Prompted by Richard Branson’s and Jeff Bezos’ sub-orbital flights last month, this post and two that follow it will provide spaceship-themed spreadsheet exercises that readers can do with their school-aged kids.
This is a departure from Naptown Numbers’ usual fare, i.e., from questioning what the press or experts have to say. But it’s consistent with our theme that although we laymen don’t know much we do know something: with nothing more than a spreadsheet program and the rudimentary algebra commonly taught in junior high, a young scholar can do actual (if admittedly simplistic) rocket science.
This first post will use spreadsheets to show that rockets’ propellant requirements tend to increase exponentially with the speeds the rockets need to reach, and it will explain without calculus—or at least without requiring any previous knowledge of it—just what the commonly misused term exponentially really means.
Although the only prerequisite is middle-school-level algebra, most kids won’t find these posts particularly easy. But working through the spreadsheets should teach them things about space flight—and math—that even a lot of adults don’t know.
Conservation of Momentum
To determine propellant requirements we start with the concept of “momentum.” Momentum is the product of mass and velocity. The momentum of a 2-kilogram (“kg”) projectile traveling at 500 meters per second (“m/s”) is (2 kg × 500 m/s =) 1000 kg‑m/s. So is the momentum of a 1‑kg projectile traveling at 1000 m/s (because 1 kg × 1000 m/s, too, equals 1000 kg‑m/s).
Physics says that total momentum is “conserved,” i.e., that it doesn’t change. In many contexts that concept isn’t very helpful, because keeping track of momentum for all interacting bodies is often too complicated. Particularly in outer-space contexts, though, useful results can sometimes be obtained even when we ignore certain interactions. So in discussing how much propellant a rocket ship needs in order to reach a given velocity we’ll initially ignore gravity.
If you’re floating stationary in space, your momentum is zero because your velocity is zero. If you then throw a 1‑kg wrench forward at 10 m/s, it acquires 10 kg‑m/s of momentum in the forward direction (1 kg × 10 m/s = 10 kg‑m/s). Since you and the wrench started out with zero momentum, though, the wrench’s momentum has to be canceled out if overall momentum is to be conserved.
This means that, if you and your space suit together weigh 100 kg, then throwing the wrench forward will cause you to float backward at 0.1 m/s (because 100 kg × 0.1 m/s = 10 kg‑m/s). Similarly, one kilogram of propellant issuing at 1000 m/s from a 1000-kg spacecraft will increase the spacecraft’s velocity by 1 m/s (because 1 kg × 1000 m/s = 1000 kg × 1 m/s).
So it may initially seem that the propellant requirement would increase only in proportion to the desired speed increase. For that same propellant velocity, that is, expelling 2 kg should increase speed by 2 m/s, 3 kg should increase it by 3 m/s, etc. And for those numbers that’s pretty close to true. As the total propellant mass gets large, though, we need to take into account an additional fact: that in practice the propellant is expelled only gradually rather than all at once.
To see why that makes a difference, consider a spacecraft that starts out with 3000 kg of propellant together with its 1000‑kg payload. Suppose that only 2000 kg of the propellant is expelled initially, leaving 1000 kg to remain with the payload. The velocity of the now‑2000 kg craft, consisting of the 1000 kg payload and the 1000 kg of remaining propellant, would thereby reach only 1000 m/s (because 2000 kg of the propellant was expelled at our assumed 1000 m/s propellant velocity).
If the remaining 1000 kg of propellant is later expelled all at once, that adds another 1000 m/s to the velocity of the (now payload-only) 1000‑kg craft, raising that velocity to only 2000 m/s, not the 3000 m/s that would have resulted if all 3000 kg had been expelled at the same time.
Of course, the 2000-m/s scenario isn’t realistic, either; increasing the velocity by 1000 m/s in less than a second would exert over a hundred times the force of gravity and therefore crush any human passengers. In real life the propellant is instead expelled gradually, so the final payload velocity would be even less than 2000 m/s.
We can see this by using a spreadsheet like the following to break the propellant expulsions into smaller bursts:
Cell A6 shows that the initial mass m is 4000 kg (3000 kg propellant + 1000 kg payload). Cell A7’s 3900 kg is the result of expelling 100 kg of propellant backward at the 1000 m/s propellant velocity u shown in D1. That expulsion thereby imparts 100,000 kg‑m/s of backward momentum to the expelled propellant.
So a forward-directed 100,000 kg‑m/s of momentum must be acquired by the remaining 3900 kg of payload and propellant. As the spreadsheet’s top-right window shows, Cell C7 therefore calculates the propellant’s momentum change (“D$1*B7”), divides it by the craft’s remaining mass (“/A7”) to obtain the negative of its resultant velocity change, and computes the craft’s new velocity, 26 m/s, by subtracting that negative velocity change from (i.e., adding a positive velocity change to) the previous velocity (“C6”).
Repeating that calculation in subsequent rows until Column A reaches the 1000‑kg payload mass yields Cell C37’s 1425 m/s approximation of the final velocity. Using smaller mass-expulsion increments to simulate continuous propulsion more closely would have made the result even smaller; the result approaches about 1386 m/s as the increment of propellant expulsion approaches zero.
