Today, in this blog post, we are going to analyze with a microeconomic lens the Troll Physics Gold Shipping Company (a company that is supposed to make a lot of profit, but is it really that?)

Comic:

Original post? 

Summary (if the comic is blocked in your school/work):

Troll buys gold in Central America, then takes advantage of a fact that Earth is wider at the equator, so the force of gravity is weaker there and stronger at the poles. Troll then travels to Alaska and sells the gold there. Since gravity is slightly stronger there, the gold will be a little heavier and Troll gets more money for it. Troll then travels back to Central America with the money and buys more gold. This cycle repeats until he is a millionaire.

Analysis:

Could Troll have made more profit (assuming everything else is free?)

First of all, Troll is missing out on a lot of profit here. Troll fails to acknowledge this equation:

${\displaystyle g\left(\theta\right)=9.78\left(1+\frac{5.27\sin^{2}\left(\theta\right)}{10^{3}}+\frac{2.34\left(\sin\left(\theta\right)\right)^{4}}{10^{5}}\right)}$.

In case you do not what this formula means, I will tell you what it means; this formula calculates the reference gravity as a function of latitude. Troll is travelling from one latitude to another in order to take advantage of the localized reference gravity here, so this formula is perfectly valid for the situation. To maximize percentage gain (and hence profit), Troll must have the largest possible final profit and the smallest possible initial investment, because when you take the ratio of those two monetary values, the resulting percentage gain is large. To obtain the largest possible final profit, Troll needs to find a place where the force of gravity is the strongest, and because of Newton's second law, the place where the force of gravity is the strongest is also the place where the reference gravity is the largest. To obtain the smallest possible initial investment, Troll needs to find a place where the force of gravity is the weakest, also known as the place where the reference gravity is the smallest. Therefore, we have turned this problem into an optimization problem. Fortunately, this problem can be done without calculus due to the bounded nature of the sine function, namely the fact that sine is entirely bounded between -1 and 1. If we take sine squared, the resulting function would be entirely bounded between zero and one. Even squaring that again to get sine to the fourth, the resulting function would still remain entirely bounded between zero and one. Since sine is 1 at 90 degrees latitude and 0 at 0 degrees latitude, and both those numbers are invariant when squared, we can conclude that Troll's best profit method would be to buy gold at the equator and sell the gold at any pole.

So how much percentage gain will Troll get away with? We can calculate the ratio of reference gravities at the equator and at the poles and we will obtain the answer. The calculation is outlined below:

${\displaystyle 100\left(\frac{9.78\left(1+\frac{5.27\sin^{2}\left(\frac{\pi}{2}\right)}{10^{3}}+\frac{2.34\left(\sin\left(\frac{\pi}{2}\right)\right)^{4}}{10^{5}}\right)}{9.78\left(1+\frac{5.27\sin^{2}\left(0\right)}{10^{3}}+\frac{2.34\left(\sin\left(0\right)\right)^{4}}{10^{5}}\right)}-1\right)=0.52934\%}$

(Note to the geologists: I used radians. Please do not hurt me.)

So, Troll only gets 0.52934% profit from this.

Of course, this is assuming that everything else is free. However, we run into problems later.

A Day in The Life of a Troll (Simple Model and Walkthrough)

Let's suppose Troll starts out with ${\displaystyle x}$ dollars and gold is fixed at ${\displaystyle g}$ dollars per troy ounce. Troll is able to buy ${\displaystyle \frac{x}{g}}$ troy ounces of gold at the equator, but Troll has zero dollars left. Fortunately, Troll was able to get a round-trip ticket to any pole in Earth, which costs ${\displaystyle a}$ dollars. Troll has ${\displaystyle -a}$ dollars at the pole, so he decides to sell all of his ${\displaystyle \frac{x}{g}}$ ounces of gold. However, due to the increased gravity at the pole, Troll was able to make ${\displaystyle kx}$ dollars, where ${\displaystyle k}$ is the ratio between the reference gravities. Troll walks away with ${\displaystyle kx-a}$ dollars. Finally, Troll uses his round-trip ticket again and does not pay anything.

The Development of a Recurrence Equation and its Non-Recursive Solution:

Troll after a while wants to find how much money he has after ${\displaystyle n}$ round-trip flights so that he can find out if he is a millionaire or not. We can help him on his task. We define a recurrence relation as follows:

${\displaystyle b_{0}=x}$, ${\displaystyle b_{n+1}=kb_{n}-a}$.

