where *G* is the gravitational constant, *M* is the mass of the body being orbited, and *r* is the distance to the center of that mass. I tested this on the ISS’s orbit and found the velocity matched quite well. Now, I’m interested in looking at orbital periods, and calculating aspects of some common orbits.

If we know the orbital velocity, *v*, we can find the orbital period, *T*, by using the velocity relationship *v* = *s*/*t*. The distance traveled is the circumference of the circle, *C* = 2π*r*.

(Of course *r* is not negative.)

This looks correct, especially because I recall that one of Kepler’s laws is that the square of of the orbital period is proportional to the cube of the semi-major axis (for a circle, the radius).

Let’s test this on the ISS’s orbit, assuming a circular orbit of 412.5 km as last time.

It’s always reassuring to see the units work out.

5560 seconds is 92.7 minutes. The Wikipedia article on the ISS lists the orbital period at 92.69 minutes, so our calculation works out.

Let’s try the reverse. What altitude would be needed for a geostationary satellite? A geostationary satellite has an orbital period of exactly one day, so that it orbits at the same rate that Earth rotates. From any point on Earth it always appears to be in the same place in the sky, and so it is especially useful for communication satellites.

First, let’s solve the orbital period equation from above for the radius, *r*:

I will square both sides. This will not include any extraneous solutions, since *T* and *r *are both positive (as are the constants).

The orbital period should be one sidereal day, or 23 hours, 56 minutes, 4 seconds (86, 164.1 s). This is slightly shorter than a standard *solar* day since we measure our days based on how long it takes for the Earth to rotate once with respect to the sun. Since the Earth is also moving a little along its orbit each day, after it completes a rotation with respect to the distant starts (the sidereal day), it must rotate a little “extra” to turn to show the same face to the sun.

This is the distance from the center of the Earth. Since geostationary satellites orbit over the equator, let’s subtract the equatorial radius of Earth, 6378.1 km = 6.3781×10^{6} m.

The altitude for geostationary orbits is 35,786 km (Wikipedia), so our answer is correct. Note that this orbit is far wider than the low earth orbit (altitude around 412 km) in which the ISS orbits.

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When I say I want to use the standard projectile motion equations, I am referring to **s** = ½**a***t*^{2} + **v**_{0}*t *+ **s _{0}**, which gives us the position

Feynman provides a diagram, but while his approach is (of course) geometrically sound, it’s not the most intuitive way for me to approach the problem. I drew my own diagram:

In this diagram, there is a circle representing the orbit, centered at *O*. Point *A* represents the starting position of the projectile. After a time *t*, it will have moved to the right. If there were no gravity, it would continue in a straight line to point *B*. However, gravity will pull it down (again, we’re going to assume it is pulled down as in the negative *y* direction, not towards the center of the planet *O*, as that would be much more complicated). The amount it falls depends only on time: the *x* and *y* movement are independent. It will take a certain amount of time for it to fall the distance *BC*. If the projectile is launched with sufficient *x* velocity, in the time it takes for the projectile to fall that distance, it will reach a horizontal distance equal to *AB* and therefore will reach point *C*, on the requisite circle.

It was tough for me to conceptualize this, so I’m going to break this problem down into four steps. **1**, find how far the projectile will fall in the *y* direction after a time *t*. **2**, find how far in the *x* direction it must travel in that time to stay on the circle. **3**, find the horizontal velocity needed to achieve the *x* distance in that time. **4**, look at very small time intervals so that our assumptions hold (in particular, that the direction of “down” doesn’t change significantly.

We know that

It has no initial *y* velocity, so *v*0,y=0.

The initial height is equal to the radius of orbit *r*.

What about the acceleration? I don’t simply want to use −*g*, since *g* represents the acceleration due to gravity at Earth’s surface, and I want to have a general expression that will hold at varying distances from Earth (or whichever planet/star). The force due to gravitation is given by Newton:

I am using *M* for the mass of the large planet, and *m* for the mass of the projectile. Of course, *M*≫*m*, otherwise we’d have to deal with the movement of both objects.

According to Newton’s second law, *F*=*ma*, which we can rearrange to *a*=*F*/*m*:

As expected, the acceleration does not depend on the mass of the projectile. Let’s put this into our equation for the position.

By the Pythagorean theorem,

I’m going to assume the clockwise direction for orbit and will not consider the *x*<0 solutions. Let’s substitute in our expression for *y* from Step 1.

