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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.
The Coriolis force is important in external ballistics for calculating the trajectories of very long-range artillery shells. The most famous historical example was the Paris gun, used by the Germans during World War I to bombard Paris from a range of about 120 km (75 mi). The Coriolis force minutely changes the trajectory of a bullet, affecting ...
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.
[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]
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 ...
It is a generalisation of the vector form, which becomes particularly useful if more than two objects are involved (such as a rocket between the Earth and the Moon). For two objects (e.g. object 2 is a rocket, object 1 the Earth), we simply write r instead of r 12 and m instead of m 2 and define the gravitational field g(r) as:
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 ...
Earth's gravity is a bit stronger at the poles than at the equator, because the Earth is not a perfect sphere, so an object at the poles is slightly closer to the center of the Earth than one at the equator; this effect combines with the centrifugal force to produce the observed weight difference. [20]