3Problems and Solutions—Review Lecture C
(There was no summary for this lecture.)
(There was no summary for this lecture.)
We’re continuing this review of how to do physics by doing a number of problems. All of the problems I chose are elaborate and complicated and difficult; I’ll leave you to do the easy problems. Also, I suffer from the disease that all professors suffer from—that is, there never seems to be enough time, and I invented more problems than undoubtedly we’ll be able to do, and therefore I’ve tried to speed things up by writing some things on the board beforehand, with the illusion that every professor has: that if he talks about more things, he’ll teach more things. Of course, there’s only a finite rate at which material can be absorbed by the human mind, yet we disregard that phenomenon, and in spite of it we go too fast. So, I think I’ll just go along slowly, and see how far we get.
3–1Satellite motion
The last problem that we were talking about was satellite motion. We were discussing the question of whether a particle that was moving perpendicular to the radius of the sun, of a planet, or any other mass $M$, at a distance $a$, and having the escape velocity at that distance, would, in fact, escape—because it’s not self-evident. It would be, if it were headed straight out, radially; but whether it would make it or not if it were headed perpendicular to the radius, is another question. (See Fig. 3-1.)
It turns out that—if we can remember some of Kepler’s laws, and add some other laws like the conservation of energy—we can figure out that if the particle didn’t escape, it would make an ellipse, and we can figure out how far away it would get, and that’s what we’re going to do now. If the perihelion of the ellipse is $a$, how far is the aphelion, $b$? (By the way, I tried to write this problem on the board, but I found I couldn’t spell “perihelion”!) (See Fig. 3-2.)
Last time we figured out the escape velocity by using the conservation of energy. (See Fig. 3-3.) \begin{equation} \begin{aligned} \textit{K.E.}+\textit{P.E.}\text{ at }a &= \textit{K.E.}+\textit{P.E.}\text{ at }\infty\\[1.5ex] \frac{mv_\text{escape}^2}{2} - \frac{GMm}{a} &= 0+0\\[1ex] \frac{v_\text{escape}^2}{2} &= \frac{GM}{a}\\[1ex] v_\text{escape} &= \sqrt{\frac{2GM}{a}} \end{aligned} \label{Eq:TIPS:3:1} \end{equation}
Now, this is the formula for the escape velocity at the radius $a$, but suppose the velocity $v_a$ is arbitrary, and we’re trying to find $b$ in terms of $v_a$. The conservation of energy tells us that the kinetic energy plus the potential energy of the particle at the perihelion must equal the kinetic energy plus the potential energy at the aphelion—and that’s what we can use to calculate $b$, at first sight: \begin{equation} \label{Eq:TIPS:3:2} \frac{mv_a^2}{2} - \frac{GMm}{a} = \frac{mv_b^2}{2} - \frac{GMm}{b}. \end{equation}
Infelizmente,^{1} however,we do not have $v_b$ so unless there’s some external machinery or analysis to obtain $v_b$, we’re never going to solve Eq. (3.2) for $b$.
But if we remember Kepler’s law of equal areas, we know that in a given time the same area is swept out at the aphelion as is swept out at the perihelion: in a short time $\Delta t$ the particle at the perihelion moves a distance $v_a\Delta t$ so the area swept out is about $a v_a\Delta t/2$ while at the aphelion, where the particle moves $v_b\Delta t$ the area swept out is about $b v_b\Delta t/2$. And so “equal areas” means that $a v_a\Delta t/2$ equals $b v_b\Delta t/2$—which means that the velocities vary inversely as the radii. (See Fig. 3-4.) \begin{equation} \begin{aligned} a v_a \Delta t/2 &= b v_b \Delta t/2\\[1ex] v_b &= \frac{a}{b}v_a \end{aligned} \label{Eq:TIPS:3:3} \end{equation}
That gives us, then, a formula for $v_b$ in terms of $v_a$ which we can substitute in Eq. (3.2). Then we will have an equation to determine $b$: \begin{equation} \label{Eq:TIPS:3:4} \frac{mv_a^2}{2} - \frac{GMm}{a} = \frac{m\bigg({\displaystyle \frac{a}{b}v_a}\bigg)^2}{2} - \frac{GMm}{b}. \end{equation} Dividing by $m$, and rearranging, we get \begin{equation} \label{Eq:TIPS:3:5} \frac{av_a^2}{2}\bigg(\frac{1}{b}\bigg)^2\!\!\!- GM\bigg(\frac{1}{b}\bigg)\!+\!\bigg(\frac{GM}{a} - \frac{v_a^2}{2}\bigg)\!=0. \end{equation}
If you look at Eq. (3.5) a while, you could say, “Well, I can multiply by $b^2$ and then it’ll be a quadratic equation in $b$,” or, if you prefer, you could look at it just the way it is, and solve the quadratic equation for $1/b$—either way. The solution for $1/b$ is \begin{equation} \begin{aligned} \frac{1}{b} &= \frac{GM}{a^2v_a^2} \pm \sqrt{\bigg(\frac{GM}{a^2v_a^2}\bigg)^2 + \frac{v_a^2/2-GM/a}{a^2v_a^2/2}}\\[1ex] &= \frac{GM}{a^2v_a^2} \pm \bigg(\frac{GM}{a^2v_a^2} - \frac{1}{a}\bigg). \end{aligned} \label{Eq:TIPS:3:6} \end{equation}
I’m not going to discuss the algebra from here on; you know how to solve a quadratic equation, and there are two solutions for $b$: one of them is $b$ equals $a$, it turns out—and that’s happy, because if you look at Eq. (3.2) you see it’s obvious that if $b$ equals $a$, the equation will match. (Of course, that doesn’t mean that $b$ is $a$.) With the other solution, we get a formula for $b$ in terms of $a$, which is given here: \begin{equation} \label{Eq:TIPS:3:7} b = \frac{a}{\displaystyle \frac{2GM}{av_a^2}-1}. \end{equation}
The question is whether we can write the formula in such a way that the relationship of $v_a$ to the escape velocity at the distance $a$ can readily be seen. Notice that by Eq. (3.1) $2GM/a$ is the square of the escape velocity, and therefore we can write the formula this way: \begin{equation} \label{Eq:TIPS:3:8} b = \frac{a}{(v_\text{escape}/v_a)^2-1}. \end{equation}
That’s the final result, and it is rather interesting. Suppose, first, that $v_a$ is less than the escape velocity. Under those circumstances, we’d expect the particle not to escape, so we should get a sensible value for $b$. And sure enough, if $v_a$ is less than $v_\text{escape}$, then $v_\text{escape}/v_a$ is greater than $1$, and the square is also greater than $1$; taking away $1$, you get some nice positive number, and $a$ divided by that number tells us $b$.