Troll then asks if there is a recurrence solution; we give Troll this formula: ${\displaystyle b_{n}=\frac{a-k^{n}\left(a-kx+x\right)}{k-1}}$.

Troll wants to be a Millionaire!

Finally, Troll has to calculate how many round-trip flights he has to do before he makes a seven-figure net worth using this method. We give Troll this function. ${\displaystyle n\left(x\right)=\operatorname{ceil}\left(\frac{\ln\left(\frac{a-1000000\left(k-1\right)}{a-kx+x}\right)}{\ln\left(k\right)}\right)}$

Troll Criterion

Troll analyzes our created function and gives it a criterion, namely, the function ${\displaystyle \left(k-1\right)k\left(a-1000000\left(k-1\right)\right)\left(a-kx+x\right)}$. If Troll founds out that this value is negative, Troll would stop. Before resuming, Troll would plug in his current net worth into ${\displaystyle x}$.

When will Troll realize his plan does not work?

Troll would stop working with his plan if the planet is a perfect sphere in rotational equillbrium, because in such a perfect sphere, the ratio of the reference gravities is exactly one. Troll would also stop if the ratio of the reference gravities is ${\displaystyle 1+\frac{a}{1000000}}$ or ${\displaystyle \frac{a+x}{x}}$.

Troll would stop if the price of the round-trip is ${\displaystyle 1000000\left(-1+k\right)}$ (5293.40 US dollars in Earth) or ${\displaystyle \left(-1+k\right)x}$.

When will Troll begin?

Troll can use this program just as long he starts with ${\displaystyle \frac{a}{-1+k}}$ US dollars or higher.

Example and Teaser:

Suppose Troll has $200 that he obtained by trolling people. As a reward for his trolling mischief, he is able to fly on an unlimited fuel aircraft with a cruise speed of 550 miles per hour for the discounted price of $1 and Troll gained an item that allows him to buy and sell at instantaneously fast speeds. Troll is also able to evade taxes and subsidies using loopholes he learned during his career at Trolling College. Travelling a round-trip for Troll would take him 22 hours and 35 minutes.

Troll calculates his criterion before he steps into the program. Seeing that it is positive, Troll commits to the program.

After 2,162 roundtrips and approximately 5.57 years later, Troll has earned $1,000,000 through his commitment.

Troll then has a son, whose name is Trollge. Trollge looks at Troll's accomplishments and decides to do things his way. So what does Trollge do?

Change the rotation of the Earth itself!

Trollge learned centripetal acceleration on his Advanced Placement Physics C Mechanics class today. He also learned that the radius of Earth is 6371.009 kilometers and the angular velocity of Earth is 0.2625161 radians per hour. Troll then knows that due to the Earth rotating, gravity is not uniform throughout the sphere. He computes that the Earth has a centripetal acceleration of 3.38778 centimeters per second squared during a problem that says that he must find the gravity at the equator and at the poles. This reminds Trollge of something, so he plays with the angular velocity. He hopes that by increasing the magnitude of angular velocity, he can make the gravity at the equator weaker and the gravity at the poles stronger, thus allowing his father to make more profit.

Trollge then watches Animaniacs and finds a way to increase the rotational kinetic energy of the Earth by creating artificial angular momentum. As such, the planet starts to spin faster. At its peak, Trollge rotates the Earth at an angular velocity of 0.00124067 radians per second by increasing the rotational kinetic energy to 1.2369×10³² Joules. As a result, Troll gets a lot of profit from his gold at the equator. If he increases it even more, the Earth might cease existing.

Other People's Opinions (From the Comment Section in the Original Post):

According to StealthAssassin, this plan would not work because one has to pay for the transportation. This comment was posted on August 9, 2012. However, we have shown that at a certain point, the fare of the transportation becomes unimportant.

Surprisingly, this post was rediscovered on March 16, 2020, by Hinoron. In it, he says that not only would have to pay for the transportation, but for the rights to transport large quantities of heavy metal. He then argues then presumably at the start of this project, one is probably not rich enough to afford their own shipping company. Finally, he refers to pay-by-weight. Obviously, in Earth, pay-by-weight percentages trivially exceed 0.52934%, so this plan would result in an equivalent k that is less than 1, and thus the plan could not be started at all.

Unlimited Profit Indeed.