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In these equations, g 0, M and R * are each single-valued constants, while ρ, L, T and h are multi-valued constants in accordance with the table below. The values used for M, g 0 and R * are in accordance with the U.S. Standard Atmosphere, 1976, and that the value for R * in particular does not agree with standard values for this constant. [2]
To make this into an equal-sided formula or equation, there needed to be a multiplying factor or constant that would give the correct force of gravity no matter the value of the masses or distance between them (the gravitational constant). Newton would need an accurate measure of this constant to prove his inverse-square law.
In addition to Poynting, measurements were made by C. V. Boys (1895) [25] and Carl Braun (1897), [26] with compatible results suggesting G = 6.66(1) × 10 −11 m 3 ⋅kg −1 ⋅s −2. The modern notation involving the constant G was introduced by Boys in 1894 [12] and becomes standard by the end of the 1890s, with values usually cited in the ...
Neither of these accounts for changes in gravity with changes in altitude, but the model with the cosine function does take into account the centrifugal relief that is produced by the rotation of the Earth. On the rotating sphere, the sum of the force of the gravitational field and the centrifugal force yields an angular deviation of approximately
Using the integral form of Gauss's Law, this formula can be extended to any pair of objects of which one is far more massive than the other — like a planet relative to any man-scale artifact. The distances between planets and between the planets and the Sun are (by many orders of magnitude) larger than the sizes of the sun and the planets.
On the other hand, the penultimate equation becomes grossly inaccurate at great distances. If an object fell 10 000 m to Earth, then the results of both equations differ by only 0.08 %; however, if it fell from geosynchronous orbit, which is 42 164 km, then the difference changes to almost 64 %.
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The gravity g′ at depth d is given by g′ = g(1 − d/R) where g is acceleration due to gravity on the surface of the Earth, d is depth and R is the radius of the Earth. If the density decreased linearly with increasing radius from a density ρ 0 at the center to ρ 1 at the surface, then ρ(r) = ρ 0 − (ρ 0 − ρ 1) r / R, and the ...