To check roughly how accurate our analysis is, a good thing to play around with is the numerical calculation we made of the orbit in the ninth lecture,^{2} to see how close the $b$ that we calculated then agrees with the $b$ we get from Eq. (3.8). Why should they not agree perfectly? Because, of course, the numerical method of integration treats time as little blobs instead of continuous, and therefore it isn’t perfect.
Anyway, that’s how we get $b$ when $v_a$ is less than $v_\text{escape}$. (Incidentally, knowing $b$ and knowing $a$, we know the semi-major axis of the ellipse, and thus we could figure out the period of the orbit from Eq. (3.2), if we wanted to.)
But the interesting thing is this: suppose, first, that $v_a$ is exactly the velocity of escape. Then $v_\text{escape}/v_a$ is $1$, and Eq. (3.8) says that then $b$ is infinite. That means that the orbit is not an ellipse; it means that the orbit goes off to infinity. (It can be shown that it is a parabola, in this special case.) So, it turns out, that if you’re anywhere near a star or a planet, and no matter what direction you’re moving, if you have the velocity of escape, you’ll escape, all right—you won’t get caught, even though you’re not pointed in the right direction.
Still another question is, what happens if $v_a$ exceeds the velocity of escape? Then $v_\text{escape}/v_a$ is less than $1$, and $b$ turns out negative—and that doesn’t mean anything; there is no real $b$. Physically, that solution looks more like this: with a very high velocity, much higher than the velocity of escape, a particle coming in is deflected—but its orbit is not an ellipse. It is, in fact, a hyperbola. So the orbits of objects moving around the sun are not only ellipses, as Kepler thought, but the generalization to higher speeds includes ellipses, parabolas, and hyperbolas. (We didn’t prove here that they are ellipses, parabolas, or hyperbolas, but that’s the answer to the problem.)
3–2Discovery of the atomic nucleus
This hyperbolic orbit business is interesting, and has a very interesting historical application, which I’d like to show you; it is illustrated in Figure 3-5. We take the limiting case of an enormously high speed, and a relatively small force. That is, the object is going by so fast that in the first approximation it goes in a straight line. (See Fig. 3-5.)
Suppose we have a nucleus with charge $+Zq_\text{el}$ (where $-q_\text{el}$ is the electron charge), and a charged particle that is moving past it at a distance $b$—an ion of some kind (it was originally done with an alpha particle), it doesn’t make any difference; you can put in your own case—let’s take a proton of mass $m$, velocity $\FLPv$, and charge $+q_\text{el}$ (for an alpha particle, it would be $+2q_\text{el}$). The proton doesn’t go quite in a straight line, but is deflected through a very small angle. The question is, what’s the angle? Now, I’m not going to do it exactly, but roughly—to get some idea of how the angle varies with $b$. (I’ll do it nonrelativistically, although it’s just as easy to take relativity into account—just a minor change that you can figure out for yourself.) Of course, the bigger $b$ is, the smaller the angle ought to be. And the question is, does the angle decrease as the square of $b$, the cube of $b$, as $b$, or what? We want to get some idea about this.
(This is, as a matter of fact, the way you start on any complicated or unfamiliar problem: you first get a rough idea; then you go back when you understand it better and do it more carefully.)
So the first rough analysis will run something like this: as the proton flies by, there are sideways forces on it from the nucleus—of course, there are forces in other directions too, but it’s the sideways force that makes it deflect so instead of going straight as it did before, it now has an upward component of velocity. In other words, it acquired some upward momentum as a result of the forces in that direction.
Now, how big is the upward force? Well, it changes as the proton goes along, but more or less, roughly, the force has to depend on $b$, and the maximum force (as the proton is passing the central position) is \begin{equation} \label{Eq:TIPS:3:9} \text{vertical force} \approx \frac{Zq_\text{el}^2}{4\pi\epsilon_0 b^2} = \frac{Ze^2}{b^2}. \end{equation} (I substituted $e^2$ for $\displaystyle \frac{q_\text{el}^2}{4\pi\epsilon_0}$ so I can write the equations quicker.^{3})
If I knew how long that force acted, I could estimate the momentum that was delivered. How long does the force act? Well, it doesn’t act when the proton is a mile away, but, roughly speaking, a force of that general order of magnitude is acting as long as the proton is in the general neighborhood. How far? More or less, when it’s passing within a distance b of the nucleus. So the time during which the force acts is of an order of magnitude of the distance $b$ divided by the speed, $v$. (See Fig. 3-6.) \begin{equation} \label{Eq:TIPS:3:10} \text{time} \approx \frac{b}{v}. \end{equation}
Newton’s law says that force equals the rate of change of the momentum—so, if we multiply the force by the time over which it’s acting, we get the change in momentum. Therefore, the vertical momentum acquired by the proton is \begin{equation} \begin{aligned} \text{vertical momentum} &= \text{vertical force} \cdot \text{time}\\[1ex] &\approx \frac{Ze^2}{b^2} \cdot \frac{b}{v} = \frac{Ze^2}{bv}. \end{aligned} \label{Eq:TIPS:3:11} \end{equation}
That’s not exactly right; ultimately, when we do an exact integration of this thing, there may be a numerical factor of $2.716$ or something—but for now, we’re just trying to find the order of magnitude as it depends on the various letters.
