A Moon Mission Without Calculus — Part 1: The Size of the Earth
So much of mathematics education is centered around eventually getting to calculus, to where often the ability to work with the material covered in pre-calculus is overlooked. We also tend to overlook the power (and limits!) of what we can do with this mathematics because calculators and computers can give us fast approximations to various answers. This is a real shame because all mathematics done before the 17th century was done without calculus or electronic computers, and we managed to do a great deal in those times! In fact, our mathematical ability would have allowed us to get to the Moon, even if we only achieved the engineering capacity in the past century.
This sequence of articles will show the mathematics of getting to the Moon, without formally using calculus. It is going to be a “paint by numbers” sort of article: along the way, I will ask questions and give their answers, but you might find it useful to work through the question yourself to actually find the answer yourself.
There are several things we will need to know to get to the Moon:
- How can we measure the size of the Earth? (This article)
- What is the strength of the Earth’s gravitational force? (Part 2)
- How do orbits work? (Part 3)
- How should we plan our mission? (Part 4)
Orbiting is often described as “falling in such a way you are always missing the ground.” Let’s think about a teapot in a perfectly circular orbit first. “Perfectly circular” means that as it circles the Earth, the teapot’s distance from the center of the Earth is constant. We can certainly measure distances above the Earth’s surface, but to measure that full distance we’ll need to know the Earth’s radius.
The most famous measurement of the Earth’s radius is that of Eratosthenes of Alexandria; while we do not know precisely what his process was, we know a simplification: In 240 BCE, he measured the angle between the Sun’s rays and “vertical” in Alexandria. He knew that at that same time in the city of Syene the Sun’s rays would be perfectly vertical, what we now call a ‘Lāhainā Noon’. From administrative measurements of the Egyptian territory (done for agricultural and taxation reasons), he also knew the physical distance between Alexandria and Syene.
The Sun angle tells us what fraction of the Earth’s circumference the vertical distance between Alexandria and Syene is; knowing the physical distance (and hoping that Syene is close to directly south of Alexandria) yields the full circumference, with the Earth’s diameter then available after dividing by pi (approximated over a thousand years prior).
What Eratosthenes might have measured, in modern units, was that a 1.5m pole cast a 19cm shadow that day and that Syene was 790km away.
Problem: Show that the line between the top of the pole and the end of its shadow (dotted) is parallel to the line between Syene and the center of the Earth.
Problem: Show that the measure of the angle θ between the pole and the hypotenuse (left) is equal to the measure of the Earth-centric angle θ between Alexandria and Syene (right).
Problem: Show that those measurements compute the radius of the Earth to be about 6,270km.
This method was, however, rather difficult to use, as you needed to travel far distances in order for the shadow length to appreciably change. Moreover, if one of the locations was not experiencing a Lāhainā noon, the computation became much more difficult.
Over a millennium later, in the 10th century, the Islamic mathematician Al-Biruni developed a much better estimate. By using an astrolabe, the Swiss Army knife of historical navigation, one can measure the angle a line connecting your location with an object in the distance makes with horizontal. A classic use of this is to measure the height of objects, like mountains, by measuring how the angle changes as you approach.
Problem: A nearby mountaintop is about 0.41 radians above the horizon. After walking 500 meters closer to it, the angle increases to about 0.47 radians. Show that the mountain is about 1505 meters tall.
Al-Biruni’s insight was that, when positioned above the Earth’s surface, the angle of the horizon drops below horizontal in such a way that depends on the curvature of the Earth!
Problem: Suppose, at the top of the 1505 meter mountain, the horizon is about 0.0218 radians below horizontal. Show the radius of the Earth is about 6,332 km.
Al-Biruni’s estimate, done with more accurate measurements than given in the problem, remained the most accurate measurement of the Earth’s radius for centuries. By the 18th century, however, humans discovered that the Earth is not a perfect sphere, and hence these styles of measurements were inherently noisy. To model the planet as an “oblate spheroid”, measurements needed to be taken along longitude lines. Large triangulation projections across western Europe, mainly in England and France, allowed for a radius computation effectively the same as what we use today: 6357km around the poles, 6371km on average.
Great, now we can make some progress in determining the velocity of our teapot. There are two parts influencing its movement: its current velocity and its acceleration due to gravity.
Problem: If the teapot is moving horizontally at a velocity v (in meters per second) and being accelerated downwards due to gravity with acceleration g, show that after t seconds the teapot will be vt meters away horizontally and gt²/2 meters lower.
Problem: If the teapot is at an altitude of 420km, the same as the International Space Station, show that after t seconds its distance from the center of the Earth will be √[(vt)² + (6791000-gt²/2)² ] meters. (Use the distance formula to compute distance from the center of the Earth)
Problem: Now assume that the teapot is in a circular orbit, meaning that its altitude does not change. Show that this means that its velocity must be equal to √[6791000g — 1/4 g² t²].
Notably, since we really want to say that its instantaneous altitude shouldn’t change, we can plug in t=0 to get v=√[6791000g]; or in general that the velocity for a circular orbit at an altitude of A kilometers is √[(6371000 + A)g].
However, whoops, we also need to know what the gravitational force will be up there in order to continue! In the next part, we will discover how to compute the force of Earth’s gravity at any height.