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How much material would have to be moved per year from mountain-tops to valleys in order to keep the Earth's rotation synchronised with UTC, thus removing the need for leap seconds to be periodically added? Would it be a feasible project to undertake in order to resolve the current disagreement about the future of the leap second?

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up vote 3 down vote accepted

That seems like a fun question!

According to Wikipedia the day is currently 2ms too long, so that's a factor of 2.31e-8. So we need to reduce the angular momentum of the earth by this factor.

To make life easy consider a mountain on the equator, with a mass $m$, treat it as a point mass and assume we manage to move it $d$ meters nearer the centre of the earth. The change in angular momentum is:

$$\Delta L = m(r_e - d)^2 - m(r_e + d)^2 = -4mr_e d$$

where $r_e$ is the radius of the Earth. Assuming the Earth is a uniform sphere it's angular momentum is:

$$L_e = \frac {2}{5} M r_e^2$$

so I get the fractional change of the angular momentum to be:

$$\frac {\Delta L}{L} = 10 \frac{d}{r_e} \frac {m}{M}$$

Bearing mind that we're modelling the mountain as a point mass, I'd say about 10km was a reasonable distance to move it, i.e. from 5km above sea level to 5km below sea level, so taking $d$ as 10km and $r_e$ as 6380km and setting the change equal to 2.31e-8 gives:

$$\frac {m}{M} = 1.5 \times 10^{-6}$$

so if the mass of the Earh is about $6 \times 10^{24}$kg, you'd need to move about $10^{19}$ kg of mountain.

For comparison, a quick Google suggests the mass of Mount Everest is of the order of $10^{15}$ to $10^{16}$ kg so that's somewhere between 1,000 and 10,000 Mount Everests.

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If you could fire Mount Everest into space, that would do the trick ... – John Rennie Feb 1 '12 at 14:59

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