The kinetic energy of a drifting tectonic plate…

…is broadly similar to the kinetic energy of me and my bike as I pedal along.

plates
Map of tectonic plates (United States Geological Survey) http://pubs.usgs.gov/publications/text/slabs.html

According the the theory of plate tectonics, the outer layer of the Earth is divided into a number of separate plates, which very slowly drift around, opening and closing oceans, causing earthquakes, and thrusting up mountain ranges.

A moving body has energy by virtue of its motion: kinetic energy. Kinetic energy is proportional to a body’s mass and to the square of its speed.

Now tectonic plates move extremely slowly: the usual comparison is with a growing fingernail. But they are also extremely heavy: tens of millions of square kilometres in area, over 100 km thick, and made of rock. I wondered how the minute speed and colossal mass play out against each other: what’s the kinetic energy of a drifting tectonic plate?

There are so many variables, that vary such a lot, that this calculation is going to be extremely approximate. But the answer is delightfully small: the kinetic energy of the tectonic plate on which I live, as observed from one of the plates next door, is about the same as the kinetic energy of me and my bike when I’m going at a reasonable pace: about 1500 joules.

Me struggling up one of the many steep roads in NW Scotland. Here, the kinetic energy of me and my bike is much less than the kinetic energy of a drifting tectonic plate. In fact the speed of me and my bike is probably less than that of a drifting tectonic plate.
Me struggling up one of the many steep roads in north-west Scotland. Here, the kinetic energy of me and my bike is much less than the kinetic energy of a drifting tectonic plate. In fact the speed of me and my bike is probably much less than that of a drifting tectonic plate ;-).

This is a fun calculation to do, but we shouldn’t get carried away thinking about the kinetic energy of tectonic plates. Plates are driven by huge forces, and their motion is resisted by equally large forces. The mechanical work done by and against these forces will dominate a plate’s energy budget in comparison to its kinetic energy.

But the calculation does provoke an interesting thought about forces and motion. I can get my bike up to full speed in, say, 10 seconds. If the Eurasian plate were as free to move as my bike, and I were to put my shoulder against it and shove as hard as I could, it would take me about 500 years to get it up to its (very tiny) full speed.

In both cases, I’m giving the moving object roughly 1500 joules of kinetic energy. How come I can give that energy to my bike in a few seconds, but to give it to the plate would take me centuries?

I’ll return to that thought in a later post.

The calculation

Depending on how you count them, there are 6-7 major tectonic plates, 10 minor plates, and many more microplates. The plates vary hugely in size, from the giant Pacific Plate with an area of 100 million km2, to the dinky New Hebridean plate, which is a hundred times smaller. The microplates are smaller still. Plates also vary a lot in speed: 10-40 mm is typical.

I’m going to be parochial, and choose the Eurasian plate for this calculation.

Let’s call the area of the plate a and its mean thickness t. Its volume is then given by at, and if its mean density is ρ, then its mass m is ρat.

A body of mass m moving at a speed v has kinetic energy ½mv2. So our plate will have kinetic energy ½ρatv2.

The area of the Eurasian plate is 67,800,000 km2 or 6.78 × 1013 m2, and its speed relative to the African plate is (the only speed I have) is given as 7-14 mm per year. We’ll use 10 mm per year, which is 3.2 × 10-10 ms-1. The thickness of tectonic plates in general varies roughly in the range 100-200 km depending upon whether we are talking about oceanic or continental lithosphere; let’s call it 150 km or 1.5× 105 m. The density of lithospheric material varies in the range 2700-2900 kg m-3; we’ll use 2800 kg m-3.

Putting all of these numbers into our formula for kinetic energy, we get a value of 1500 joules (to 2 significant figures, which the precision of the input data certainly doesn’t warrant).

Now for me and my bike. I weigh about 57 kg, my bike is probably about 10 kg. Suppose I’m riding at 15 mph, which is 6.7 ms-1. My kinetic energy is almost exactly…

…1500 joules!