The horizontal momentum that the particle has when it comes out is, for all intents and purposes, the same as when it went in, which is $mv$: \begin{equation} \label{Eq:TIPS:3:12} \text{horizontal momentum} = mv. \end{equation} (This is the only thing you need to change to take relativity into account.)
Now, then, what is the angle of deflection? Well, we know the “up” momentum is $Ze^2/bv$ and the “sideways” momentum is $mv$, and the proportion of “up” to “sideways” is the tangent of the angle—or, practically, the angle itself, since it’s so small. (See Fig. 3-7.) \begin{equation} \label{Eq:TIPS:3:13} \theta \approx \frac{Ze^2}{bv}{\Large/}{mv} = \frac{Ze^2}{bmv^2} \end{equation}
Eq. (3.13) shows how the angle depends on the velocity, on the mass, on the charge, and on the so-called “impact parameter”—the distance $b$. When you actually calculate $\theta$ by integrating the force instead of just estimating it, it turns out that there is indeed a numerical factor missing, and that factor is exactly $2$. I don’t know whether you’ve gotten that far in integrations or not: if you can’t do it, all right; it’s not essential, but the correct angle is \begin{equation} \label{Eq:TIPS:3:14} \theta = \frac{2Ze^2}{bmv^2}. \end{equation} (Actually, you can work the formula out exactly for any hyperbolic orbit, but never mind: you can understand everything for this case, for small angles. Of course Eq. (3.14) is not true when the angles get to $30$ or $50$ degrees; then we’ve made too rough an approximation.)
Now, this has a very interesting application in the history of physics—it is the way Rutherford discovered that the atom has a nucleus. He had a very simple idea: by making an arrangement in which alpha particles from a radioactive source would go through a slit—so he knew that they were going in a definite direction—and letting them impinge on a zinc sulfide screen, he could see scintillations in a single spot right behind the slit. But if he put a gold foil between the slit and the screen, the scintillations would sometimes appear elsewhere! (See Fig. 3-8.)
Of course, the reason was, the alpha particles coming past the little nuclei in the gold foil were deflected. By measuring the angles of deflection and using Eq. (3.14) in reverse, Rutherford was able to obtain the distances, $b$, required to produce that much deflection. The great surprise was, these distances were very much smaller than an atom. Before Rutherford made this experiment it was believed that the positive charge of the atom was not concentrated at a point in the center, but distributed uniformly throughout. Under those circumstances, the alpha particle could never get the big force needed to make the observed deflections, because if it were outside the atom it wouldn’t be close enough to the charge, and if it were inside the atom there’d be as much charge above it as below it, and that wouldn’t produce enough force. So it was demonstrated by the large deflections that there were sources of strong electric force inside the atom, and then it was guessed that there must be a central point where all the positive charges are, and by observing the deflections as far out as possible, and how many times they occurred, one could obtain an estimate of how small $b$ might be, and ultimately obtain the size of the nucleus—and the size of the nucleus turned out to be $10^{-5}$ times smaller than the atom! This was the way that it was discovered that nuclei exist.
3–3The fundamental rocket equation
Now, the next problem I want to talk about is completely different: it has to do with rocket propulsion, and I’m going to take a rocket floating around in empty space first—forgetting all about gravity, and so on. The rocket’s built to hold a lot of fuel; it’s got some kind of engine by which it squirts fuel out the back—and from the point of view of the rocket, it’s always squirting it out at the same speed. It doesn’t turn on and off; we start it, and it just keeps squirting stuff out the rear end until it runs out. We’ll suppose that the stuff is squirted out at a rate of (that’s mass per second), and that it goes out at velocity $\FLPu$. (See Fig. 3-9.)
You might say, “Aren’t those the same thing? You know the mass per second; isn’t that the velocity?”
No. I can dump a certain amount of mass per second by taking a great big lump of stuff and putting it quietly out each time, or I can take the same mass and throw it out each time. So, you see, they’re two independent ideas.
Now, the question is, how much velocity will the rocket accumulate after a time? Suppose, for instance, that it uses up $90$ percent of its weight: that is, when it’s finished using all its fuel the mass of the shell that’s left is one-tenth as great as the mass of the whole thing loaded before it started. What speed will the rocket acquire?
Anybody in his right mind would say that it is impossible to get any faster than the speed $u$, but that’s not true, as you’ll see in a moment. Maybe you’ll say that’s perfectly obvious; well, all right. But it is, in fact, true for the following reason.
Let’s look at the rocket at any moment, moving at any speed at all. If we move along with the rocket and watch for a time $\Delta t$, what do we see? Well, there’s a certain mass $\Delta m$ that goes out—which is, of course, the rocket’s rate of loss $\mu$ times the time $\Delta t$. And the velocity that this mass comes out at is $\FLPu$. (See Fig. 3-10.)
