# Changing a planets rotation speed

1. Dec 9, 2010

### hidlAP2010

Is it possible to speed up or slow down the speed of a planets rotation by attaching jet engines (or some other device) along the equator of the planet?

2. Dec 9, 2010

### dacruick

yes but infinitesimally so.

3. Dec 9, 2010

### jambaugh

If you look at the length of a solar day, it is not a constant but varies somewhat randomly due to the change in winds throughout the seasons. Now considering the atmosphere, your scheme of attaching jets will in effect change the winds but friction will eventually overcome the effect.

To permanently change the rotation speed your jets need to exhaust reaction mass (e.g. gas) into space. This way you affect the solid/liquid Earth plus its gaseous atmosphere together in the same direction.

4. Dec 9, 2010

### K^2

What? First of all, the mass of the atmosphere is a little less than 10-6 of the Earth's mass. Secondly, the highest winds don't reach a 3rd of the Earth's rotation speed at the equator. And finally, and most importantly, the winds always blow perpendicular to pressure gradient, and that means that the average wind velocity in any cyclone, storm, or tornado is zero.

There are additional prevalent winds due to Coriolis effect, but these also net zero if you take the entire atmosphere.

There is absolutely nothing happening in the atmosphere that can possibly change the length of solar day by any measurable amount. Tectonic drift makes a bigger difference.

5. Dec 9, 2010

### Staff: Mentor

Sure, by imparting an overall angular velocity to the atmosphere you could change
the angular velocity of the Earth. This works by conservation of angular momentum;
the total angular momentum of the Earth-Atmosphere system is constant. Beware of
high winds, though!

Eventually, friction between the atmosphere and Earth's surface would restore the original
state once the engines stopped driving energy into the atmosphere.

To get an idea of the magnitude of the problem, a little mathematics:

The moment of inertia of a homogeneous sphere divided by its mass and the square
of its radius is 2/5. That of a hollow spherical shell is 2/3. So, representing the Earth
as a homogeneous sphere of mass Me and radius R, the atmosphere as a hollow
spherical shell of mass Ma and radius R (it's a very thin skin on the surface of the Earth),
and the initial rotational velocity of both as W0, we have:

I = ((2/5)*Me*R^2 + (2/3)*Ma*R^2)*Wo

This will be a constant.

Now, if we allow the components (Earth and Atmosphere) to have different angular
velocities (We and Wa), the sum of their contributions of angular momentum must still

I = (2/5)*Me*R^2*We + (2/3)*Ma*R^2*Wa

Rearranging to solve for we,

We = (I - (2/3)*Ma*R^2*Wa)/(*2/5)*Me*R^2)

In order to see how We varies with Wa, we can differentiate the above expression for We
with respect to the variable Wa:

dWe/dWa = -(5/3)*Ma/Me

and given known values for the masses of the Earth and Atmosphere:

Me = 5.97 x 10^24 kg
Ma = 5.27 x 10^18 kg

dWe/dWa = -1.47 x 10^-6

That's a very small effect.

6. Dec 9, 2010

### Staff: Mentor

The state of the atmosphere does have an effect on the rotation rate of the Earth.
The main seasonal effect is due to the overall expansion and contraction of the
atmosphere with differing heat content, which shifts its moment of inertia and hence
its angular momentum.

The different heat content is due to the asymmetry of the land mass and water
distribution between the Northern and Southern hemispheres, which influences the
overall albedo.

See, for example, the graph "Atmospheric excitation over the current year" at the
site:

http://hpiers.obspm.fr/eop-pc/" [Broken]

Last edited by a moderator: May 5, 2017
7. Dec 9, 2010

### K^2

The expansion/contraction is balanced by the two hemispheres. Same with water content.

8. Dec 9, 2010

### Staff: Mentor

The overall atmosphere height varies, changing the its moment of inertia. Furthermore,
the jet streams, which moves a significant amount of air mass around, shifts significantly
over the course of a year.