The closeness of these two values is unmitigated luck*, and we shouldn’t be  seduced by the coincidence. Just varying the speed of the plate in the range 7-14 mm would cause a 4-fold change in kinetic energy, and there’s the variability in plate thickness and rock density to take into account as well. The choice of bike speed was arbitrary, I guessed the mass of the bike, and I’ve since realised that I didn’t account for the fact that the wheels of my bike rotate as well as translate.

However, what we can say is that the kinetic energy of a drifting continent is definitely on a human scale, which leads to a new question:

Suppose the Eurasian plate were as free to move as my bicycle, and that I put my shoulder against it and shoved, how long would it take me to get it up to speed?

From the figures above, the mass of the plate is 2.85 × 1022 kg. If I can push with a force equal to my own weight (about 560 newtons) then by Newton’s 2nd Law I can give it an acceleration of about 1.96 × 10-20 ms-2. Rearranging the equation of motion v = at, where v is the final speed, a is the acceleration, and t is the time, then t = v/a. Inserting the values for v and a, we get t = 1.6 × 1010 seconds, or about 500 years.

 

* I didn’t tweak my assumptions: what you see above really is the very first version of the calculation!

A dish designed by nature

Paraboloid 1

When you stir a cup of tea, the surface of the rotating liquid develops a dip in the middle. The faster you stir, the deeper the dip. But the liquid surface is quite uneven; to get a smooth surface, throw the spoon away and spin the whole cup continuously. Once the liquid inside has caught up with the cup, and everything is turning at the same speed, the liquid surface forms a beautiful smooth curve known as a paraboloid of revolution.

Sarah McLeary and I are applying this idea to make thin paraboloidal plaster shells. We spin a bucket on a potter’s wheel (Sarah is a potter), and pour plaster into it. The plaster rapidly flows until its surface forms a deep paraboloidal curve, and then sets. We now use this cast, still spinning, as a mould, and cast a thin layer of plaster inside it, to make a paraboloidal shell.

That’s our first attempt above. It’s about 20 cm across and 3 mm thick. It might look like a part of a sphere, but in a profile view it’s easy to see that the curvature is tightest at the base and gradually decreases up the sides, as you’d expect for a paraboloid.

I love the fact that we didn’t decide what shape this shell was going to be: physics did.

There’s lots of experimentation ahead. When we’ve got the hang of it, I’ll explain our methods in more detail. But as this picture of our third attempt shows, we haven’t quite cracked it yet.

Paraboloid 3 cracked

 

 

 

Kettles, cars, and energy

Be careful about how much water you put in your kettle and ‘do your bit’ to save energy!

That’s what we’re often told, and boiling only as much water as you need for your cup of tea can only be a good thing. But how much impact does it really have on your overall energy consumption?

Consider this: if you’re driving along in a car at 50 mph, you’re using enough energy to make a cup of tea every few seconds.

1, 2, 3, tea, 1, 2, 3, tea, 1, 2, 3, tea…

As most people don’t think twice about driving their cars for an extra few minutes, never mind an extra few seconds, it’s clear that the energy savings made by being frugal with your kettle are very small compared to your overall energy use.

Does this mean that we shouldn’t be careful with our kettles after all? No. It all matters. But having boiled exactly one cup’s worth of water for your refreshing cuppa, you shouldn’t put your feet up and think that your energy consumption issues are sorted. You’ve much bigger fish to fry.

The calculation

Imagine that you’re travelling in a car that’s doing 50 miles per hour, with a rate of fuel consumption of 50 miles to the gallon (the units are customary in a UK motoring context). This is a realistic situation; if anything it’s optimistic in terms of energy consumption.

In this scenario, your car will burn a gallon of petrol, that’s 4.5 litres approximately, every hour.