This is rather complex, and I’m not sure if I should multiply this out or if something will otherwise cancel.

That is an unfortunately complicated expression. I don’t see how I can simplify this, aside from factoring out the first term to get

The position of *x* is given simply by

We’re assuming over this short period that the acceleration is downward, that there is no acceleration in the *x* direction.

The initial velocity is entirely in the *x* direction (recall that the *y* component was 0 above), so we can write *v*0,*x*=*v*0

And the projectile starts from a position of *x* = 0:

This is the basic distance-velocity-time equation, and is the usual partner to *y* projectile motion equation above. We solve for velocity:

and plug in our expression for *x* from Step 2 above:

Note that √(*t*²) = |*t*|, so if we assume *t* is positive, we can bring it into the radical without worrying about a plus/minus sign. I am interested in the time moving forward, not the time in the past when it could have gone from this position up to be at the “apex” at time=0.

I don’t think I can further simplify this. Time for Step 4.

Feynman used one second for his rough calculation. I could do the same, but I’m favoring a more mathematically rigorous approach. Why stop at one second? We could evaluate at one millisecond, or one nanosecond. I’ll try taking the *limit* as *t* goes to 0. I hope this results in a viable expression, and doesn’t just indicate that in an infinitesimal time, the projectile will hardly drop so any velocity works or something like that.

We can plug in our expression for *v*0 from Step 3.

This is a wonderful result: it looks just like the type of answer I was hoping for, with an appropriate amount of complexity (it “feels right”). Of course, there is probably a more straightforward way to derive it. Let’s test it a real-world example, the orbit of the International Space Station.

The ISS doesn’t have a perfectly circular orbit. According to Wikipedia, the perigee and apogee (minimum and maximum distances) are 409 and 416 km, respectively, so I’m just going to take 412.5 km as an average distance (I don’t really know what the appropriate “average” distance for an elliptical orbit would be). This altitude will be on top of the radius of Earth, about 6371 km. Plugging these values in, along with the mass of Earth and the gravitational constant, gives

I am relieved that the units worked out.

According to Wikipedia, the ISS orbits at an average velocity of 7.66 km/s, or 7.66×10³ m/s. Our approach worked! To be honest, I wasn’t expecting this degree of precision. This is fantastic!

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In this thought experiment, a cannon at the top of a tall mountain fires a cannonball at increasing velocities, until eventually it moves so fast that it achieves orbit.

For low speeds, the cannonball will fall in the familiar parabolic arc (curve A). But, as you can see from the diagram above, if launched with greater speed, it will fall a greater distance than it would have on a flat Earth, since the Earth’s surface has curved away from under it (curve B). If it were launched with sufficient speed, Earth’s surface will curve away from it as fast as it’s falling, and it will never reach the ground. It would be perpetually falling — it would be in orbit.

I programmed a simulation in Python. First let’s review projectile motion. For constant acceleration, **s** = ½**a***t*^{2} + **v _{0}**

I will divide up the trajectory into short time intervals and assume that the direction and magnitude of the force/acceleration don’t change over an interval. Also, I will assume the Earth is perfectly spherical, and there is no air resistance. (If not for the atmosphere, we could launch something into orbit from a tall building — and hope it doesn’t hit another tall building!) For visual clarity, I am going to launch from a height of 10% of the Earth’s radius, which works out to 10% of 6367 km or 637 km. Note that Earth’s highest mountain, Mount Everest, is only 8.8 km high, and the International Space Station orbits between 409 km and 416 km, so we’re a bit beyond that.

I’m setting the origin (0, 0) at the center of the Earth. The force due to gravity is

where *m*_{E} is the mass of the Earth and *m*_{p} is the mass of the projectile, and **r** is the distance to the projectile. For the projectile, **F**=*m*_{p}**a**, or **a** = **F**/*m*_{p}. We can divide the force by the mass of the projectile to get acceleration (which is therefore not dependent on the mass of the projectile).

At each step, we’ll take the coordinates of the projectile (*x*, *y*) and calculate the distance *r*. Using the above equation, we can get the magnitude of the acceleration, and multiply by *x*/*r* and *y*/*r* to get the components of acceleration in the *x* and *y* directions, respectively. We’ll use the *x*– and *y*-components of acceleration, velocity, and position to get the new position coordinates and the components of acceleration to get the new velocity.

