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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. The gravity depends only on the mass inside the sphere of radius r. All the contributions from outside cancel out as a consequence of the inverse-square law of gravitation. Another ...
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 ...
Although the symbol É¡ is sometimes used for standard gravity, É¡ (without a suffix) can also mean the local acceleration due to local gravity and centrifugal acceleration, which varies depending on one's position on Earth (see Earth's gravity). The symbol É¡ should not be confused with G, the gravitational constant, or g, the symbol for gram.
Usually, the relationship between mass and weight on Earth is highly proportional; objects that are a hundred times more massive than a one-liter bottle of soda almost always weigh a hundred times more—approximately 1,000 newtons, which is the weight one would expect on Earth from an object with a mass slightly greater than 100 kilograms.
One g is the force per unit mass due to gravity at the Earth's surface and is the standard gravity (symbol: g n), defined as 9.806 65 metres per second squared, [5] or equivalently 9.806 65 newtons of force per kilogram of mass.
Instead, the result was originally expressed as the relative density of Earth, [5] or equivalently the mass of Earth. His experiment gave the first accurate values for these geophysical constants. The experiment was devised sometime before 1783 by geologist John Michell , [ 6 ] [ 7 ] who constructed a torsion balance apparatus for it.
At a fixed point on the surface, the magnitude of Earth's gravity results from combined effect of gravitation and the centrifugal force from Earth's rotation. [2] [3] At different points on Earth's surface, the free fall acceleration ranges from 9.764 to 9.834 m/s 2 (32.03 to 32.26 ft/s 2), [4] depending on altitude, latitude, and longitude.
Non-zero coefficients C n m, S n m correspond to a lack of rotational symmetry around the polar axis for the mass distribution of Earth, i.e. to a "tri-axiality" of Earth. For large values of n the coefficients above (that are divided by r ( n + 1) in ( 9 )) take very large values when for example kilometers and seconds are used as units.