Random Walk in One Dimension, Part 1

As I continue reading the Feynman Lectures on Physics, I’m intrigued by the discussion of the random walk in Chapter 6: “Probability,” Section 6–3: “The random walk.” Feynman discusses a random walk in one dimension, where at each step an object moves one unit either forward or backward at random. In Figure 6–5, he draws a graph tracing the distance moved over 30 steps, for 3 trials. I remember seeing graphs like this in my textbooks when I was in school, studying them and trying to appreciate them. And I realize that now technology has advanced so much that I can explore these further on my home computer.

Problem: Graph a random walk in one dimension, following more trials over longer periods.

Continue reading “Random Walk in One Dimension, Part 1”


Rocket Propulsion, Part 1

The kinetic energy of an object (the energy it has due to its motion) is equal to ½mv2; that is, half the product of its mass and the square of its velocity. But a question I saw recently pointed out some interesting consequences of this, especially in regards to travel in space.

We often think about constant forces accelerating objects. If the force is constant, so is the acceleration (by F=ma). And if the acceleration is constant, then the velocity increases at a constant rate. It would seem that it would take a constant supply of energy to supply a constant force (that is, the total energy used would increase linearly). But the kinetic energy of the object would increase quadratically. If it took x amount of energy to increase the velocity by a certain amount, it would take 4x to double that and 9x to triple that. Where is this extra energy coming from?

Let’s construct a scenario and put in some test numbers. I’m going to stick to velocities well below the speed of light and analyze the situation according to classical mechanics, not taking relativity into account. We’ll start with summarizing the basic physics involved. If you’re already familiar with the basics of momentum, feel free to skip ahead.

Of course, forces can’t be applied arbitrarily. There has to be a source. Even if a craft has plenty of power, conservation of momentum affects how it can move. Recall that (linear) momentum is the product of mass and velocity (p = mv); each object has a momentum. The sum of the momenta of a closed system (m1v1 + m2v2 + m3v3 + …) is conserved and cannot change; the total momentum of all the objects in the system before an event must be equal to the total momentum afterwards. For an isolated spacecraft, mivimfvf. If a spacecraft is initially at rest, then vi = 0 and so momentum p = 0. If it is to start moving with a positive velocity vf, then it will have a positive momentum mvf. Of course, 0≠mvf. (Or even if it’s not at rest, in order to accelerate, the velocity on the right will be higher than on the left, and therefore, so will the momentum). We either need to add a positive mv term to the left side of the equation or a negative mv term to the right (and since mass is positive, it will have to be an object with positive velocity before or with negative velocity after).

A positive term on the left would represent a second object with positive velocity, flying toward the spacecraft. This could be a laser beam on Earth fired towards the craft, particles from the solar wind pushing a solar sail, and so on. However, it’s not the type of propulsion we are looking for, since it has to be produced by an external source and can only push the craft away from the source.

The other way for momentum to be conserved is to add a negative mv term on the right side of the equation. That is, something must be pushed backwards. On Earth, this could be air (for an airplane), or the Earth itself (for a car, or for people walking). In the near vacuum of space, there is not much to push against, and so any mass pushed backward must come from the spacecraft itself. This is called reaction mass. Let’s designate the craft mass 1, and the propellant mass 2. The momentum equation would thus be (m1 + m2)vim1v1f + m2v2f.

In this example, the questioner suggested the object be a spacecraft with an ion thruster, which uses an electric field to accelerate ions out the rear of the craft, pushing the craft forward. These use only small amounts of mass, and generate a roughly constant thrust. This sounds perfect for our initial exploration, since we’ll get a constant force and acceleration and the mass of the craft shouldn’t change significantly.

Finally, it’s time to put in some test numbers. We’ll start with a spacecraft mass of 1000 kg to start — of course, it will get lighter over time. It will start at rest. Let’s assume it can eject 1 g (=0.001 kg) of mass each second at -1000 m/s. To make it easier, we’ll break this down into 1-second intervals. At t=0, the craft is 1000 kg and is moving at 0 m/s. After the first second, it ejects 1 g of mass at -1000 m/s. This is a momentum of -1 kg·m/s. The craft now still has a mass of 1000 kg (technically 999.999 kg), and since the initial momentum was 0, it must have a momentum of 1 kg·m/s. Dividing by the new mass gives a velocity of 0.001 m/s, and a kinetic energy of 0.0005 J = 0.500 mJ.

