A Moon Mission Without Calculus
This is a consolidation of the recent four-part series for the purpose of submission to the 3Blue1Brown contest.
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 all but allowed us to get to the Moon, even if we only achieved the engineering capacity in the past century.
This article will show the mathematics of getting to the Moon, without formally using calculus. It is going to be a “paint by numbers” sort of deal: 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? (Part 1)
- 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.
Part 1: The Size of the Earth
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 their 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 projects 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 a height of H kilometers is √[(6371000 + H)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.
Part 2: Computing Gravity
Thankfully, during that time when the measurement of Earth’s radius was being refined, so too was our understanding of gravity. Towards the end of the 17th century, Isaac Newton developed the Law of Universal Gravitation, g = GM/r².
The acceleration due to gravity, g, can be computed by knowing the Gravitational Constant, G; the mass of the Earth, M; and the distance to the center of the Earth, r. Unfortunately, so far we don’t know G or M — and computing the mass of the Earth accurately is well beyond our capacity in this article (In real life, we actually compute the mass of the Earth via this gravitational formula, estimating the value of G through extremely delicate experiments).
Thankfully, it suffices to compute the value of the product GM, since neither G nor M will change throughout our spaceflight. If we can measure the gravitational acceleration at the surface of the Earth, where r=6371000, then we can use Newton’s Law to solve for GM. Unfortunately, the obvious experiment of “drop a ball from a height and measure the time needed to fall” is too imprecise due to the effects of air resistance, imprecision in starting and stopping the timer, and the heights required. In Galileo’s famous experiment dropping items off the Tower of Pisa, the objects would have taken about 3 seconds to hit the ground. With human reaction times of about a quarter-second, reaction time could introduce as much as 16% relative error!
Instead, let’s look at a pendulum. Pendula are helpful because of their small and oscillatory nature: we can easily re-run experiments several times with different parameters to make different measurements. Moreover, the movement of a pendulum is caused by gravity, and hence if we can determine how a pendulum actually moves, we might be able to determine the value of gravitational acceleration.
Problem: Consider a pendulum with length L and a heavy mass on the end. If the pendulum cord makes an angle θ with vertical, show that the acceleration of the bob is -g sin(θ).
Tracking the 2D position of the pendulum bob is a little tedious, so let’s instead track the angle the cord makes with vertical. The angular acceleration is computed by dividing the linear acceleration by the radius, yielding -g sin(θ)/L. We hence have a formula for the angular acceleration in terms of the angle itself.
We are about to brush very close to using calculus, which we shall avoid as follows:
If we think of the angular velocity as the “change in angle over time”, we might approximate it as follows:
The angular acceleration is similarly the “change in angular velocity over time”, or the “change in the change in angle over time, over time”, which can be approximated as
While we’re approximating velocities and accelerations, we might as well approximate our trigonometric functions. The small-angle approximations of sin(x) ≈ x and cos(x) ≈ 1-x²/2 have been known for millennia, with one small catch that requires us to progress our timeline: these small-angle approximations only hold as stated when the angles are measured in radians. It is not clear when the radian entered into widespread use, with the quantity first being used by Roger Cotes in 1714 but the term ‘radian’ not occurring until 1873.
Problem: Show that cos(kt) approximately solves the following difference equation (use the sum of cosines formulas and the small-angle approximations)
Knowing this, we realize that the following equation approximately satisfies our formula for angular acceleration, and so approximately models the movement of the pendulum (calculus allows us to be slightly more precise, but still inexact in this case).
Problem: Show that the period of a pendulum with cord length L is 2π √[L/g].
Problem: A “seconds pendulum”, a pendulum with a period of 2 seconds, has a length of 0.9936 meters. Show that gravitational acceleration at the pendulum is g=9.806 m/s².
It is actually not a coincidence that the length of a seconds pendulum is so close to 1 meter; in fact, at one point the meter was defined to be the length of a seconds pendulum. However, variations in Earth’s surface gravity made this unreliable over long distances, and so the definition was changed. Defining the meter in this way reveals another non-coincidence: that g is approximately π² — with L=1, this is forced by our requirement that the seconds pendulum have a period of 2 seconds.
Problem: Recalling that the radius of the Earth is about 6371 km, show that the value of GM is about 3.980 × 10¹⁴
It turns out that the true value of GM is about 3.986 × 10¹⁴, so we aren’t too far off! The difference comes because the Earth’s radius is meant to be the ‘average’ radius, and 9.806 the ‘average’ gravity, however since the radius is squared in this computation the averaging does not cancel out exactly. We will use the more exact value from here out.
Having finally computed GM, we can determine orbital velocity.
Problem: Show that at 420km the gravitational acceleration is 8.64 m/s², and hence the orbital velocity of our orbiting teapot is 7660 m/s. Derive the following formula for circular orbital velocity with radius R: v=√[GM/R]
Because GM is so large, the orbital velocity is really quite fast — over 27,000 km/h or 17,000 mph; about 300 times faster than highway speed limits!
