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Is there a good way to directly measure a moment of inertia of the Earth, or say, other planet?

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ppl - "directly measure" does not mean estimate or calculate. –  Pavel Radzivilovsky Sep 27 '13 at 13:13
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3 Answers 3

I assume you mean its moment of inertia around some axis which goes through the centre of the planet? Approximating Earth as a perfect sphere, this shouldn't be too hard in principle to derive without using any direct measurements, using the definition of the moment of inertia: $$I=\int r^2\ \mbox{d}m=\int r^2\rho(r)\ \mbox{d}V=4\pi\int_0^{r_E} r^4\rho(r)\ \mbox{d}r$$. All you need is the radial mass density distribution $\rho(r)$, which can be found here. It seems this distribution is not nice and continuous, so you should break up the integral in parts that correspond to relatively smooth curves in the density distribution function (which you should each approximate), and add all parts up.

Alternatively, if you only want a ballpark estimate, you can try to find a function which is a reasonable approximation for the whole range from $r=0$ to $r=r_E$ and use that.

I realize that this might not be exactly what you're looking for, but I thought it was too neat not to mention.

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If you mean measure rather than calculate, then it can be done by studying how the Earth's rotation responds to external forces. fibonatic mentions one approach, but I think a better estimate is possible from measuring the precession of the Earth.

The maths involved are a bit complicated. The Wikipedia article I linked gives a description of the procedure in this section, and a somewhat simpler account is given in this article.

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This. The leading terms of the torque come from the tidal influences of the sun and the moon on the equitorial bulge, and those alone get you reasonably close. –  dmckee Sep 27 '13 at 14:13
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You can approximate it by assuming the planet is a homogeneous sphere, but that would not be accurate.

One other way that came to mind, at least for Earth, is to use the rate at which it tidal locks to the Moon. Since due to this the moon moves a little bit further away from us into a higher orbit, so its orbital energy is increased. However most of Earths energy loss is transformed into heat energy: -3.321 TW and only +0.121 TW is tranfered to the Moon. The dissipation of Earths energy by tidal friction averages about 3.75 TW. The current rate of Earths angular deceleration is 1.4 ms/day/century which translates to about 3.73e-22 rad/s^2. Using these values solve the follow equation allows us to find a value for Earths moment of inertia. $$ P=I\alpha\omega $$ $$ I=\frac{P}{\alpha\omega}=\frac{3.75e12}{3.73e-22*7.27e-5}=1.38e38 kg m^2 $$

However the rate at which Earth's angular velocity changes varies quite a bit, for example the movement of the Earth's crust relative to its core, changes in mantle convection, and any other events or processes that cause a significant redistribution of mass change the Earth's moment of inertia.

So my calculation might not be accurate, since I did a quick search on Earth's moment of inertia, which returned a value of 8e37 kg m$^2$ which is quite a bit lower.

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