With 3600 seconds in an hour, that’s 4.5L/3600s = 0.00125 litres per second. The density of petrol is about 0.8 kg per litre, so the rate of petrol consumption comes to 0.00125 L s-1 × 0.8 kg L-1 = 0.001 kilograms per second.

The energy density of hydrocarbon fuels is about 46 megajoules per kilogram, so the rate of use of energy by your car is 0.001 kg s-1 × 46×106 J kg-1 = 46000 joules per second (watts).

Now let’s think about the tea. The energy E required to produce a temperature change of ΔT in a mass m of a substance whose specific heat capacity is C is

E = m C ΔT

A typical mug of tea contains about 250 ml, or 0.25 kg, of water. For water, C = 4200 J kg-1 K-1. The temperature change is about 90 K, if the water comes out of the tap at about 10°C and is raised to boiling point. So the amount of energy required to heat the water for a cup of tea is 0.25 kg × 4200 J kg-1 K-1× 90 K = 94500 joules.

As the car is using 46000 joules per second, it follows that the car is using enough energy to heat the water for a cup of tea every two seconds.

Now to be fair to the car, we need to recognise that in real life there is more energy used to heat the water for the tea than actually goes into the water. There are heat losses in energy generation and transmission, and from the kettle itself.

Things get very complicated here. A thermal electricity generating plant (coal, oil, gas, or nuclear) converts energy in the fuel to electrical energy with about 35-45% efficiency. But in the UK, about 20% of our electricity is generated from renewables, where the concept of efficiency is harder to pin down. For example, I could use some oil to heat my water directly instead of doing it indirectly by generating electricity, but I couldn’t do the same with the wind. I’m going to assume that 80% of the electricity that I used was generated in thermal plants at 40% efficiency, and 20% of it was generated at 100% efficiency, giving an overall efficiency of 52%. But be aware that this is a very rough figure, and that there is no right answer.

According to this document, about 93% of the energy in generated electricity makes it unscathed through the transmission and distribution systems to the end user.

I did an experiment and estimated the efficiency of my kettle to be about 88%. I give details later.

This gives an overall efficiency of boiling water as 52% × 93% × 88% = 43%. So instead of using 94500 joules to boil water for a cup of tea, we’re actually using 94500J/0.43 = 220000 joules, which means that the car’s rate of energy use is equivalent to a cup of tea roughly every 5 seconds. Because of all the uncertainties involved (not least the size of a cup of tea), we should treat this figure as very approximate.

Efficiency of my kettle

My kettle took 180 s to raise 1 litre of water from 17°C to boiling.

The kettle is rated at 1850-2200 W for supply voltages in the range 220-240 V. When the kettle was running, I measured the supply voltage to be 242 V, so I assumed that the kettle was operating at the top of its power range, ie 2200 W.

The heat supplied by the kettle element was therefore 2200 W × 180 s = 396000 J.

Using the same equation as earlier, the energy required to raise the temperature of 1 kg of water by 83 K is 1 kg × 4200 J kg-1 K-1 × 83 K = 348600 J.

The kettle was therefore heating the water with efficiency 348600/396000 = 0.88 or 88%.

Walking bass

How fast does a plucked guitar string move? It’s a complete blur, so surely it’s travelling at a terrific speed. 50 miles per hour? 100 miles per hour? What do you think?

"Vibrating guitar string" by jar [0] is licensed under CC BY 2.0
“Vibrating guitar string” by jar[0]. See end of post for full attribution.

Walking pace. A typical speed for the middle of a guitar string given a good twang is walking pace. And that’s the middle of the string. Near its ends, it’ll be moving much more slowly.

How can that be so? Well, although the string is going back and forth hundreds of times a second, it’s only travelling a few millimetres on each trip. So the distance that it travels in each second isn’t as much as you might expect. It certainly isn’t as far as I expected.

If you think that the string moves slowly, what about the body of the guitar? The string itself radiates very little sound into the air; its job is to set the body of the guitar vibrating. The body of the guitar, with its much larger area, is much more effective than the string at setting air into motion. Yet we can’t even see the body vibrating. At what snail’s pace must it be moving?