Also, because of the different overall atmospheric heat content in different seasons
(again due to different land mass and water distributions in the hemispheres), the
atmosphere will contain more water moisture when it is hotter, and thus be more
massive and have larger angular momentum.

9. Dec 9, 2010

### K^2

If there is sufficient temperature change, the thermal expansion of the continents themselves is going to make a bigger difference.

The oceanic mass is also significantly greater than atmospheric. You'd have to talk about shifting currents before you talk about the atmosphere.

I see absolutely nothing in your argument to point to atmosphere being the cause of the seasonal changes in day cycle. The graph on the site you posted suggests that the error in measurement is on the order of 0.5ms with oscillations on order of 2ms. If the atmospheric mass is less than 100th of oceanic, how are you planning to distinguish effects?

10. Dec 9, 2010

### RocketSci5KN

Unless you physically threw mass out at escape velocity, you wouldn't change the rotation rate of the earth. What you could do is transport mega amounts of rock from the equator to the poles, or visa versa, to allow for extremely small changes. Note that major earthquakes can slightly affect the length of a 'day'. Of course, no one will help you pay for this ;->

11. Dec 9, 2010

### Staff: Mentor

The thermal expansion rate for the crust and water is far smaller than that of a gas
(the atmosphere). Further, the average temperature of the crust below a few centimeters
and the oceans below a few meters depth does not vary significantly (except perhaps where ocean currents carry water between depth levels, the so-called "elevators").

Further, any expansion of the Earth's surface is going to exacerbate the atmospheric change, pushing the whole atmosphere further from the Earth's center.

I do not see where you obtained the information regarding the measurement error for
the graph presented at

http://hpiers.obspm.fr/eop-pc/" [Broken]

One would need to find the author's data reduction method to determine that. Simply
eyeballing the graph, which is not raw data and does not contain error bars, will not
tell you.

I should also point out that nowhere did I affirm that changes in the atmosphere is the
only thing affecting the rotation rate of the planet. Anything that can affect the overall
moment of inertia of the system or its components can have an effect. This includes
seasonal variations in ice packs, snow cover, etc.

Last edited by a moderator: May 5, 2017
12. Dec 9, 2010

### K^2

See these high frequency oscillations? That's your measurement error. Whether these are real oscillations or problems with measurement method is irrelevant. They prevent you from determining the seasonal dependence more accurately.

The seasonal dependence is just barely greater than these oscillations. So if you want to prove that atmosphere provides a measurable change of solar day, you need to prove that atmospheric effect is one of the major effects.

Go.

13. Dec 9, 2010

### Staff: Mentor

http://www.jstor.org/pss/74779

http://www.nasa.gov/centers/goddard/news/topstory/2003/0210rotation.html

and so on.

Last edited by a moderator: Apr 25, 2017
14. Dec 9, 2010

### Staff: Mentor

I disagree. If the graph depicts the rotational effects due to atmospheric effects alone,
then the seasonal variation stands out clearly.

You can look at longer time intervals for this sort of data, too, and the same seasonal
variations stand out on the background "noise". This site allows you to produce graphs
for various periods:

http://hpiers.obspm.fr/eop-pc/index.php?index=excitactive&lang=en"

Here's such a graph for the years 2000 - 2010

http://hpiers.obspm.fr/eop-pc/analysis/excitactive1.php?IB=1&term=1&AAM=1&option=1&dimx=600&dimy=450&langue=1&sel_option1=1&choix=3&trend=1&filter=Select+band+above&P0=1&tr=95&spec=0&freqmin=-10&freqmax=10&choixspec=4&chi_g=1&chi_f=1&TC=433&QC=100&SUBMIT=Submit+request&an1=2000&mois1=1&jour1=1&an2=2010&mois2=12&jour2=31

and one for just 2009:

http://hpiers.obspm.fr/eop-pc/analysis/excitactive1.php?IB=1&term=1&AAM=1&OAM=1&option=1&dimx=600&dimy=450&langue=1&sel_option1=1&choix=3&trend=1&filter=Select+band+above&P0=1&tr=95&spec=0&freqmin=-10&freqmax=10&choixspec=4&chi_g=1&chi_f=1&TC=433&QC=100&SUBMIT=Submit+request&an1=2009&mois1=1&jour1=1&an2=2009&mois2=12&jour2=31

Pick any year for (for which data is recorded) and you'll see the same clear variations.