Here is the animation. I used a Δ*t* of 1 second, and ran the program for 100,000 simulated seconds (at each launch velocity), or 27.8 hours. It started out with a launch velocity of 0 m/s (the trajectory is straight down), then increased to end up at 16,000 m/s (around the speed of *New Horizons,* the fastest spacecraft launched to date).

I’ll discuss my observations in a subsequent post. For now, note that the projectile is essentially falling continuously, but never reaching the ground. This is why people in orbiting spacecraft experience weightlessness. It is *not* because there is no gravity — on the contrary, gravity is what is keeping them in orbit. But they are falling, and since they and the spacecraft are falling at the same rate, there is nothing to push them against the craft.

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First, let’s look at the outcome of 1000 trials for 2000 steps. This is the same graph as from the previous post.

Now, let’s look at the final positions of all the trajectories after the 2000 steps. I’ll plot a histogram of the frequency of finding an object at that position.

As you can see, most objects end up near the origin (distance = 0), and it drops off towards a probability of 0 in either direction. The shape is familiar: it is the so-called “bell curve,” or normal distribution. The dashed line is a plot of the normal distribution with mean = 0 and standard deviation = √(2000 )(the number of steps).

Let’s compare how this distribution changes over time — I’ll select 500, 1000, and 2000 steps as the points at which to graph the distance distribution, indicated by the red, green, and blue lines below.

The blue line of course represents the final positions, already graphed above. Here are all three distributions, superimposed on the same graph:

Each histogram is paired with the predicted distribution. You can see that after 500 steps (red), most objects are clustered near the origin. After 1000 steps (green), they are more spread out, and after 2000 steps (blue, also seen above), they are even more spread out. The spread increases as the square root of the number of steps, which can again be seen visually on this graph from the prior post:

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I was able to code this in Python/Numpy/Matplotlib. First, I tried running it with 10 trials, again for 30 steps. The walks are in blue, partially transparent so that it is easier to see when multiple walkers are in the same place. The yellow line traces the mean, and the red lines trace one standard deviation from the mean.

The dashed yellow line is *y*=0 (the expected mean), and the dashed red lines are ±√N. There is reasonable agreement, with the final mean -0.6 (expected 0) and standard deviation 5.80 (expected 5.48).

Let’s try this again, with a lot more trials — say, 1000.

Note that the vertical axis is compressed a bit to fit in the outliers. The calculated mean and standard deviation (0.21 and 5.38) closely match the predicted ones. You can also see that the chance that a walker will continue moving away from the origin for each step drops over time (the chance that it will move +1 each time for *n* steps is 1/2^{n}, and so is the chance that it will move −*n*). By the time 11 steps have been taken, there is less than one chance in a thousand that a walker would be at the maximum distance (1/2^{11},multiplied by 2 since it could go either always forward or always backward).

Finally, let’s let this go for a lot longer: 2000 steps.

Now *this* looks really cool, and the density distribution is readily apparent. The mean is 2.078, (predicted=0), and standard deviation is 46.23 (predicted=44.72).

Here’s another run, with the predicted mean ± 1, 2, and 3 standard deviations, to illustrate the “68–95–99.7 rule” — that for normally distributed data, 68.3% are within one standard deviation, 95.5% are within two, and 99.7% are within 3 standard deviations of the mean.

To round this out, here are two more graphs. First is a run without any statistical curves added.

And finally, here is a version with equal scales for the *x*– and *y*-axes. The gray lines represent the theoretical maximum and minimum (if the random walker moved in the same direction for every step).

It’s neat to see that even though the positions spread a bit over time, they still stay pretty close to a net distance of 0. The probability that a walker could follow one of the gray lines all the way out to 2000 steps is 2/2^{2000}, about 1.74×10^{−602}, or 1 chance in 5.74×10^{601}. There are only an estimated 10^{80} atoms in the universe, so even if each atom participated in this random walk, by the time we got to 266 steps or so, chances are that no atom would still be on the gray line, taking every step in the same direction.