Now, the moment after this mass is thrown back, how fast is the rocket moving forward? The speed at which it’s moving forward must be such that the total momentum is conserved. That is to say, it picks up a little speed $\Delta v$, in such a manner that, if the mass of the rocket shell and remaining fuel at that instant is $m$, then $m$ times $\Delta v$ matches the outgoing momentum during that time, which is $\Delta m$ times $u$. And that’s all there is to the theory of rockets; that’s the fundamental rocket equation: \begin{equation} \label{Eq:TIPS:3:15} m \Delta v = u \Delta m. \end{equation}
We could put in $\mu\Delta t$ for $\Delta m$ and by fiddling around, find out how long it takes to get up to a given velocity,^{4} but our problem is to find the final velocity, and we can do that directly from Eq. (3.15): \begin{equation} \begin{gathered} \frac{\Delta v}{\Delta m} = \frac{u}{m}\kern{1ex}\\[1ex] dv = u \frac{dm}{m}. \end{gathered} \label{Eq:TIPS:3:16} \end{equation}
In order to find the velocity that the rocket acquires, starting from rest, you integrate $u(dm/m)$ from the initial mass to the final mass. Now, $u$ was assumed constant, so it can be taken outside the integral, and we have, therefore, \begin{equation} \label{Eq:TIPS:3:17} v = u\kern{-1ex}{\large\intop_{m_\text{initial}}^{m_\text{final}}\kern{-1.5ex}}\lower.5ex\frac{dm}{m}. \end{equation}
The integral of $dm/m$ may or may not be known to you; let’s suppose that it isn’t. You say, “$1/m$ is such a simple function, I must know the derivative: I’ll fiddle around with differentiating things until I find it.”
But it turns out you can’t find anything that’s simple—in terms of $m$, powers of $m$, and things like that—which, when you differentiate it, gives $1/m$. So, not knowing how to do it that way, we’ll do it a different way. We’ll do it by numerical integration.
Remember: Whenever you’re stuck in a mathematical analysis, you can always do it by arithmetic!
3–4A numerical integration
Let’s suppose that the initial mass is $10$, and take as a simple approximation that we drop one unit of mass at a time. Furthermore, let’s measure all the velocities in terms of the unit $u$, because then we will have simply $\Delta v = \Delta m/m$.
We want to find the total accumulated velocity. Well, let’s see: during the first dropping of one unit of mass, how much speed is acquired? Well, that’s easy; it’s \begin{equation*} \Delta v = \frac{\Delta m}{m} = \frac{1}{10}. \end{equation*}
But that isn’t exactly right, because while you’re spitting one unit of mass out, the mass that’s reacting is not $10$; when you’re all finished spitting it out, it’s only $9$. You see, after $\Delta m$ is shot out, the mass of the rocket is only $m - \Delta m$ so maybe it would be better to put \begin{equation*} \Delta v = \frac{\Delta m}{m - \Delta m} = \frac{1}{9}. \end{equation*}
But that isn’t exactly right either. It would be true if the rocket were really throwing out blobs, but it’s not—it’s dumping mass continuously. At the beginning the mass of the rocket is $10$. At the end of the one unit going out, the mass is only $9$—so on average, it’s more or less like $9.5$. During the time the first unit is dropped, we’ll say that is the effective average inertia that reacts against the $\Delta m = 1$, so that the rocket receives an impulse $\Delta v$ equal to $1/9.5$: \begin{equation*} \Delta v \approx \frac{\Delta m}{m - \Delta m/2} = \frac{1}{9.5}. \end{equation*}
It helps to put these halves in, because then you need fewer steps to get high accuracy. Of course, it still isn’t exact. If we wanted to do it more carefully, we could use smaller blobs of mass, like $\Delta m = 1/10$, and do much more analysis. But we’ll do it roughly, with $\Delta m = 1$ and keep on going.
Now the mass of the rocket is only $9$. We drop another unit off the rear end of the thing, and we find next that $\Delta v$ is $\dots1/9$? No. $\dots1/8$? No! It’s $\Delta v = 1/8.5$ because the mass has been continuously changing from $9$ to $8$, and on the average it was roughly $8.5$. For the next unit we get $\Delta v = 1/7.5$, and so we discover that the answer is the sum of $1/9.5$, $1/8.5$, $1/7.5$, $1/6.5$, ta, ta, ta, ta, tum—to the end. With the last step we go from $2$ units of mass down to $1$, on the average the mass is $1.5$, and we’re left with one unit of mass.
Finally, we calculate all these ratios (which takes only a moment to do; these numbers are all honest; it’s easy to figure them out) and merely add them together to get the answer, $2.268$, which means that the final velocity $v$ is $2.268$ times faster than the velocity of the exhaust $u$. That’s the answer to this one—nothin’ to it! \begin{equation} \label{Eq:TIPS:3:18} \matrix{ 1/9.5 & 0.106 \cr 1/8.5 & 0.118 \cr 1/7.5 & 0.133 \cr 1/6.5 & 0.154 \cr 1/5.5 & 0.182 \cr 1/4.5 & 0.222 \cr 1/3.5 & 0.286 \cr 1/2.5 & 0.400 \cr 1/1.5 & \underline{\;0.667\;} \cr & 2.268 } \kern 1.5em v \approx 2.268 u \end{equation}
Now you might say, “I don’t like the accuracy here—this is a little sloppy. It’s all very well to say, ‘in the first step the mass changes from $10$ to $9$, so it’s about $9.5$’. But in the last step, it changes from $2$ to $1$ and you’ve taken that all on the average of $1.5$. Wouldn’t it be better to split the last step, dropping half a unit at a time, to get a little bit better accuracy?” (This is a technical point of arithmetic.)
a technical point of arithmetic.) Let’s see. While the first half a unit goes out, the mass drops from $2$ to $1.5$; on average it’s $1.75$, so I take $1/1.75$ times a half unit for my $\Delta m/m$. Then I do the same thing for the second half a unit; the mass drops from $1.5$ to $1$, averaging $1.25$: \begin{equation*} \Delta v \approx \frac{0.5}{(2+1.5)/2} + \frac{0.5}{(1.5+1)/2} = \frac{0.5}{1.75} + \frac{0.5}{1.25} = 0.686. \end{equation*} \begin{equation*} \begin{aligned} \Delta v &\approx \frac{0.5}{(2+1.5)/2} + \frac{0.5}{(1.5+1)/2} = \frac{0.5}{1.75} + \frac{0.5}{1.25}\\[1ex] &= 0.686. \end{aligned} \end{equation*}
So you can make an improvement in the last step—you can improve all the rest of them too, the same way, if you want to go to the trouble—and it comes out $0.686$ instead of $0.667$, which means that our answer was a little bit low. When you figure it out better it comes out $v \approx 2.287u$. The last digit is really not reliable, but our estimate is pretty close, and the exact answer isn’t going to be far from $2.3$.
