I was recently watching an episode of The Universe, the show from the History Channel, which detailed ten different ways to destroy the Earth (Season 4 Episode 6). One of the proposed destruction techniques was to instantly stop the rotation of the Earth. This would cause everything on the surface of the Earth to be violently thrown as the ground beneath it instantly stands still. It would be quite a violent stop. For some perspective, in New England, where I am located, you would be thrown at a velocity of about 800 miles an hour to the East. It is very unlikely that anybody could survive an impact with anything at that speed. That is not to mention that every building would be thrown at that velocity as well. However, the contributors to the show, professional astrophysicists, claimed that if you could survive the initial stop it is still unlikely that you could survive the aftermath. Specifically, the atmosphere would not be stopped, so there would be windspeeds of up to 1000 miles per hour as it continued rotating at its initial velocity. The show claims that this wind would create so much energy that it would heat up the atmosphere enough to melt rock. This claim seemed a bit too dramatic to be true, so I decided to do some calculations to see how credible it is. Brace yourselves, quite a bit of math is ahead as we fact-check…The Universe!

The first thing we need to do is find the kinetic energy of the atmosphere as it rotates around the Earth. In order to do this we need a few pieces of information: the density of the atmosphere as a function of elevation and the rotational speed of the atmosphere as a function of elevation and latitude. The first part of this is relatively difficult to figure out on your own, so I turned to some experts, the United States Navy. Specifically I used some data from the NRLMSISE-00 model produced by the Naval Research Laboratory. As can be seen on the graph of the model, the density of the atmosphere is approximately a straight line from density ρ=10^{-3} altitude A=0 to ρ=10^{-12} A=150km.

We can easily turn this into an equation:

ρ(A) = 10^{-3A/50000}

where A is in meters and ρ is in kg/m^{3}. With this in hand, now we can determine the velocity of the atmosphere at different points on the Earth. This is easy enough to do, we just need the rotational speed of the Earth (ѡ) which is just 2π radians every 24 hours. Also, the velocity is going to depend on the latitude of the point of interest. If we define an angle θ that is zero at the north pole and π at the south pole, then the velocity is given by:

v(r,θ) = r ѡ sinθ

where r is the distance from the center of the Earth (R) in meters. If we rewrite the density in terms of r we get:

ρ(A) = 10^{-3(r-R)/50000}

The kinetic energy of a mass is usually given by

KE = m v^{2}/2

In our case, m will be replaced by the density as a function of r multiplied by a differential volume element and v will be replaced by v(r,θ) from above. This gives us an infinitesimal chunk of the kinetic energy, to get the total we need to integrate over the volume of the atmosphere:

Working out this volume integral gives a total kinetic energy of 2.66542*10^{23} Joules. This is a whole lot of energy to be sure, but there is also a whole lot of atmosphere to absorb that energy so that any one spot does not heat up too much. In fact, the next thing we need to know is the total mass of the atmosphere. We have just done this integral except we do not want the 1/2 or the velocity squared term:

where the 1000 in the front is to calculate the mass in grams. This volume integral gives a value for the mass of the atmosphere as 3.70849*10^{21} grams.

Armed with this information, we can begin to calculate the change in temperature of the atmosphere caused by the conversion of this energy into heat. Specifically, what we want to use to convert from energy into a change in temperature is the specific heat of air. Specific heat, C_{p}, has units of energy/(mass*change in temperature). So, by rearranging these quantities we can get

change in temperature = energy/(mass*C_{p})

For air, C_{p} has the value of 1.01 Joules/(grams*Kelvin). So, by plugging in the numbers from above, we calculate a change in temperature of Earth’s atmosphere of 71.2K. This means that the temperature is raised by 71.2 degrees Celcius. So, if it is 70 degrees Fahrenheit outside, after the rise in temperature, it would be just about 200 degrees Fahrenheit outside afterwards. Definitely much too hot to really survive a stroll in the park, but not hot enough to melt rock.

However, maybe there is more to the story. Using the above values assumes that the heat is evenly spread throughout the atmosphere. In reality, the heat would all be generated near the surface. We can add a layer of complexity to our model calculation to try to account for this. Let’s look at a plot of the density of the atmosphere without the aid of a logarithmic scale as above.

As it is easy to see, the vast majority of the air in our atmosphere is very close to the surface of the Earth. So, for our improved model, lets distribute the kinetic energy according to an exponential decay with a decay length of 800 meters. This decay length is equal to the average elevation of the Earth above sea level. Also, we should include a sin^{2}θ because the kinetic energy is proportional to velocity squared. This gives us a distribution for the kinetic energy of:

where γ is the proportionality constant. This new equation means that there is much more energy at the surface than farther out in the atmosphere. In order to find the proportionality constant that would preserve our total kinetic energy, we need to solve the following equation:

Solving this gives a value of 977415 for the proportionality constant. Now that we have a new expression for our energy, we can look again at our equation for the change in temperature. We have an energy term and a mass term. However, both of these values in our new model vary depending on what part of the atmosphere we are looking at. So, I am going to coarse grain the model a bit and look at 10 meter thick layers of the atmosphere and evaluate our new model one layer at a time. This will give us a series of points that should show us a trend of what the change in temperature is like as a function of elevation. For example, the layer starting at an elevation of 1000 meters would be given by:

This can be evaluated at different start and end points for the integrals, giving a value for each slice of the atmosphere. When we do this and plot the results we get:

This plot is not particularly informative due to the rapid drop in the change in temperature on the left hand side of the graph. To get a better sense of what is happening close to the surface of the Earth, let’s zoom in:

This graph is much more informative. For instance, it is easy to see that at sea level, you can expect a change in temperature of 640 Kelvin. This means that if it was a pleasant 70 degrees Fahrenheit outside like earlier, after the increase in temperature it would be a toasty 1221 degrees Fahrenheit. That translates to 660 degrees Celsius. Now, if we compare that to the typical melting point of rock, somewhere around 700 degrees Celsius, we can see that the atmosphere’s temperature is close to being able to melt rock, but not quite there. Of course, there is quite a bit of uncertainty in this calculation, the models used and the assumptions made may not be entirely accurate. So it is possible that if we were to stop the Earth from spinning, the atmosphere could heat up to the point of melting rock. However, nobody could really survive long enough to find out how hot the atmosphere gets in that scenario anyway.

So, there you have it, The Universe may or may not have exaggerated its claim. It is sitting happily in a region of uncertainty. I hope you found this exercise as rewarding as I have and feel free to comment with suggestions and/or comments.

Could you explain why you decided not to factor in the Flavin Effect? Thanks.

Well, you see, the Flavin Effect only applies in cases of extreme Sylvester Stallone involvement. That is because we all know that wherever he turns up his wife, Jennifer Flavin, is sure to follow. As far as I know the Flavin effect would only apply to this problem if Stallone were to somehow be the cause of the end of Earth’s rotation. If that were the case, Stallone would surely be the solution to the problem after a dramatic story takes place which would later be turned into a blockbuster action movie.

Riveting. Did they happen to posit just what kind of apocalyptic event might stop the earth in its tracks?

The only thing that could stop the earth abruptly would be a massive impact. It would take a huge asteroid the size of a small planet such as Mars in order to have enough energy to stop the rotation of Earth. However, such a large impact would have much more severe problems than stopping the rotation of the planet. The scenario was really an exercise in scientists’ “what if” part of their brain; my favorite part of the brain.