**Read more in Part 2.**

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A couple of connections have occurred to me. One is that ½*mv*² is ½*v*² multiplied by a constant (*m*). The form ½*v*² reminds me of the integral of a variable with respect to itself:

Perhaps I can do something similar with velocity:

In fact, I can multiply this by the constant *m*:

and *mv* is the momentum, *p*:

Since we know the kinetic energy is zero when velocity (and momentum) = 0,

This is very neat, but I don’t really understand what , the integral of momentum with respect to velocity, should represent. But the *v dv* above gives me an idea. I know that *v dv *= *a* *dx* (for movement in the *x* direction). It is easy to prove this. Start with the following expression, which is of course equal to itself.

I don’t have an intuitive sense of what *a dx* represents any more than of what *v dv* represents. But what if I multiply the *m* back in?

Of course, *F*=*ma *by Newton’s second law.

This is fantastic. Force times displacement (in the same direction) is equal to work. And by the work-energy principle the work done on an object equals the increase in its kinetic energy. Carrying the whole chain of reasoning:

where the constant is the prior kinetic energy of the object, and where I assumed an initial starting velocity (and kinetic energy) of zero.

I thought about it another way, as well. Intuitively, I might think that energy requirements or transfers depend on force times time *Ft*, or . But this is *impulse*, not work — which we’ve noted earlier is force times displacement *Fx*, or . Part of the reason for this discrepancy is that you cannot arbitrarily apply forces to objects. As I explored in the Rocket Propulsion posts, a rocket engine using a steady amount of fuel does provide a constant force to accelerate the spacecraft, but most of the work is done on accelerating the exhaust. The work done on the spacecraft and therefore the spacecraft’s kinetic energy does increase quadratically, but it is only a small fraction of the total kinetic energy. Another example of a steady force is an object falling in a constant gravitational field, as we’ve been exploring in the prior post. But an accelerating object of course falls faster and faster — it drops 1, then 4, then 9, then 16 units of distance from its original height, even as its velocity increases linearly. Therefore, it is losing potential energy at an increasing rate, and so by conservation of energy, the kinetic energy will increase at an increasing rate.

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**Problem: **Given the formula for gravitational potential energy, can I use the principle of conservation of energy (of a projectile) to derive the equation for its kinetic energy?

The total energy of an object should be the sum of its potential energy and kinetic energy: *E* = *U* + *K*, and it should remain constant.

If we pick the right times, we can make this much simpler. If an object is launched vertically (from, say, height = 0), it initially has kinetic energy but no potential energy (*U*_{1} = 0). At the peak of its trajectory, it will be motionless for an instant — it will have potential energy but no kinetic energy (*K*_{2} = 0). Therefore,

In a constant gravitational field, the potential energy is *U*=*mgy*, where *m *is the mass, *g* is the acceleration due to gravity, and *y* is the height. At the second time point,

where *y*_{2} is the maximum height it will reach. We can find this from the equations of projectile motion, given constant acceleration *a*:

The acceleration *a* is the acceleration due to gravity, oriented down:

And the initial height was zero:

The velocity is the derivative of position (with respect to time):

At the peak, the velocity and therefore kinetic energy will be zero.

As an aside, this should have been intuitively obvious: acceleration is change in velocity over time, so time should be change in velocity over acceleration.

If the acceleration is constant, then *a* = *ā* (the instantaneous acceleration equals the mean acceleration).

Let’s substitute this back into the projectile equation.

We can now substitute this back into our equation for *K*_{1}:

This is indeed the formula for (nonrelativistic) kinetic energy.

The calculation was straightforward, but unfortunately, it didn’t really give me an intuitive understanding of the kinetic energy equation. I have some ideas, which I’ll explore in another post.

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I would like to make this approach a little more rigorous. In particular, I assumed a drive that ejected a constant amount of mass each second, and looked at the behavior over the first minute. I am hoping that this approximates the continuous case, but I’ll want to verify this by using very small time intervals, if needed. And I also want to explore the behavior of the rocket over a much longer period of time. For this, I will need more than a spreadsheet — I will write a program in Python. Also, I assumed that the drive ejected the mass at a constant velocity relative to the craft. There is no guarantee that this will take the same amount of energy over time, since the mass of the craft changes. Let’s work it out more rigorously.

To make the math simpler, for each time interval, let’s choose a reference frame moving with the craft, so the initial velocity = 0. The craft will consume a certain amount of energy *E* to push apart the reaction mass and craft. Let’s assume that the drive is perfectly efficient and that all the energy consumed goes to increasing the kinetic energy of the objects. Let *m*_{1} be the mass of the craft (after the mass is ejected), and *m*_{2} be the reaction mass ejected. The final velocities will be *v*_{1} and *v*_{2}, respectively. Momentum (*p*=*mv*) must be conserved, so the initial and final momentum must be the same:

Kinetic energy (*E*_{K} = ½*mv*²) is not conserved; we are adding an amount of energy *E* each time interval.