Now, I must tell you, that because the integral $\int_1^x dm/m$ is such a simple function and comes up in so many problems, people have made tables of it and given it a name: it’s called the natural logarithm, $\ln(x)$. And if you ever look up $\ln(10)$ in a table of natural logs, you will find it’s actually $2.302585$: \begin{equation} \label{Eq:TIPS:3:19} v = u \int_1^{10} \frac{dm}{m} = \ln(10)u = 2.302585\,u. \end{equation}
You can get that many digits of accuracy by the same technique we used, provided you use a much finer spacing like $\Delta m = 1/1,\!000$ or so, instead of $1$—and that’s precisely what’s been done.
Anyway, we did pretty well in no time at all, without knowing anything, and without looking in tables. So, I keep emphasizing that in emergencies you can always do arithmetic.
3–5Chemical rockets
Now, this question of rocket propulsion is interesting. You’ll notice, first of all, that the speed that is finally acquired is proportional to $u$, the speed of the exhaust. Therefore all kinds of effort has been put into trying to get the exhaust gases to go out as fast as possible. If you burn hydrogen peroxide with this and that, or oxygen with hydrogen or something, then you get a certain chemical energy generated per gram of fuel. And if you design the nozzles and whatnot correctly, you can get a high percentage of that chemical energy to go into the outgoing velocity. But you can’t get more than $100$ percent, naturally, and so there’s an upper limit for a given fuel as to what speed can be acquired by the most ideal design with a given mass ratio, because there’s an upper limit to the value of $u$ that can be acquired from a given chemical reaction.
Consider two reactions, $a$ and $b$, which have the same energy per atom liberated, but atoms of different masses, $m_a$ and $m_b$. Then, if $u_a$ and $u_b$ are the exhaust velocities, we have \begin{equation} \label{Eq:TIPS:3:20} \frac{m_a u_a^2}{2} = \frac{m_b u_b^2}{2}. \end{equation}
The velocities will therefore be higher for the reaction with the lighter atom, because whenever $m_a < m_b$, Eq. (3.20) implies that $u_a > u_b$. That’s why most of the fuels used in rockets are light materials. The engineers would like to burn helium with hydrogen, but unfortunately that mixture doesn’t burn, so for instance, they make do with oxygen and hydrogen.
3–6Ion propulsion rockets
Instead of using chemical reactions, another proposal is to make a device by which you ionize atoms, and accelerate them electrically. Then you can get a terrific velocity, because you can accelerate the ions as much as you want. And so I have another problem here for you.
Suppose we have a so-called ion propulsion rocket. Out of the rear end we are going to squirt cesium ions, accelerated by an electrostatic accelerator. The ions start at the front of the rocket, and a voltage $V_0$ has been applied between the front and the rear end—in our particular problem, it’s not an unreasonable voltage—I took $V_0 = 200,\!000$ volts.
Now, the problem is, what thrust is this going to produce? It’s a different problem than we had before, which was to find how fast would the rocket go. This time, we would like to know what force is produced if the rocket is held in a test stand. (See Fig. 3-11.)
The way it works is this: Suppose that in a time $\Delta t$ the rocket were to shoot an amount of mass $\Delta m = \mu \Delta t$ at velocity $u$. Then the momentum going out is $(\mu \Delta t)u$; since action equals reaction, that much momentum is being poured into the rocket. In the other problem the rocket was in space, and so it took off. This time, it’s held by the test stand, and the momentum per second that is acquired by the ions is the force that must be applied to hold the rocket in place. The total amount of momentum per second acquired by the ions is $(\mu \Delta t)u/\Delta t$. So the thrust force of the rocket is simply $\mu u$, the mass per second that is liberated times the velocity at which it goes out. And therefore all I have to do is figure out for my cesium ion what mass per second would go out, and at what velocity: \begin{equation} \begin{aligned} \text{thrust} &= \frac{\Delta (\text{momentum out})}{\Delta t}\\[1.5ex] &= (\mu \Delta t)u/\Delta t\\[1ex] &= \mu u. \end{aligned} \label{Eq:TIPS:3:21} \end{equation}
We work out the velocity of the ions first, as follows: the kinetic energy of a cesium ion coming out of the rocket is equal to its charge times the voltage difference across the accelerator. That’s what voltage is: it’s like potential energy, just like field is like force—you just have to multiply by the charge to get the potential energy difference.
The cesium ion is univalent—it has one electron charge—so \begin{equation} \begin{aligned} \frac{m_{\text{Cs}^+}u^2}{2} &= q_\text{el}V_0\\[1ex] u &= \sqrt{2V_0\frac{q_\text{el}}{m_{\text{Cs}^+}}}\\[1ex] \end{aligned} \label{Eq:TIPS:3:22} \end{equation}
Now, let’s figure out this $q_\text{el}/m_{\text{Cs}^+}$. The charge per mole^{5} is that famous number $96,\!500$ coulombs per mole. The mass per mole is what’s called the atomic weight, and if you look it up in the periodic table, for cesium it’s $0.133$ kilograms per mole.
