Synchronising the Earth's rotation via mass redistribution 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?
 A: 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.
