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A mass distribution can be modeled as a measure. This allows point masses, line masses, surface masses, as well as masses given by a volume density function. Alternatively the latter can be generalized to a distribution. For example, a point mass is represented by a delta function defined in 3-dimensional space.
In the case of a disk seen face-on, area density for a given area of the disk is defined as column density: that is, either as the mass of substance per unit area integrated along the vertical path that goes through the disk (line-of-sight), from the bottom to the top of the medium:
So when water molecules (water vapor) are added to a given volume of air, the dry air molecules must decrease by the same number, to keep the pressure from increasing or temperature from decreasing. Hence the mass per unit volume of the gas (its density) decreases. The density of humid air may be calculated by treating it as a mixture of ideal ...
= molar mass of Earth's air: 0.0289644 kg/mol The value of subscript b ranges from 0 to 6 in accordance with each of seven successive layers of the atmosphere shown in the table below. The reference value for ρ b for b = 0 is the defined sea level value, ρ 0 = 1.2250 kg/m 3 or 0.0023768908 slug/ft 3 .
The "mass of the earth in gravitational measure" is stated as "9.81996×6370980 2" in The New Volumes of the Encyclopaedia Britannica (Vol. 25, 1902) with a "logarithm of earth's mass" given as "14.600522" [3.985 86 × 10 14]. This is the gravitational parameter in m 3 ·s −2 (modern value 3.986 00 × 10 14) and not the absolute mass.
In terms of density, m = ρV, where ρ is the volumetric mass density, V is the volume occupied by the mass. This energy can be released by the processes of nuclear fission (~ 0.1%), nuclear fusion (~ 1%), or the annihilation of some or all of the matter in the volume V by matter–antimatter collisions (100%). [citation needed]
At Earth, this energy is passing through a sphere with a radius of a 0, the distance between the Earth and the Sun, and the irradiance (received power per unit area) is given by = The Earth has a radius of R ⊕ , and therefore has a cross-section of π R ⊕ 2 {\displaystyle \pi R_{\oplus }^{2}} .
Solve this equation for the coordinates R to obtain = (), Where M is the total mass in the volume. If a continuous mass distribution has uniform density , which means that ρ is constant, then the center of mass is the same as the centroid of the volume.