surface. Essentially, all its saying is that, instead of the gravitational acceleration changing at each instance the Earth pulls us down, the problem wants us to assume that gravity is going to pull down an object at $$32 ft/s$$ no matter what distance. So, all that means is that we have to use the standard work formula to calculate the work accomplished:
$$W = F * d = 3200 * (500 * 5280) = 8448 * 10^6.$$