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Analogously to the way that electromagnetic fields are related to the distribution of charges and currents via Maxwell's equations, the EFE relate the spacetime geometry to the distribution of mass–energy, momentum and stress, that is, they determine the metric tensor of spacetime for a given arrangement of stress–energy–momentum in the ...
The surface of a sphere can be completely described by two dimensions, since no matter how rough the surface may appear to be, it is still only a surface, which is the two-dimensional outside border of a volume. Even the surface of the Earth, which is fractal in complexity, is still only a two-dimensional boundary along the outside of a volume. [3]
However, the repulsive electromagnetic pressures resulting from protons being tightly squeezed inside atomic nuclei are typically on the order of 10 28 atm ≈ 10 33 Pa ≈ 10 33 kg·s −2 m −1. This amounts to about 1% of the nuclear mass density of approximately 10 18 kg/m 3 (after factoring in c 2 ≈ 9×10 16 m 2 s −2). [19]
For example, w = 0 describes a matter-dominated universe, where the pressure is negligible with respect to the mass density. From the generic solution one easily sees that in a matter-dominated universe the scale factor goes as a ( t ) ∝ t 2 / 3 {\displaystyle a(t)\propto t^{2/3}} matter-dominated Another important example is the case of a ...
The density at the center is the same as in the PREM, but the surface density is chosen so that the mass of the sphere equals the mass of the real Earth. See also: Shell theorem An approximate value for gravity at a distance r from the center of the Earth can be obtained by assuming that the Earth's density is spherically symmetric.
Cavendish's stated aim was the "weighing of Earth", that is, determining the average density of Earth and the Earth's mass. His result, ρ 🜨 = 5.448(33) g⋅cm −3, corresponds to value of G = 6.74(4) × 10 −11 m 3 ⋅kg −1 ⋅s −2. It is surprisingly accurate, about 1% above the modern value (comparable to the claimed relative ...
The actual Hill radius for the Earth-Moon pair is on the order of 60,000 km (i.e., extending less than one-sixth the distance of the 378,000 km between the Moon and the Earth). [ 9 ] In the Earth-Sun example, the Earth ( 5.97 × 10 24 kg ) orbits the Sun ( 1.99 × 10 30 kg ) at a distance of 149.6 million km, or one astronomical unit (AU).
The standard gravitational parameter μ of a celestial body is the product of the gravitational constant G and the mass M of that body. For two bodies, the parameter may be expressed as G ( m 1 + m 2 ) , or as GM when one body is much larger than the other: μ = G ( M + m ) ≈ G M . {\displaystyle \mu =G(M+m)\approx GM.}