You say, “What about these moles? I want to get rid of them!”
They’re already gotten rid of: all we need is the ratio between the charge and the mass. I can measure that in one atom, or in one mole of atoms, and it’s the same ratio. So we get for the outgoing speed \begin{equation} \begin{aligned} u &= \sqrt{2V_0\frac{q_\text{el}}{m_{\text{Cs}^+}}} = \sqrt{400,\!000 \cdot \frac{96,\!500}{0.133}}\\[1ex] &\approx 5.387 \times 10^5 \text{m/sec}. \end{aligned} \label{Eq:TIPS:3:23} \end{equation}
Incidentally, $5 \times 10^5$ m/s is much faster than you can ever get by a chemical reaction. Chemical reactions correspond to voltages of the order of one volt, and so this ion propulsion rocket provides $200,\!000$ times more energy than a chemical rocket.
Now, that’s fine, but we don’t want just the velocity; we want the thrust. And so we have to multiply the velocity by the mass per second, $\mu$. I want to give the answer in terms of the current of electricity that is pouring out of the rocket—because of course, that’s proportional to the mass per second. So, I want to find out how much thrust there is per ampere of current.
Suppose that one ampere is going out: how much mass is that? That’s one coulomb per second, or $1/96,\!500$ moles per second, because that’s how many coulombs are in a mole. But one mole weighs $0.133$ kilograms, so it’s $0.133/96,\!500$ kilograms per second, and that’s the rate of flow of the mass: \begin{equation} \begin{aligned} 1\text{ ampere} &= 1\text{ coulomb/sec} \rightarrow \frac{1}{96,\!500}\text{ mole/sec}\\[1ex] \mu &= \bigg(\frac{1}{96,\!500}\text{ mole/sec}\bigg)\cdot(0.133\text{ kg/mole})\\[1.25ex] &= 1.378 \times 10^{-6}\text{ kg/sec}. \end{aligned} \label{Eq:TIPS:3:24} \end{equation} \begin{equation} \begin{gathered} 1\text{ ampere} = 1\text{ coulomb/sec}\\ \rightarrow \frac{1}{96,\!500}\text{ mole/sec}\\[1.25ex] \mu = \bigg(\frac{1}{96,\!500}\text{ mole/sec}\bigg)\!\cdot\!(0.133\text{ kg/mole})\\[.75ex] = 1.378 \times 10^{-6}\text{ kg/sec}. \end{gathered} \label{Eq:TIPS:3:24} \end{equation}
I multiply $\mu$ by the speed, $u$, to find the thrust per ampere, and the result is \begin{equation} \begin{aligned} \text{thrust per ampere} &= \mu u = (1.378 \times 10^{-6})\cdot(5.387 \times 10^5).\\[1ex] &\approx 0.74\text{ newtons/ampere}. \end{aligned} \label{Eq:TIPS:3:25} \end{equation} \begin{equation} \begin{aligned} \text{thrust }&\text{per ampere} = \mu u\\[1ex] &= (1.378 \times 10^{-6})\cdot(5.387 \times 10^5).\\[1ex] &\approx 0.74\text{ newtons/ampere}. \end{aligned} \label{Eq:TIPS:3:25} \end{equation}
So, we get less than three-quarters of a newton per ampere—that’s very poor, lousy, low. An ampere isn’t a hell of a lot of current, but $100$ amperes or $1,\!000$ amperes is quite a job, and it still hardly gives any push. It’s hard to get a reasonable amount of ions.
Now let’s figure out how much energy is being consumed. When the current is $1$ ampere, $1$ coulomb of charge per second is dropping through a potential of $200,\!000$ volts. To get the energy (in joules) I multiply the charge by the voltage because volts, really, are nothing but energy per unit charge (joules/coulumb). Therefore $1\times200,\!000$ joules per second is consumed, which is $200,\!000$ watts: \begin{equation} \label{Eq:TIPS:3:26} 1\text{ coulomb/sec}\times200,\!000\text{ volts} = 200,\!000\text{ watts}. \end{equation} \begin{equation} \begin{gathered} 1\text{ coulomb/sec}\times200,\!000\text{ volts}\\[.5ex] = 200,\!000\text{ watts}. \end{gathered} \label{Eq:TIPS:3:26} \end{equation}
We get only $0.74$ newtons out of $200,\!000$ watts, which is a pretty punk machine, from an energetic standpoint. The thrust to power ratio is only $3.7\times10^{-6}$ newtons per watt—which is very, very weak: \begin{equation} \label{Eq:TIPS:3:27} \text{thrust/power} \approx \frac{0.74}{200,\!000} = 3.7\times10^{-6}\text{ newtons/watt}. \end{equation} \begin{equation} \begin{aligned} \text{thrust/power} &\approx \frac{0.74}{200,\!000}\\[1ex] &= 3.7\!\times\!10^{-6}\text{ newtons/watt}. \label{Eq:TIPS:3:27} \end{aligned} \end{equation}
So, although it’s a nice idea, it takes an awful lot of energy to get anywhere in this thing!
3–7Photon propulsion rockets
Another rocket has been proposed on the basis that the faster you can push the exhaust out the better things are, and so why not push out photons—they’re the fastest thing on Earth—shoot light out the back! You get out there at the rear end of the rocket, you turn on a flashlight, and you get a push! However, you can appreciate that you can pour an awful lot of light out without getting much of a push: you know from experience that when you turn on a flashlight, you don’t find yourself thrown off your feet; even if you turn on a $100$-watt bulb and put a focuser on it, you don’t feel a damn thing! So it’s very unlikely that we’re going to get much push per watt. Nevertheless, let’s try to figure out the thrust-to-power ratio for a photon rocket.