Remember also that the air molecules on which the guitar body acts are already travelling at something like 500 metres per second. Isn’t it astonishing that the sub-pedestrian movements of the guitar affect the movement of the air molecules enough to produce a sound that we can easily hear?

The calculation

Suppose that we look near the centre of the string, where its movement is the greatest. Shortly after being plucked, the width of the blur that we see is going to be something like 5 mm. So for every complete oscillation, the string does a round trip of about 10 mm.

The frequencies of the strings on a standard 6-string acoustic guitar are (to the nearest whole number) 82, 110, 147, 196, 247 and 330 hertz (one hertz is one oscillation per second). If we multiply these frequencies by the 10 mm round trip, it tells us how far the centre of each string travels in one second, that is, its average speed. I’ve converted these speeds into metres per second. For comparison, a brisk walk at 4 mph is about 1.8 ms-1.

Note nameMean speed of middle of string in metres per second (to 2 sig. fig.)
E0.82
A1.1
D1.5
G2.0
B2.5
E3.3

Our brisk walk is right in the middle of this range. And remember, we’ve done the calculation for the part of each string that’s moving the most. Near its ends, each string will be moving much more slowly than this.

Complication 1 – how long is the round trip really?

So far, we’ve assumed that each part of the string does a simple back-and-forth movement along a straight line, but if you carefully watch a vibrating guitar string you’ll see that the string often moves in an irregular but roughly elliptical orbit. The wire-wound lower strings show this most clearly; you can see a hint of it in the image at the top. This makes the round-trip distance a bit longer than the 10 mm that we used in the calculation earlier. Does this affect the string’s average speed much?

We’ll take the extreme case where each part of the string moves at constant speed in a circle of diameter 5 mm rather than along a straight line 5 mm long. The circumference of this circle will be 5 \pi mm, or about 15 mm. So the speeds of the strings (in this rather unlikely extreme case) will be about 50% greater than the ones listed above. They are still hardly impressive.

Complication 2 – the peak speed

So far, we’ve calculated the mean (average) speed of the string over its round trip. However, unless it’s moving in a perfect circle, its speed changes constantly, and its peak speed will be higher than its mean speed. How much higher?

Imagine that part of the string is vibrating back and forth along a straight line in the simplest possible way. At one end of the movement the string is stationary as it changes direction. It then speeds up, reaching its peak speed at the centre of its range of movement. Then it slows down until it reaches a halt again at the other end of the movement and changes direction again. How do we calculate the peak speed if we know the time taken for the round trip?

You’ll need to know a bit of maths for the next bit. The simplest vibration of the string is where each part undergoes simple harmonic motion, that is, where its position varies sinusoidally with time. This means in turn that the velocity of the string also varies sinusoidally in time. So we need to ask: how does the mean value of a sinusoid compare to its peak value?

graph cropped 1Consider the function y = \sin \theta, for half a cycle, that is, for theta from 0 to \pi radians. We construct a rectangle between these limits, such the rectangle’s area equals the area under the sine curve. The mean value of the sine function is the height of the rectangle, which its area divided by its width (which is \pi). So we need to work out the area a under the sine curve between 0 and \pi, which we can do by integration:

a = \int\limits_0^\pi \sin \theta \,d\theta

= \left[-\cos \theta\right]_0^\pi

= \left[-(-1)-(-1)\right]

= 2

We constructed the rectangle to have area a. Its width is \pi, so its height, and therefore the mean value of the sine function, is a/\pi = 2/\pi. The height to the peak of the sine curve is 1, so the peak value of the sine function is \pi/2 times its mean value. This means that the peak speed of our guitar string is \pi/2 times, or 50% more than, its average speed. Again, nothing to write home about.

