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The gravitational effects of the Moon and the Sun (also the cause of the tides) have a very small effect on the apparent strength of Earth's gravity, depending on their relative positions; typical variations are 2 μm/s 2 (0.2 mGal) over the course of a day.
The gravity anomaly at a location on the Earth's surface is the difference between the observed value of gravity and the value predicted by a theoretical model. If the Earth were an ideal oblate spheroid of uniform density, then the gravity measured at every point on its surface would be given precisely by a simple algebraic expression.
For such problems, the rotation of the Earth would be immaterial unless variations with longitude are modeled. Also, the variation in gravity with altitude becomes important, especially for highly elliptical orbits. The Earth Gravitational Model 1996 contains 130,676 coefficients that refine the model of the Earth's gravitational field.
Gravitational field strength within the Earth Gravity field near the surface of the Earth – an object is shown accelerating toward the surface If the bodies in question have spatial extent (as opposed to being point masses), then the gravitational force between them is calculated by summing the contributions of the notional point masses that ...
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.
Gravity is usually measured in units of acceleration.In the SI system of units, the standard unit of acceleration is metres per second squared (m/s 2).Other units include the cgs gal (sometimes known as a galileo, in either case with symbol Gal), which equals 1 centimetre per second squared, and the g (g n), equal to 9.80665 m/s 2.
The actual Hill radius for the Earth-Moon pair is on the order of 60,000 km (i.e., extending less than one-sixth the distance of the 378,000 km between the Moon and the Earth). [9] In the Earth-Sun example, the Earth (5.97 × 10 24 kg) orbits the Sun (1.99 × 10 30 kg) at a distance of 149.6 million km, or one astronomical unit (AU). The Hill ...
The above equation describes the Earth's gravitational potential, not the geoid itself, at location ,,, the co-ordinate being the geocentric radius, i.e., distance from the Earth's centre. The geoid is a particular equipotential surface, [ 27 ] and is somewhat involved to compute.