Now that we know how gravity works, we’ve started to understand how circular orbits work. However, there’s more about orbits we’d like to know: how long does it take an object to complete an orbit? How do elliptical orbits work?
Part 3: Orbital Mechanics
Now, we’re going to need to figure out a little bit of how objects move in orbit, particularly non-circular orbits. To get a foothold, let’s look at two fundamental laws of Physics: Conservation of Angular Momentum and Conservation of Energy. For simplicity, we will consider the massless versions of these laws.
The Conservation of Angular Momentum was first described in its full generality by Newton, however, we will only need a particular case known as Kepler’s Second Law. The idea is that for any orbiting object with fixed mass, there is a quantity A=ωr, where ω is the angular velocity and r is the distance from what it is orbiting. If no outside forces are applied (like igniting a rocket engine), the only way for ω to increase is for r to decrease. This is often described through an analogy of a figure skater, who spins much faster with arms tucked in at the chest compared to having them extended. In terms of Kepler’s Second Law, it is described as a conserved area, as “A line segment joining an orbiting object and the object it’s orbiting sweeps out equal areas during equal intervals of time.”
The Conservation of Energy originates from Émilie du Châtelet, which for our purposes says that for any orbiting object with fixed mass there is a quantity E (energy density) such that E=-GM/r + v²/2 is constant, assuming no energy is put into the system (like igniting a rocket engine; we are actually conserving energy per unit mass). In this equation, r is the distance between the object and what it is orbiting and v is its velocity; the first term is the object’s potential energy. Similar to the gh term in introductory physics, it has a different form here since we can’t assume g is constant anymore. The second term is the usual form of kinetic energy, with the mass term ignored.
One particularly counterintuitive fact about orbits is that if you go faster, it can take longer to complete an orbit. Let’s see why that is the case! From this point forward, we’ll be using r to denote the radius of the orbit, a value that we previously have written 6371000+H.
Problem: Show that √[GMr] represents the (massless) angular momentum of an object in a circular orbit of radius r.
Problem: Recall that the circumference of a circle of radius r is 2π r. Derive the following formula for the orbital period T, or how long it takes to complete an orbit, of a circular orbit of radius r:
This is usually written in the form of Kepler’s Third Law, which we have just derived:
For educational reasons, we have actually reversed chronological order. Kepler’s Third Law was discovered first, in 1619, whereas Newton’s Law of Universal Gravitation was published in 1687. In reality, Kepler recognized this “square-cube law” empirically — one of the first uses of a log-log plot — from which the inverse-square law of gravitation can be deduced.
Problem: Derive the following formula for the orbital period T of a circular orbit with velocity v: T = 2πGM/v³
Problem: The International Space Station orbits at about 420km, and so has an orbital speed of about 7.66 km/s. Assuming it has a circular orbit, show that it takes a little less than 93 minutes to complete its orbit.
The ISS computation shows that if we know the orbital radius of an object, we can compute its orbital period. This also works in reverse, which is lucky for us — our goal is to go to the Moon!
Problem: The Moon takes 27.321 days to complete an orbit. Assuming the Moon has a circular orbit, show that the Moon is about 383,000km away.
Great! Without even needing to go to space, we can compute the distance to the Moon… assuming it has a circular orbit. However, what we’ve found is what’s called the “semimajor axis”: the average of the closest and furthest points of the orbit.
This section began by mentioning that there are some counterintuitive aspects to orbital dynamics, like how going faster increases your orbital period. To be sure, this is not the case when comparing circular orbits:
Problem: Show that the orbital period of an circular orbit at 1km/s is much slower than that of a circular orbit of 2km/s.
However, if an object in a circular orbit increases its velocity (like by igniting a rocket engine), its orbit will become elliptical. Let’s work out what happens if it speeds up:
Problem: A spacecraft is orbiting the Earth in an elliptical orbit such that at its closest point (periapsis) it is 420km above the surface and traveling 8 km/s; 0.34 km/s faster than the ISS at the same altitude. Show that the potential energy is -GM/6791000, and the kinetic energy is about 3.2 × 10⁷ J.
Problem: By the Conservation of Energy, at the spacecraft’s highest point (apoapsis) its total energy must be conserved. Show that the spacecraft’s velocity vₐ at its apoapsis can be written in terms of its height rₐ at its apoapsis in the following way, by solving the Energy equation for vₐ²
Problem: Since at the periapsis and apoapsis the spacecraft’s velocity has no vertical component, it is simple to convert from orbital velocity to angular velocity. Use Conservation of Angular Momentum to derive the following relationship between the velocity and altitude of the spacecraft at is apoapsis: rₐ = 5.433 × 10¹⁰ / vₐ.
Problem: Show that the velocity of the spacecraft at its apoapsis is 6673 m/s, and hence its altitude above the Earth’s surface is 8140 km.