I’m primarily interested in *v*_{1}, the final velocity of the craft, so I’ll solve the momentum equation above for *v*_{2}

and substitute it into the kinetic energy equation:

I am specifying that the propellant is ejected in the negative direction (*v*_{2 }< 0 ), so logically and by the momentum equation above, the craft will move in the positive direction (*v*_{1} > 0); therefore, we can discard the negative solution.

This answer looks reasonable. Dividing energy by mass (*m*_{1}), doubling, and taking the square root are just the inverses of the kinetic energy equation. The is the fraction of the original mass that was ejected.

At last, I’m ready to code this into my program. The code will reevaluate the variables each timestep Δ*t*. Given a predefined rate of propellant use (in mass/time), we know how much is used each interval: multiply the rate of propellant use by the timestep to get *m*_{2}. We can subtract that from the ship’s prior mass to get its new mass, *m*_{1}. We’ll use the velocity equation derived just above to derive its velocity increase (recall that I defined those equations in a reference frame where the craft was starting at rest), so we’ll add that to the craft’s prior velocity to get its new velocity. The energy *E* added each step is the predefined rate of energy production multiplied by the timestep.

Let’s use initial conditions similar to the example in the last post. I’ll start with a 1,000-kg craft. It will eject 1 g/s, and it will be powered by a 500-W drive (since in the last post, the kinetic energy increased by 5,000,005 mJ / 10 s = 5,000.005 J / 10 s = 500.0005 W). Running the simulation for 60 seconds yields a velocity of 0.0600 m/s and a kinetic energy of 1.80 J, just as in the prior example. The approach seems sound. Let’s run it for longer. I’ll say that 10% of the mass of the craft is propellant. Let’s see what happens:

The drive will work for 100,000 s (which makes sense since we eject 1 g/s, and we started with a reaction mass of 100,000 g — 10% of 1,000,000 g), or 27.8 h. It will reach a velocity of 105 m/s (235 mph, 378 km/h). The acceleration looks constant. Is it? I’ll add in a line in the program to calculate the acceleration, by taking the new velocity minus the old velocity, and dividing by the timestep: *a* = (*v*_{f} – *v*_{i}) / Δ*t*.

Interesting. I actually thought the acceleration would be roughly constant over this period. It starts off with an acceleration of 1.00×10^{−3} m/s², and ends with an acceleration of 1.11×10^{−3} m/s². I assume that this rate of increase is because the craft is 10% lighter by the end, so a constant thrust will accelerate it more. We can calculate the force by taking the acceleration and multiplying by the mass at that time (*F*=*m _{1}a*).

Well, it certainly looks constant. That matches what I’ve read about ion thrusters. It starts and ends at 1.00 N of thrust. That means that the increasing acceleration is due to the progressive lightening of the craft. Our 500-W drive is able to produce 1 N of thrust when ejecting 1 g/s. That’s interesting. I wonder how it depends on the parameters (it doesn’t seem to depend much on the mass of the ship), and if we can derive a mathematical relationship — I’ll investigate that later. Finally, let’s explore the issue that sparked my interest in this problem in the first place, the kinetic energy. It just takes another line of code to calculate the kinetic energy as *E*_{K} = ½*m _{1}v_{1}*²

As expected, it looks parabolic. Of course, the velocity is not increasing linearly, as we know the acceleration is actually increasing over time. But the mass is decreasing, too. It reaches a kinetic energy of 5.00 MJ (5.00×10^{6} J). We already know that this must only be a small portion of the total kinetic energy. Let’s find the kinetic energy of the propellant. This is slightly more complex to code, since each at each time step we need to add in the (unchanging) kinetic energy of the new bit of propellant that’s been ejected, rather than looking at the changing velocity and mass of the craft at that instant. And I’ll track the sum of the two kinetic energies as well.

