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The standard acceleration of gravity or standard acceleration of free fall, often called simply standard gravity and denoted by ɡ 0 or ɡ n, is the nominal gravitational acceleration of an object in a vacuum near the surface of the Earth. It is a constant defined by standard as 9.806 65 m/s 2 (about 32.174 05 ft/s 2).
Gravity on the Earth's surface varies by around 0.7%, from 9.7639 m/s 2 on the Nevado Huascarán mountain in Peru to 9.8337 m/s 2 at the surface of the Arctic Ocean. [6] In large cities, it ranges from 9.7806 m/s 2 [7] in Kuala Lumpur, Mexico City, and Singapore to 9.825 m/s 2 in Oslo and Helsinki.
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.
The data is in good agreement with the predicted fall time of /, where h is the height and g is the free-fall acceleration due to gravity. Near the surface of the Earth, an object in free fall in a vacuum will accelerate at approximately 9.8 m/s 2, independent of its mass.
Near the surface of the Earth, the acceleration due to gravity g = 9.807 m/s 2 (metres per second squared, which might be thought of as "metres per second, per second"; or 32.18 ft/s 2 as "feet per second per second") approximately. A coherent set of units for g, d, t and v is essential.
Now, imagine a distant rocket ship that has escaped Earth's gravitational field and is accelerating at a constant 9.8 m/s 2 due to thrust from its rockets; objects in the rocket ship that are unconstrained will move towards the floor with an acceleration of 9.8 m/s 2. This example shows one way that a uniformly accelerating reference frame is ...
For example, the equation above gives the acceleration at 9.820 m/s 2, when GM = 3.986 × 10 14 m 3 /s 2, and R = 6.371 × 10 6 m. The centripetal radius is r = R cos(φ), and the centripetal time unit is approximately (day / 2 π), reduces this, for r = 5 × 10 6 metres, to 9.79379 m/s 2, which is closer to the observed value. [citation needed]
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 farthest from the center of the Earth. The force of gravity varies with latitude, and the resultant acceleration increases from about 9.780 m/s 2 at the Equator to about 9.832 m/s 2 at the poles ...