For the benefit of readers who are familiar with differential equations, we set forth the way in which separation of variables was used to arrive at that 1386 m/s value:
What those differential equations imply is that initial mass grows exponentially as a function of final velocity.
Since the remainder of this post is probably too densely mathematical for many people’s tastes, some readers may want to skip to the end, where Equation 1 expresses the implied exponential relationship mathematically and Fig. 1 illustrates it. But those who soldier on will get a spreadsheet demonstration of the exponential relationship as well as an explanation of what the commonly misused term exponentially actually means.
“Exponentially”
Sheet 2 below reverses Sheet 1’s mass-values column to illustrate the following (forbiddingly abstract) definition of exponentially:
A quantity changes exponentially with a variable on which it depends if the quantity’s rate of change with respect to that variable is the product of some proportionality constant and the quantity itself.
Its top-right window displays the formula used in column B to calculate the velocities that can be reached by starting out with corresponding mass values from Column A:
Column E contains the rate Δm/Δv at which the initial mass m required by a desired final velocity v changes with that velocity, while Column F holds the products of a constant, namely, -1/u, and the required initial mass. Column E equals Column F, so according to our definition the required initial mass is an exponential function of the desired final velocity.
Why “Exponentially”?
That relationship is called “exponential” because it’s a characteristic of functions that can be expressed as (possibly some constant multiple of) a constant base raised to a variable power—which is conventionally displayed as an exponent. In the following identity
for example, the exponent 3 indicates that the base 10 is raised to the third power.
The property that makes the function-value change caused by an exponent change proportional to the function value itself is that in multiplication the exponents add:
To see why the proportionality specified in our definition results from this property, compare the following two exponent changes:
Independently of the function value we start with, changing the exponent by 2 changes the result by 99 times the starting value: the change in the function value is proportional to the original function value. And, as the definition requires, so is the rate at which the function changes per unit change in the variable exponent.
To be more precise, let’s call the exponent x, the function value y, and their respective changes Δx and Δy. Then the rate referred to in the definition is the limit, call it y´, that the ratio Δy/Δx approaches as Δx approaches zero: y is an exponential function of x if the ratio y´/ y is a constant. If it is, we’ll call that constant the “proportionality constant.”
To estimate the proportionality constant in the case where the base a is 10, the following spreadsheet’s Cell E5 divides the y value from Cell E1 into the Δy/Δx value in E4 that Cell D4 calls y´. Although Cell D4 isn’t precisely correct in saying that y´ equals Cell E4’s Δy/Δx value—remember, y´ is the limit of Δy/Δx as Δx approaches zero—it’s pretty close if the Δx value in B3 is very small:
So Δy/Δx’s ratio in Cell E5 to the original y value in E1 closely approximates the true proportionality constant. That true proportionality constant is what’s known as the base a’s natural logarithm (“ln(a)”), which as we can see in the top-of-the-spreadsheet formula window is calculated by Cell E6 to show that the approximation in E5 is quite close to the theoretical value.
Euler’s Number
We’ve now seen that with respect to some variable the rate of change of a function defined as a constant raised to that variable is proportional to the function itself, e.g.,
and Sheet 2 showed that as a function of final velocity the ratio of a rocket’s initial-to-final-mass ratio exhibits this relationship.
To connect those observations mathematically we’ll make one further observation: that if we changed Sheet 3’s base-containing cell B1 from 10 to 2 the basic relationship would persist but the proportionality constant would be lower:
That is, for base 2 the proportionality constant is about 0.693 instead of 2.303 as it is for base 10.
So somewhere between 2 and 10 there must be a base value, call it e, such that the proportionality constant exactly equals 1, i.e., such that the function value’s rate of change is numerically equal to the function value itself:
That base e, which is called Euler’s number, is transcendental but approximated by 2.71828, as entering that approximation into Cell B1 would verify.
The Rocket Equation
And Euler’s number can be used to express the relationship between a final velocity and the initial mass required to reach it, as Sheet 4 below will show us.
Column A contains the result μ of normalizing Sheet 2’s initial-mass values by the payload mass. Column C contains a value w that equals the negative of the ratio v/u that the final velocity v bears to the propellant velocity u. And Column F contains the normalized initial mass μ’s rate of change Δμ/Δw with respect to w.
Whereas in Sheet 2 m’s rate of change Δm/Δv equaled a constant (-1/u) times the initial mass m, μ’s rate of change Δμ/Δw in Sheet 4 equals μ itself: here the proportionality constant is 1. From this we can infer that normalized mass μ equals Euler’s number e raised to the power w = -v/u:
(That exponent’s negative sign is an artifact of our having above taken propellant velocity as a negative value in the discussion above. This convention is not typical.)
Fig. 1 depicts that relationship:
Next Time
In the next post we’ll show our young scholars how to calculate orbital velocities, and we’ll use Equation 1 to calculate a lower limit on orbital flights’ propellant requirements. In the post after that we’ll turn our attention to sub-orbital-flight trajectories and see that our results closely match the actual trajectories of Messrs. Branson and Bezos. We’ll also compare those sub-orbital flights’ propellant requirements with requirements for orbital flights.