(Rather pleasingly, this ratio of \pi/2 is the same ratio that we got earlier in Complication 1. It means that the peak speed of a particle doing simple harmonic motion with a given amplitude r and period t is exactly the same as the (constant) speed of a particle moving in a circular orbit of radius r with period t. This isn’t a coincidence. It arises because the circular motion can be considered as two linear simple harmonic motions at right angles to each other.)

Credits

Vibrating guitar string” by jar [0] is licensed under CC BY 2.0

Mathematical typesetting by QuickLaTeX

A creative atmosphere

It’s sometimes said that if you sit an immortal monkey in front of an equally durable typewriter and leave it to tap randomly away at the keys, then eventually it will produce the entire text of Richard III (or any other Shakespeare play of your choice), completely by chance. All you have to do is wait long enough.

I was thinking about this one day, and also thinking about air molecules. In the room I’m sitting in at the moment, there are at least 1,000,000,000,000,000,000,000,000,000 air molecules, all frantically dashing around bumping into each other. How often, I wondered, do little clusters of these molecules fleetingly arrange themselves, by chance, in arrangements that we would regard as being somehow regular or remarkable?

It’s not possible for me to directly observe air molecules, so instead I used my computer to make a 2-dimensional simulation of some atoms of gas doing what atoms of gas do. I set it going, and waited…and waited…and waited…

I promise you that my simulation didn’t involve any secret forces drawing atoms towards certain positions. The movements and collisions of the atoms all occurred in accordance with the laws of mechanics. But I did cheat a little. Can you figure out how?

Molecules and atoms

In case you’re fidgeting on your chair wondering why I started talking about air molecules but finished talking about gas atoms, let me explain.

Nearly all of the air is nitrogen and oxygen. Nitrogen and oxygen atoms are essentially spherical. But in the air, nitrogen atoms are bonded together in pairs to form nitrogen molecules that are, roughly speaking, a stubby rod shape. The same goes for oxygen. Now a collision between two moving rods is much more complicated than a collision between two spheres, because the rods can spin end-over-end in a way that spheres can’t. In fact, I’m not sure that I know how to do the calculations. As the point of the video could be made just as well using atoms rather than molecules, I did the simulation using atoms. If you like, you can think of them as atoms of helium or argon, which do go around on their own.

If I suddenly became weightless, what would happen to me?

Suppose that you were standing perfectly still, and gravity suddenly stopped operating on your body. What would happen? Nothing much, you might think, apart from a queasy feeling of weightlessness. After all, an object won’t start to move unless a force acts on it, and no force is acting on your body.

nogravity1
Diagram of the Earth, looking straight down on the north pole.

However, when you stand “still”, in fact you’re travelling in a very large circle at rather high speed, as the Earth turns on its axis and carries you with it. Newton’s 1st Law tells us that things travel in straight lines unless a sideways force acts upon them. The force that keeps tugging you to make you travel in a circle rather than in a straight line is gravity. This means that the moment gravity stops acting on you, you’ll start moving along a straight line (the red line in the diagram) while the ground continues to move in a circular path underneath you (the blue line).

The consequence is that you’ll lose contact with the Earth and float upwards, serenely or otherwise. At least, that’s what it will look like to earthbound observers. But what’s really happening is that the ground is accelerating downwards away from you as it moves on its curved path. Try to remember that as you watch your footprints receding beneath you.

I wondered how fast this would all happen. The answer is: remarkably quickly. I give the geometry later, for those who are interested, but here are some example results for a person standing in Edinburgh, on a latitude of 56° N. Just for now, we’ll pretend that there isn’t any air.

  • After 1 second your feet will be 5 millimetres off the ground.
  • After 10 seconds you’ll be 53 centimetres off the ground.
  • After a minute, you’ll be 19 metres up
  • After an hour you’ll be at an altitude of over 68 km (though, being half frozen to death by now, you may be losing interest).

The lower your latitude, the quicker your ascent. At the equator, you’ll rise about three times as fast as in Edinburgh, and at the poles, you won’t lose contact with the ground at all.

