Planck length is closer than the provided choices, in my guessy opinion. The answer really does not need any calculation. We do need to ignore obfuscative red herring(s). And to make sure that our angulars are momenting the right way.
The less than 1.0E-4 people are 2/3 as I post. I do not know the answer, but guessed less than 1 meter (more than 1cm) which is mighty small range (99cm) to hit. I based my guess on fact that the angular momentum difference between two "geostationary" orbits only 10s of cm different in circular altitudes, must be "dam small" also. So for me it boiled down to which "dam small" is roughly the same as the "dam small" change in Earth's angular momentum associated with putting much, much, larger (than these "dam smalls") angular momentum into orbit. I can hardly wait to learn what answer is, but as usual I am much too lazy to calculate it when I thought of this way to get someone else to do it. Please Register or Log in to view the hidden image! The 1cm to 0.1mm range (a 1cm band, essentially*) was too small a range for me to chose it. I predict few will chose it. (Unless it is the correct answer and we have a few cheaters among us.) Please Register or Log in to view the hidden image! The 1.0E-4 range is "open bottomed" (if I may coin a phrase, without being accused of some new form of preversion. Please Register or Log in to view the hidden image! ) but it is after all only a band of o.1mm and I did not like that narrow choice. I must admit I do admire the courage of the guy/gal who went for "greater than 1000M." (I can no longer see voter IDs that I saw before voting, but I guess this is necessary as if written under the color bars there might not be room. - Is there a way to see who is guessing what still?) --------------------------------------- *"essentially" added by edit as I do not want to be jumped on again for poor science methods etc.
As only 10 days before poll closes, feel free to give the angular momentum of Earth. (In Kg meter^2 / second.) That is the only part requiring any real effort, at least as I would do it. If you give that to me, I will calculate the answer too. - I am lazy, but not that lazy and also very courious to know the answer.
Here you go. The moment of inertia of the earth - a solid sphere - is: I = 2/5 M<sub>E</sub> R<sub>E</sub><sup>2</sup> = 9.7212563816824800E+37 Kg m<sup>2</sup> (I kept the satellite a point mass: M<sub>S</sub> d<sub>S</sub><sup>2</sup>) The angular momentum is of course I x w but if you do it as a ratio of w<sub>final</sub> to w<sub>initial</sub> to find the change (as I did) you don't really need w. You will need a precision calculator. I found some on the web. Excel only keeps 15 digit precision... It'll be interesting to compare results and methods!
Thanks. Yes it will be. I have not done calculation yet, but method I have in mind does not requre much accuracy (No small differences between two big, almost equal, numbers. - I think, my old slide rule will get answer to two places OK.) I will just call I = 9.72E37 Kg M^2 and suggest others do the same (unless some one thinks SL's number is significantly wrong.) Surely it is not known to all those significant figures - I.e. was yours the summer or winter value? Please Register or Log in to view the hidden image! Perhaps someone will be kind enough to give us the nominal geostationary distance from center of Earth (Not altitude) so all can use that as a standard value as well. Every one should of course follow SL and assume satellite is point mass, without rotation, I might add.
I suspect sl's number is high. Maybe as much as 20 or 30 percent. Remember the earth's core is a lot denser than the mantle and crust. I believe the formula he's using is for a sphere of uniform density.
kevinalm, you are probably right, but I was unable to find a reference for an empirically determined moment of inertia or spin angular momentum. In such cases an approximation like this is the best you can do (like the famous spherical cows approximation). Since the various answers are generally separated by 2 orders of magnitude even a 30% error should be ok. However, if you are able to find an empirical reference please post it and we will gladly use that. -Dale
I also could only find the form of I for a uniform solid sphere. I also don't think 20% - 30% matters (and I personally don't think the earth is that non-uniform).
Agreed. I'm an engineer so these kinds of approximations/assumptions are just standard operating procedure. I was planning on using the solid uniform sphere approximation myself whenever I did the calculation. I don't think you could make a better approximation, only a measured value of the spin would be preferable. -Dale
Any ideas as to how the Earth's angular momentum (or actually the moment of inertia, I, as rotation rate is very well known) can be measured? Is any method accurate enough to actually mesure the seasonal difference as ice above sea level melts and runs into ocean, etc?
Well, the density of most rock is somewhere around 2 to 3 g/cc and heavy metals like Fe and Ni around 5 or 6 g/cc. Actually I'm not suggesting that we should try to correct the constant density model, just that we be aware of the limitations. SL's figure should be taken as an upper bound for the Earth's rotational inertia.
I think that the precession of the equinoxes should give a measure of moment of inertia. Hang on, I'll ask Google Please Register or Log in to view the hidden image!
Here are all the numbers I used: M<sub>E</sub> = 5.9742 x 10<sup>24</sup>kg M<sub>S</sub> = 1000kg (given) R<sub>E</sub> = 6378 x 10<sup>3</sup>m D<sub>s</sub> = 35786 x 10<sup>3</sup>m (above the surface, not earth centered. You can figure it that way if you want...) All courtesy Google, the oracle of our age.
OK, this site (Precession, by Dr Jeffrey Barker, Associate Professor of Geophysics at Penn. State U.) says that because we know the Earth's precession and the torque applied by the Moon reasonably accurately, we can pin down the Earth's Moment of Inertia to 8.07 E37 kg-m^2, or about 17% less than the uniform sphere estimate (kudos to kevinalm!) But like others said, for the purpose of the exercise, I think that a nice round figure of 10^38 kg.m^2 will be fine for our purposes today.
I note that we're assuming that all fuel expended from the launching rocket returns to Earth, otherwise more torque is applied to the satellite and less (or none!) to Earth. Is this reasonable?