We’ll use the momentum equation from above. We know the mass of the craft and the propellant, and the initial velocity. The propellant velocity vpf (the propellant velocity) is vi – 1000 (starting at the velocity of the craft, and being ejected at a relative -1000 m/s). If mc is the mass of the craft, mp is the mass of the propellant, and mc’ = mcmp is the new mass of the craft, then

m_c v_i = m_{c^\prime}v_f + m_p v_{pf}

m_c v_i - m_p v_{pf} = m_{c^\prime} v_f

\frac{m_c v_i - m_p v_{pf}}{m_{c^\prime}} = v_f

Or put simply, we need to find the initial momentum, subtract the final momentum of the propellant to get the final momentum of the craft, and divide by the new mass to find the final velocity (at each step).

Continuing in this fashion, at t = 2, the mass is down to 999.998 kg, velocity is 0.002 m/s, and kinetic energy is 2.00 mJ. Let’s look at the first minute:

Time / s Mass / kg Velocity / (m/s) Acceleration / (m/s²) Kinetic energy / mJ
0 1000 0 0 0
10 1000 0.0100 0.00100 50
20 1000 0.0200 0.00100 200
30 1000 0.0300 0.00100 450
40 1000 0.0400 0.00100 800
50 1000 0.0500 0.00100 1250
60 1000 0.0600 0.00100 1800

Over this minute, we can see that the craft stays roughly the same mass (it drops to 999.94 kg) and has roughly constant acceleration (it actually increases from 0.001000000 to 0.00100006 m/s²). But the kinetic energy is going up quadratically, as one would expect with constant acceleration and subsequent linearly increasing velocity. What are we missing? Of course, we’ve ignored the propellant that’s been ejected. It may have very low mass compared to the craft, but it has proportionally high velocity, and remember that kinetic energy is ½mv2. In that first second when the 1,000,000-g craft reached 0.001 m/s and a kinetic energy of 0.500 mJ, we ejected 1 g of reaction mass at -1000 m/s. With this high velocity, this mass has a kinetic energy of 500,000 mJ! That is one million times the kinetic energy of the craft. The combined kinetic energy is 500,000.5 mJ. Let’s look at what happens over the first minute, remembering that we have to add up the kinetic energy of each block of mass that’s been ejected up to that point.

Time / s Kinetic energy / mJ
Spacecraft Propellant Total
0 0 0 0
10 50 4,999,955 5,000,005
20 200 9,999,810 10,000,010
30 450 14,999,565 15,000,015
40 800 19,999,220 20,000,020
50 1250 24,998,775 25,000,025
60 1800 29,998,230 30,000,030

This, then, is the resolution to the paradox. The vast majority of the kinetic energy goes to the propellant, not to the spacecraft. The spacecraft does accelerate and its kinetic energy does go up quadratically, but it is only a tiny fraction of the total kinetic energy (which in this example, goes up by 5,000,005 mJ every 10 seconds).

Of course, we used a model where the amount of mass ejected was very small relative to the craft. What will happen over longer periods, or if we use more mass? Any quadratically increasing function will eventually pass a linear one. In order to explore these issues, we’ll have to make our model a bit more rigorous (in particular, we have to clarify how our drive works). I’ll look at these in the next post.

Individual Action is Not Enough

The Canadian chapter of the World Wildlife Federation produced this cool video/commercial:

Most people try, to at least some degree, to take steps to help the environment. And these small changes, when summed across the whole population, are significant. But still, the collective action of individuals can only do so much — government and industry need to be on board, too. Unfortunately, in the United States, leadership from the federal government has been lacking (and at times, actively impedes) so state and local governments and industry have had to take their own steps. There is some hope, though, that this will change when the Obama administration takes office (I hope to write more on this in a later post).