Why on earth should I be interested in a situation that is contrived and physically impossible? It’s because it brings home the fact that each of us is constantly moving along a curved path as the Earth rotates. On the equator, it takes only 10 seconds for our trajectory to deviate from a straight line by 1.7 metres (during which time we’ve travelled 465 metres).

Why did I pretend that there isn’t any air? Because the presence of air muddies the waters by adding another force: the upward force of our buoyancy in the air.  At low altitudes this force isn’t negligible: it slightly more than doubles the first three figures above. As you rise further and the air gets thinner, it matters less and less. I left it out because I wanted to make the effect of the Earth’s rotation clear.

The question that I’ve just answered is a trimmed-down version of a question that my friend Malcolm and I occupied ourselves with once when we were on a rather long and boring tramp along a glen at the end of a camping trip in the Cairngorms in Scotland. The question we asked then was: what would we observe if gravity suddenly stopped operating altogether? I may return to that subject in a future post.

The geometry

Let the centre of the Earth be at O, the origin, and let the Earth’s radius be r and its angular velocity about its own axis be \omega. You are standing at latitude \phi and are therefore a distance r \cos \phi from the Earth’s axis. Your linear velocity as you stand still on the rotating Earth will be \omega r \cos \phi, tangential to the Earth’s surface.

nogravity3Suppose that gravity stops acting on you at time zero, when you are at point Q. With no gravity acting on you, will now travel in a straight line tangential to the Earth’s surface. The diagram shows the situation from a suitable vantage point, looking sideways on to your direction of travel. We are not looking down on the north pole.

After a time t, you will have travelled a distance \omega t r \cos \phi, to point P.

Your altitude is the distance PR, where R is the point on the Earth’s surface for which P is directly overhead.  R lies on OP, the line from P to the centre of the Earth.  The length of OP is given directly by Pythagoras’ Theorem in triangle OPQ: it’s \sqrt{(\omega t r \cos \phi)^2 + r^2}. As OP=PR+r, the altitude of point P is OP - r. So

PR = ((\omega t r \cos \phi)^2 + r^2)^{1/2} - r

All we need now is \omega = 2\pi/86400 \text{ s}^{-1}, because the Earth does one full rotation in 86400 seconds, and r=6.4\times10^6 \text{ m}, because that’s how big the Earth is. We can now choose \phi and calculate PR for any value of t.

 

nogravity4We can check this answer in two ways. Firstly, we can use the very useful intersecting chords theorem to calculate the distance marked h in the diagram on the right. For small values of t, where h \ll r, then h should be approximately equal to the your altitude PR. For values of t of 1, 10, or 60 seconds, h and PR agree to 4 significant figures. As we expect, as t increases, the agreement gets less good: h and PR differ by about 2% after 1 hour.

The second check is to differentiate the expression for PR twice with respect to time. The first differentation gives us an expression for the rate of change of PR with respect to time, that is, your rate of gain of altitude:

\frac{dPR}{dt} = \frac{\omega^2rt}{  (1+\omega^2t^2)^{1/2}}

(Note: to keep things clear, I’ve omitted \cos \phi, which merely accompanies r everywhere and doesn’t change the conclusions.) Where t=0, and your motion is purely tangential, this expression for your speed away from the ground should be zero, and where t is very large (\omega t \gg 1), and your motion is purely radial, your speed away from the ground should be \omega r. Both are true.

The second differentiation gives us an expression for your upward radial acceleration:

\frac{d^2PR}{dt^2} = \frac{\omega^2r}{(1+\omega^2t^2)^{3/2}}

As it’s not you accelerating, but the ground that is accelerating away from you as it continues on its circular path, this expression for your upwards acceleration should, where t=0 and your path is tangential to the surface, become the same as that for the centripetal acceleration of the ground, \omega^2 r, which it does. In addition, when \omega t r \gg r, and your path is almost radial, the expression for your acceleration should approach zero, which it does.