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[1] [2] The acceleration of a body near the surface of the Earth is due to the combined effects of gravity and centrifugal acceleration from the rotation of the Earth (but the latter is small enough to be negligible for most purposes); the total (the apparent gravity) is about 0.5% greater at the poles than at the Equator. [3] [4]
Physics Today is the membership magazine of the American Institute of Physics. First published in May 1948, it is issued on a monthly schedule, and is provided to the members of ten physics societies, including the American Physical Society. It is also available to non-members as a paid annual subscription.
[4] [5] This indicated the acceleration of gravity was less at Cayenne than at Paris. Pendulum gravimeters began to be taken on voyages to remote parts of the world, and it was slowly discovered that gravity increases smoothly with increasing latitude, gravitational acceleration being about 0.5% greater at the poles than at the equator.
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 ...
The g-force acting on an object under acceleration can be much greater than 1 g, for example, the dragster pictured at top right can exert a horizontal g-force of 5.3 when accelerating. The g-force acting on an object under acceleration may be downwards, for example when cresting a sharp hill on a roller coaster.
G is quite difficult to measure because gravity is much weaker than other fundamental forces, and an experimental apparatus cannot be separated from the gravitational influence of other bodies. Measurements with pendulums were made by Francesco Carlini (1821, 4.39 g/cm 3 ), Edward Sabine (1827, 4.77 g/cm 3 ), Carlo Ignazio Giulio (1841, 4.95 g ...
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 ...
The planet Earth has a rather slight equatorial bulge; its equatorial diameter is about 43 km (27 mi) greater than its polar diameter, with a difference of about 1 ⁄ 298 of the equatorial diameter. If Earth were scaled down to a globe with an equatorial diameter of 1 metre (3.3 ft), that difference would be only 3 mm (0.12 in).