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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 ...
When the rotational component is included (as above), the gravity at the equator is about 0.53% less than that at the poles, with gravity at the poles being unaffected by the rotation. So the rotational component of change due to latitude (0.35%) is about twice as significant as the mass attraction change due to latitude (0.18%), but both ...
All that was needed to obtain a numerical value for standard gravity was now to measure the gravitational strength at the International Bureau. This task was given to Gilbert Étienne Defforges of the Geographic Service of the French Army. The value he found, based on measurements taken in March and April 1888, was 9.80991(5) m⋅s −2. [6]
A conventional standard value is defined exactly as 9.80665 m/s² (about 32.1740 ft/s²). Locations of significant variation from this value are known as gravity anomalies. This does not take into account other effects, such as buoyancy or drag.
The difference of 0.0178 m/s 2 between the gravitational acceleration at the poles and the true gravitational acceleration at the Equator is because objects located on the Equator are about 21 km (13 mi) further away from the center of mass of the Earth than at the poles, which corresponds to a smaller gravitational acceleration.
The force of gravity is weakest at the equator because of the centrifugal force caused by the Earth's rotation and because points on the equator are furthest from the center of the Earth. The force of gravity varies with latitude and increases from about 9.780 m/s 2 at the Equator to about 9.832 m/s 2 at the poles. [80] [81]
According to the New York Times, here's exactly how to play Strands: Find theme words to fill the board. Theme words stay highlighted in blue when found.
[20] [21] As a result, gravity increases from the equator to the poles. By applying Clairaut's theorem, Laplace found from 15 gravity values that f = 1/330. A modern estimate is 1/298.25642. [22] See Figure of the Earth for more detail. For a detailed account of the construction of the reference Earth model of geodesy, see Chatfield. [23]