Zisin (Journal of the Seismological Society of Japan. 2nd ser.)

Density and κ/ρ of the silicate portion of stony meteorites were calculated and the comparison with the Earth's mantle were made for several simplified models of the origin of Earth.
It was shown that both density and compressibility of the silicate portion of stony meteorites decrease with the content of metallic portion, and that the stony meteorites are not good representatives of the Earth's mantle, when a polymorphic transition of crystal lattices would not be assumed.

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The amount of energy E which is sent out from the “origin” of a large earthquake in the form of elastic waves is so enormous that it will be difficult to conceive of this much amount having been stored up within a small confined volume of the earth's crust until the outbreak takes place. Since the material of which the earth's crust is made up has the limit of strength which is finite, a huge volume of the earth's crust must be needed for this much energy to be stored up in it in order that at no part within the volume the stress should exceed this limit of strength. Let us call this volume the “earthquake volume.”
The writer postulates that E is given by
E=1/2ex²V=9/2ex²d³,
where e, x and d are the effective elastic constant, ultimate strain, and thickness of the earth's crust respectively and V is the earthquake volume d×3d×3d. Putting e=5×10¹¹-10¹², x=10^-4-2×10^-4, d=4×10⁶-5×10⁶, in this formula, E is found to be 1.4×10²⁴-2.3×10²⁵. The 1955 formula of B. GUTENBERG and C. RICHTER, log E=11.8+1.5M, gives E=5.0×10²⁴ for M=8.6, which is the largest magnitude listed in “Seismicity of the Earth.” The two estimated values for E agree very well.
According to T. UTSU and A. SEKI, the area A of aftershock occurrences is related to the magnitude M of the main shock as follows:
log A=M+6.
If M is eliminated between this formula and log E=11.8+1.5M, we get
E=6×10²×A1·5.
If the aftershock area is the earthquake volume projected on the earth's surface, E can be written as follows.
E=1/6ex²A1·5.
If we put e=5×10¹¹, x=10^-4, into this formula we get
E=8×10²×A1·5
which agrees well with E=6×10²×A1·5.

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