Here's a graph for 2009 that shows just the effects due to "Oceanic excitation"

http://hpiers.obspm.fr/eop-pc/analysis/excitactive1.php?IB=1&term=1&OAM=1&option=1&dimx=600&dimy=450&langue=1&sel_option1=1&choix=3&trend=1&filter=Select+band+above&P0=1&tr=95&spec=0&freqmin=-10&freqmax=10&choixspec=4&chi_g=1&chi_f=1&TC=433&QC=100&SUBMIT=Submit+request&an1=2009&mois1=1&jour1=1&an2=2009&mois2=12&jour2=31

Last edited by a moderator: Apr 25, 2017
15. Dec 9, 2010

### D H

Staff Emeritus
Late in the game, but this is wrong. Borek and gneill are correct in this regard, K^2. There has been *no* observable effect from the Chilean earthquake. The seasonal signature in length of day is quite obvious. Things like large El Nino events stick out like sore thumbs.

16. Dec 9, 2010

### K^2

Absolutely none of that tells me that it's atmospheric. Yes, there is a correlation with weather. Good support there, especially with the El Nino confirming it. But the mass of the air is still tiny compared to the mass of the water and land masses that are being affected.

Alright, lets say that we can throw land out of equation. Top 15m of world's oceans alone has greater mass than all of Earth's atmosphere. And you are telling me that winds are responsible for observed effects, and not oceanic currents? I need to see some good models backed by hard evidence to believe that.

Edit: And I'm going to see how much one would need to warm up the entire planet by to get the solar day changed by 1ms due to thermal expansion of the atmosphere. I suspect the number to be fairly large.

17. Dec 9, 2010

### dacruick

Last edited: Dec 9, 2010
18. Dec 9, 2010

### Staff: Mentor

You mean besides the fact that the scientists who produced the graphs
state that it is?

As was already pointed out, the coefficient of expansion of water is far smaller than
that of air. The oceans also have a significant thermal mass which tends to greatly
smooth out short cycles. The atmosphere is much more sensitive to heat content
variations. I would imagine that land effects, such as snow and ice coverage, would
exceed those of water expansion.

That should be an interesting calculation. It would be nice if you would post the details.

19. Dec 9, 2010

### K^2

I'm also a scientist, and I also can make graphs. I see a correlation between condition of atmosphere and planet rotation. So yes, I agree, there has to be a weather-relation. Interesting. Didn't know that. Good. What I don't see is any evidence that air currents have any direct effect on Earth's rotation.

You tell me severe storms don't affect oceanic currents? Didn't El Nino screw up Gulf Stream? Now consider how much water is flowing through Gulf Stream and what that's going to do to Earth's rotation. Now THAT is a significant change.

Show me some evidence that suggests that air currents can have direct effect.

Of course.

20. Dec 9, 2010

### D H

Staff Emeritus
Wrong calculation. The right calculation is to see how much you would need to cool the atmosphere of the northern hemisphere, but maintaining the same pressure, to change length of day by 1 ms.

Simplifying things somewhat, the northern hemisphere is largely land while the southern hemisphere is largely water. While Antarctica does get considerably colder than the Arctic, Antarctica is very high (mean elevation = 2500 meters). Putting these two items together means that atmospheric mass moves from equatorial regions toward the north pole in northern hemisphere winter and back toward the equator in northern hemisphere summer.