Each photon we throw out the back carries a certain momentum $p$, and a certain energy $E$, and the relationship, for photons, is that the energy is the momentum times the speed of light: \begin{equation} \label{Eq:TIPS:3:28} E = pc. \end{equation}
So for a photon the momentum per energy is equal to $1/c$. That means that, no matter how many photons we use, the momentum we throw out per second has a definite ratio to the energy we throw out per second—and that ratio is unique and fixed; it’s $1$ over the speed of light.
But the momentum per second thrown out is the force needed to hold the rocket in place, while the energy per second thrown out is the power of the engine generating the photons. So the thrust-to-power ratio is also $1/c$ (c being $3\times10^8$), or $3.3\times10^{-9}$ newtons per watt, which is a thousand times worse than the cesium ion accelerator, and a million times worse than a chemical engine! These are some of the points of rocket design.
I am showing you all these rather complicated semi-new things so you can appreciate that you have learned something, and that you can now understand a great deal of what goes on in the world.
3–8An electrostatic proton beam deflector
Now, the next problem that I cooked up, to show you how you can do things, is the following. In the Kellogg Laboratory,^{6} we have a Van de Graaff generator that generates protons at $2$ million volts. The potential difference is generated electrostatically by a moving belt. The protons drop through this potential, pick up a lot of energy, and come out in a beam.
Suppose, for certain experimental reasons, we would like the protons to come out at a different angle, so that we need to deflect them. Now, the most practical way to do this is with a magnet; nevertheless, we can also work out how it can be done electrically—they have been made that way—and that’s what we’re going to do now.
We take a pair of curved plates that are very close together compared to the radius of their curvature—say they’re about $d=1$ cm apart, separated by insulators. The plates are curved in a circle, and we put as high a voltage as we can across them, from a voltage supply, so that we get an electric field in between that deflects the beam radially, around the circle. (See Fig. 3-12.)
In fact, if you put much more than $20$ kilovolts across a $1$ cm gap in a vacuum, you have breakdown troubles—whenever there is a little leak, dirt gets in and it’s very hard to keep it from sparking over—so let’s say we put $20$ kilovolts across the plates. (However, I’m not going to do this problem with numbers; I’m just explaining it all with the numbers, so I’ll call the voltage across the plates $V_\text{p}$.) Now, we would like to know: to what radius of curvature do we have to bend the plates so that $2$ MeV protons will be deflected between them?
This simply depends on the centripetal force. If $m$ is the mass of a proton, then Eq. (2.17) tells us that $mv^2/R$ equals the force that’s needed to pull it in. And the force that we have pulling it in is the charge of the proton—which is again our famous $q_\text{el}$—multiplied by the electric field $\emf$ that’s in between the plates: \begin{equation} \label{Eq:TIPS:3:29} q_\text{el}\emf = m\frac{v^2}{R}. \end{equation}
This equation is Newton’s law: you have force equals mass times acceleration. In order to use it, however, you’ve got to know the velocity of the protons coming out of the Van de Graaf generator.
Now, information on the velocity of the protons comes from our knowledge of how much potential they have fallen through—$2$ million volts—which I’ll call $V_0$. The conservation of energy tells us that the kinetic energy of the proton, $mv^2/2$, equals the charge of the proton multiplied by the voltage through which it has fallen. We can calculate $v^2$ directly from this: \begin{equation} \begin{aligned} \frac{mv^2}{2} &= q_\text{el}V_0\\[1ex] v^2 &= \frac{2q_\text{el}V_0}{m} \end{aligned} \label{Eq:TIPS:3:30} \end{equation}
When I substitute $v^2$ from Eq. (3.30) into Eq. (3.29), I get \begin{equation} \begin{gathered} q_\text{el}\emf = m\frac{\displaystyle \bigg(\frac{2q_\text{el}V_0}{m}\bigg)}{R} = \frac{2q_\text{el}V_0}{R}\\[1em] R = \frac{2V_0}{\emf}. \end{gathered} \label{Eq:TIPS:3:31} \end{equation}
So if I knew what the electric field between the plates was, I could easily find the radius—because of this simple relationship between the electric field, the voltage at which the protons started, and the curvature of the plates.
Well, what is the electric field? If the plates don’t bend too much, the electric field is approximately the same everywhere between them. And when I put a voltage across the plates, there’s an energy difference between a charge on one plate and a charge on the other. The energy difference per unit charge is the voltage difference—that’s what voltage means. Now, if I carried a charge $q$ from one plate to the other through a constant electric field $\emf$ the force on the charge would be $q\emf$ and the energy difference would be $q\emf d$, where $d$ is the distance between the plates. By multiplying force times distance I get energy—or by multiplying field times distance, I get potential. So the voltage on the plates is $\emf d$: \begin{equation} \begin{gathered} V_\text{p} = \frac{\text{energy difference}}{\text{charge}} = \frac{q\emf d}{q} = \emf d\\[1em] \emf = \frac{V_\text{p}}{d}. \end{gathered} \label{Eq:TIPS:3:32} \end{equation}
I have therefore substituted $\emf$ from Eq. (3.32) into Eq. (3.31) and by fiddling around, I get the formula for the radius—it’s $2V_0/V_\text{p}$ times the distance between the plates: \begin{equation} \label{Eq:TIPS:3:33} R = \frac{2V_0}{(V_\text{p}/d)} = 2\frac{V_0}{V_\text{p}}d. \end{equation}
In our particular problem, the ratio of $V_0$ to $V_\text{p}$—$2$ million volts to $20$ kilovolts—is $100$ to $1$, and $d=1$ centimeter. Therefore the radius of curvature should be $200$ cm, or $2$ meters.