I notice several interesting things about this graph. As expected, the kinetic energy of the exhaust is much more than the kinetic energy of the spacecraft (note the 10-fold increase in scale from the previous graph). The total kinetic energy does appear to be increasing linearly: the kinetic energy of the exhaust ends at 4.50×10^{7} J, and after adding in the 0.500×10^{7} J of spacecraft kinetic energy, we get a total of 5.00×10^{7} J. In a way, this is a relief, because we’ve run our 500-W power source for 1.00×10^{5} s, and multiplying the two gives 5.00×10^{7} J. This gives me some confidence that the coding is correct. I also note that the kinetic energy of the exhaust increases, but that rate of increase slows over time (it curves away from the total line). Of course, this is to be expected, because the spacecraft’s kinetic energy is increasing at an increasing rate (and there is a fixed amount of total energy), but I’m still interested in understanding why this is happening. Let’s look at the exhaust velocity (*v*_{2} in our derivation above).

The exhaust velocity starts and ends at -1000 m/s (for graphing purposes, I flipped the sign to make it positive). This is also a relief, because our engine power was based on the earlier model which used a craft that ejected propellant at -1000 m/s. If the exhaust velocity is constant, and we’re ejecting a constant amount of mass each second, why is the rate of kinetic energy dropping off? It’s because the constant exhaust velocity is relative to the spacecraft. But the velocity is less, relative to our rest observational frame. To begin with, if the craft is motionless and ejecting propellant at -1000 m/s, we’ll also measure that velocity. But if the craft were moving at, say, 10 m/s, then the propellant would be ejected at -990 m/s relative to us. It would have less kinetic energy than the propellant that was previously ejected. (From the observer’s point of view, the propellant had been moving forward at 10 m/s, so even though it was pushed out with the same energy, it didn’t achieve as high a speed moving backward.)

If the spacecraft’s kinetic energy were truly parabolic, it would eventually cross any arbitrary line, including the line of total kinetic energy. Since this presumably can’t happen, I’ll want to explore what actually does occur — in the next post.

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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 4*x* to double that and 9*x* 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 (*m*_{1}*v*_{1} + *m*_{2}*v*_{2} + *m*_{3}*v*_{3} + …) 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, *m*_{i}*v*_{i} = *m*_{f}*v*_{f}. If a spacecraft is initially at rest, then *v*_{i} = 0 and so momentum *p* = 0. If it is to start moving with a positive velocity *v*_{f}, then it will have a positive momentum *mv*_{f}. Of course, 0≠*mv*_{f}. (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 (*m*_{1} + *m*_{2})*v*_{i} = *m*_{1}*v*_{1f }+ *m*_{2}*v*_{2f}.

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 *v*_{pf} (the propellant velocity) is *v*_{i} – 1000 (starting at the velocity of the craft, and being ejected at a relative -1000 m/s). If *m*_{c} is the mass of the craft, *m*_{p} is the mass of the propellant, and *m*_{c’} = *m*_{c}* − **m*_{p} is the new mass of the craft, then

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 ½*mv*^{2}. 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.

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I had initially stated I wanted to start with *x*(0) = 0, to keep it simple. But I didn’t end up needing this restriction. My equation for velocity used the initial *y _{i}*, which we got from plugging in the initial value of

I also had started with an initial velocity of zero. That did make a difference. But looking back on part 1, I don’t think it would complicate the equation too much, and if it’s zero it will be an extra term to just drop out. Let’s go back to the conservation of energy equation, and keep *v _{i}* in this time.

I’m leaving the ± in this time. Strictly speaking, I’m not treating this as the magnitude of the vector, since magnitudes must be positive. Rather, I want to consider a velocity vector that can point ether forwards or backwards along the direction of the curve. I’m going to allow motion in both directions, not just forward.

Recall the graph showing components of the velocity vector:

Now I want to find the *x*-component. As I discovered last time, I don’t need to bother with the *y*-component — once I find an equation for *x*(*t*), I can use that directly to obtain *y*(*t*). The *x-*component will be

Since I will have *y* and *y*′ in terms of *x*, I would need to rearrange to solve the differential equation. Let’s see how far I can take the general case:

where I did not include a constant of integration on the right side, since it can be absorbed into the constant that the left integral will produce.

So, the general approach should be as follows: Given our equation *y*(*x*), find *y*′(*x*). Plug in those expressions, plug in the initial velocity, and plug in the initial height *y*[*x*(0)]. Integrate, and solve for *x* in terms of *t* to get *x*(*t*), then plug that into *y*(*x*) to get *y*(*t*). I’ll test if this approach can actually work in the next post.

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