While we’ve only really dealt with circular orbits so far, much of our work can be applied to elliptical orbits. Here, instead of using the radius of the circular orbit, we parameterize by the semimajor axis, or the average of the apoapsis and periapsis lengths. For example, our spacecraft has a semimajor axis of 7,466 km.
Problem: Let a be the semimajor axis of an ellipse, and note that the period of an elliptical orbit is equal to that of a circular orbit with radius equal to the semimajor axis, as below. Show that the period of our spacecraft’s orbit is just about 107 minutes.
Notice how, even though the spacecraft is going faster than the ISS at its periapsis, it takes longer to complete an orbit! Similarly, the spacecraft is going slower than needed for a circular orbit at its apoapsis, but its orbital period is shorter!
Problem: Derive the following relationship between velocity and height at periapsis, vₚ and rₚ, and velocity and height at apoapsis, vₐ and rₐ:
Problem: Derive the vis-viva equation: At any point in an elliptical orbit with semimajor axis a, the velocity v satisfies v² = GM(2/r — 1/a). (Hint: Remember that 2a = rₚ + rₐ, and use Conservation of Energy).
With the ability to compute orbital speeds and orbital periods for arbitrary orbits, we not only know how far away the Moon is, but can finally plan out our mission! In the next part, we’ll piece everything together and prepare to launch!
Part 4: Reaching the Moon!
We finally know enough to put together a crude Moon mission! Let’s collect everything we’ve learned:
Thanks to Eratosthenes, Al-Biruni, and the triangulation of England and France, we know the Earth’s radius to be about 6,371 km
Thanks to Newton and classical trigonometry we know the Earth’s gravitational parameter GM is about 3.986 × 10¹⁴
Thanks to Newton, Kepler, and du Châtelet, we have determined several properties of orbital mechanics, that
- A circular orbit with radius r has orbital velocity v = √[GM/r]
- An elliptical orbit with semimajor axis a (average of the periapsis or lowest point of the orbit and the apoapsis or highest point) has a period given by
- An object on an elliptical orbit with semimajor axis a has a velocity v which depends on its current distance from the orbiting object, r, in the following way:
This also allowed us to determine that the Moon orbits at about 383,000km.
Now, let’s plan our Moon mission in reverse order:
Problem: Our spacecraft is traveling in an elliptical orbit to the Moon, with periapsis (lowest point) in Low Earth Orbit at 6,791km and apoapsis (highest point) in Lunar Orbit at 383,000km. Show that it takes a little less than 5 days to travel from periapsis to apoapsis.
Problem: Show that the Moon will travel about 1.14 radians through its orbit during this time, and hence we should set our periapsis to be about 2 radians behind the Moon.
Problem: Show that our spacecraft should be traveling at 10.74 km/s at its periapsis of 6,791 km to achieve this elliptical orbit, which is 3.079 km/s faster than needed for a circular orbit at that altitude.
Problem: Our spacecraft starts on the surface of the Earth, traveling 465.1m/s due to the rotation of the Earth. Show that we must increase our velocity by at least 7570m/s to reach Low Earth Orbit at 420km in altitude, and that this part of the trip will take about 44 minutes. Ignore the effect of air resistance for this computation.
Problem: When our spacecraft reaches Low Earth Orbit that way, show that it will be traveling 122 m/s slower than needed for a circular orbit at that altitude.
Hence, we have the following mission plan: We launch our rocket from the surface and quickly increase its velocity by 7570 m/s, reaching Low Earth Orbit about 44 minutes later, disregarding (substantial) effects of air resistance. Once there, we increase our velocity by 122 m/s to 7660 m/s, circularizing our orbit. We now wait in orbit until the Moon is in the right position — 2 radians ahead of us — and increase our velocity by 3,079 m/s in order to raise our apoapsis to Lunar orbit. A little less than 5 days later, we arrive! While we spend most of the time en route to the Moon, certainly most of the effort is spent just getting into space! That’s why the Saturn V rockets had to be so huge, especially compared to the size of the orbiter.
You may have noticed a slight incongruity here, in how this mission plan corresponds to real-life Moon missions. And indeed, we have overlooked a large factor in our mission! I will end this series with a question that won’t immediately be answered: we’ve spent so much time developing the mathematics, that it’s important to look for things we might have missed!
Problem: Our computations predicted it would take a little under 5 days to reach the Moon, which would mean a little under 10 days to complete a round trip. Apollo 10 traveled to the Moon and back in a little over 8 days. What aspects of the Apollo 10 mission did we fail to account for in our computations, and why would they result in such a large discrepancy of nearly two days?
I hope you enjoyed this series and our trip to the Moon, and moreover, I hope this series has instilled an appreciation for the trigonometry and algebra that are often out-shined by calculus in school curricula. Mathematics did not start in the 17th century; some of the tools we used in this series have been known for millennia, developed independently by cultures across the world because they are so powerful and useful for tackling incredibly complex problems!
If you have any questions about the material in this series, please don’t hesitate to reach out!