An assumption that’s been made here is that the electric field between the plates is constant. If the electric field isn’t constant, how good is our deflector? Pretty good anyway, because with a $2$-meter radius, the plates are almost flat, so the field is nearly constant, and if we’ve got the beam right in the middle, it’s just right. But even if we don’t, it’s very good because if the field is too strong on one side, it’ll be too weak on the other, and those things will compensate, nearly. In other words, by using the field near the middle, we’re getting an excellent estimate: even if it’s not perfect, it’s damn close for such dimensions; at $R/d = 200\text{ to }1$, it’s almost exact.
3–9Determining the mass of the pi meson
I have no more time, but I’ll ask you to stay just a minute extra, so I can tell you about one more problem: this is historically the way the mass of the pi meson ($\pi$) was determined. In fact, the pi meson was first discovered on photographic plates in which there were tracks of mu mesons^{7} ($\mu$): some unknown particle had come in and stopped, and where it stopped, there was a little track coming off whose properties were found to be those of a mu meson. (Mu mesons were known before, but the pi meson was just discovered from these pictures.) It was presumed that a neutrino ($\nu$) went off in the opposite direction (leaving no track, because it is neutral). (See Fig. 3-13.)
The rest energy of the $\mu$ was known to be $105$ MeV, and its kinetic energy was found from the properties of the track to be $4.5$ MeV. Supposing all that, how can we find the mass of the $\pi$? (See Fig. 3-14.)
Let’s suppose that the $\pi$ is at rest, and that it disintegrates into a $\mu$ and a neutrino. We know the rest energy of the $\mu$, as well as the kinetic energy of the $\mu$ and therefore the total energy of the $\mu$. But we also need to know the energy of the neutrino because, by relativity, the mass of the $\pi$ times c squared is its energy, and all that energy goes into the $\mu$ and the neutrino. You see, the $\pi$ disappears, and the $\mu$ and the neutrino are left, and by the conservation of energy, the energy of the $\pi$ must be the energy of the $\mu$ plus the energy of the neutrino: \begin{equation} \label{Eq:TIPS:3:34} E_\pi = E_\mu + E_\nu. \end{equation}
So we need to calculate both the energy of the $\mu$ and the energy of the neutrino. The energy of the $\mu$ is easy; it’s practically given: it’s $4.5$ MeV kinetic, added to the rest energy—so you get $E_\mu=109.5$ MeV.
Now what’s the energy of the neutrino? That’s the hard one. But by the conservation of momentum, we know the momentum of the neutrino because it’s exactly equal and opposite to the momentum of the $\mu$—and that’s the key. You see, I’m running it backwards here: if we knew the momentum of the neutrino, we could probably figure out its energy. So, let’s try.
We calculate the momentum of the $\mu$ from the formula $E^2 = m^2c^4 + p^2c^2$, choosing a system of units for which $c=1$ so that $E^2 = m^2 + p^2$. Then, for the momentum of the $\mu$ we get \begin{equation} \label{Eq:TIPS:3:35} p_\mu = \sqrt{E_\mu^2-m_\mu^2} = \sqrt{(109.5)^2-(105)^2} \approx 31\text{ MeV}. \end{equation} \begin{equation} \begin{aligned} p_\mu &= \sqrt{E_\mu^2-m_\mu^2} = \sqrt{(109.5)^2-(105)^2}\\[1ex] &\approx 31\text{ MeV}. \end{aligned} \label{Eq:TIPS:3:35} \end{equation}
But the momentum of the neutrino is equal and opposite, so—not worrying about signs, only magnitude—the momentum of the neutrino is also $31$ MeV.
What about its energy?
Because the neutrino has zero rest mass, its energy equals its momentum times $c$. We talked about that for the “photon rocket.” For this problem we let $c=1$, so the energy of the neutrino is the same as its momentum, $31$ MeV.
Well, we’re all finished: the energy of the $\mu$ is $109.5$ MeV, the energy of the neutrino is $31$ MeV, so the total energy liberated in the reaction was $140.5$ MeV—all given by the rest mass of the $\pi$: \begin{equation} \label{Eq:TIPS:3:36} m_\pi = E_\mu + E_\nu \approx 109.5 + 31 = 140.5\text{ MeV}. \end{equation} \begin{equation} \begin{aligned} m_\pi &= E_\mu + E_\nu\\[1ex] &\approx 109.5 + 31 = 140.5\text{ MeV}. \end{aligned} \label{Eq:TIPS:3:36} \end{equation} And this is the way that the mass of the $\pi$ was originally determined.
That’s all I have time for. Thank you.
See you next term. Best of luck!
- “Unfortunately,” in Brazilian Portuguese. ↩
- See FLP Vol. I, Section 9–7. ↩
- This historical convention is introduced in FLP Vol. I, Section 32–2. Today, the letter $e$ in this context would typically be reserved for the charge on an electron. ↩
- If the rocket starts at time $t=0$ with mass $m=m_0$, and $\mu = dm/dt$ is constant, then $m = m_0 - \mu t$, and Eq. (3.16) becomes $dv = u\mu dt/(m_0-\mu t)$. Integrating yields $v=-u\ln[1-(\mu t/m_0)]$, and solving for $t$ gives the time required to reach speed $v$: $t(v) = (m_0/\mu)(1-e^{-v/u})$. ↩
- One mole equals $6.02 \times 10^{23}$ atoms. ↩
- The Kellogg Radiation Laboratory at Caltech performs experiments in nuclear physics, particle physics, and astrophysics. ↩
- “Mu meson” is an obsolete term for a muon, an elementary particle with the same charge as an electron but approximately $207$ times the mass (and which in fact isn’t a meson at all in the modern meaning of the word “meson”). ↩