Standard gravity

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Standard Gravity (usually algebraically written in italics as the lower case symbol g or gn or g0) is the name for the nominal acceleration due to gravity at the Earths surface at sea level.

It is equal to 9.80665  m·s−2 (approx. 32.174 ft·s−2).

The value of g defined above is an arbitrary midrange value on the Earth, approximately equal to the sea level acceleration of free fall at a geodetic latitude of about 45.5°; it is larger in magnitude than the average sea level acceleration on Earth, which is about 9.797 645 m·s−2.

The units of acceleration due to gravity, meters per second squared, are interchangeable with newtons per kilogram. The quantity, 9.806 65, stays the same. These alternate units may be more helpful when considering problems involving pressure due to gravity, or weight.

In practice the actual gravity on the Earth varies according to location; but for weights and measures and many calculation purposes this figure is used.

The standard acceleration of free fall is properly written as gn (sometimes g0) to distinguish it from the local value of g that varies with position.

The symbol g is properly written in lowercase and italic, to distinguish it from the symbol G, the gravitational constant, which is always written in uppercase; and from g, the abbreviation for gram, which is not italicized. The conventional value was established by the 3rd CGPM (1901, CR 70).

[edit] Variations of Earth's gravity

Gravity varies by altitude, latitude and local variation.

On the earth's surface, the gravity will depend on the location at which it is measured, and is smaller at lower latitudes, for two reasons.

The first is that in a rotating non-inertial or accelerated reference frame, as is the case on the surface of the earth, there appears a 'fictitious' centrifugal force acting in a direction perpendicular to the axis of rotation. The gravitational force on a body is partially offset by this centrifugal force, reducing its weight. This effect is smallest at the poles, where the gravitational force and the centrifugal force are orthogonal, and largest at the equator. This effect on its own would result in a range of values of g from 9.789 m·s−2 at the equator to 9.832 m·s−2 at the poles [1] .

The second reason is that the Earth's equatorial bulge (itself also caused by centrifugal force), causing objects at the equator to be farther from the planet's centre than objects at the poles. Because the force due to gravitational attraction between two bodies (the Earth and the object being weighed) varies inversely with the square of the distance between them, objects at the equator experience a weaker gravitational pull than objects at the poles.

The combined result of these two effects is that g is 0.052 m·s−2 more, hence the force due to gravity of an object is 0.5 % more, at the poles than at the equator.

If the terrain is at sea level, we can estimate g:

g_{\phi}=9.780 327 \left( 1+0.0053024\sin^2 \phi-0.0000058\sin^2 2\phi \right)

where

gφ = acceleration in m·s−2 at latitude φ

This is the International Gravity Formula 1967, the 1967 Geodetic Reference System Formula, Helmert's equation or Clairault's formula.

The first correction to this formula is the free air correction (FAC), which accounts for heights above sea level. Gravity decreases with height, at a rate which near the surface of the Earth is such that linear extrapolation would give zero gravity at a height of one half the radius of the Earth, i.e. the rate is 9.8 m·s−2 per 3200 km. Thus:

g_{\phi}=9.780 327 \left( 1+0.0053024\sin^2 \phi-0.0000058\sin^2 2\phi \right) - 3.086 \times 10^{-6}h

where

h = height in meters above sea level

For flat terrain above sea level a second term is added, for the gravity due to the extra mass; for this purpose the extra mass can be approximated by an infinite horizontal slab, and we get 2πG times the mass per unit area, i.e. 4.2×10-10 m3·s−2·kg−1 (0.042 μGal·kg−1·m2)) (the Bouguer correction). For a mean rock density of 2.67 g·cm−3 this gives 1.1×10-6 s−2 (0.11 mGal·m−1). Combined with the free-air correction this means a reduction of gravity at the surface of ca. 2 µm·s−2 (0.20 mGal) for every meter of elevation of the terrain. (The two effects would cancel at a surface rock density of 4/3 times the average density of the whole Earth.)

For the gravity below the surface we have to apply the free-air correction as well as a double Bouguer correction. With the infinite slab model this is because moving the point of observation below the slab changes the gravity due to it to its opposite. Alternatively, we can consider a spherically symmetrical Earth and subtract from the mass of the Earth that of the shell outside the point of observation, because that does not cause gravity inside. This gives the same result.

Local variations in both the terrain and the subsurface cause further variations; the gravitational geophysical methods are based on these: the small variations are measured, the effect of the topography and other known factors is subtracted, and from the resulting variations conclusions are drawn. See also physical geodesy and gravity anomaly.

Helmert's equation may be written equivalently to the version above as either:

gφ = 9.8061999 − 0.0259296cos(2φ) + 0.0000567cos2(2φ)

or

gφ = 9.780327 + 0.0516323sin2(φ) + 0.0002269sin4(φ)

An alternate formula for g as a function of latitude is the WGS (World Geodetic System) 84 Ellipsoidal Gravity Formula:

g_{\phi}= 9.7803267714 ~ \frac {1 + 0.00193185138639\sin^2\phi}{\sqrt(1 - 0.00669437999013\sin^2\phi)}

A spot check comparing results from the WGS-84 formula with those from Helmert's equation (using increments 10 degrees of latitude starting with zero) indicated that they produce values which differ by less than 1e-6 m/s2.

[edit] Calculated value of g

Given the law of universal gravitation, g is merely a collection of factors in that equation:

F = G \frac{m_1 m_2}{r^2}=\left(G \frac{m_1}{r^2}\right) m_2 where g is the bracketed factor and thus:
g=G \frac {m_1}{r^2}

To find the acceleration due to gravity at sea level you can plug in values of G and the mass (in kilograms) and radius (in meters) of the Earth to obtain the calculated value of g:

g=G \frac {m_1}{r^2}=(6.6742 \times 10^{-11}) \frac{5.9736 \times 10^{24}}{(6.37101 \times 10^6)^2}=9.822 \mbox{ m.s}^{-2}

This agrees approximately with the measured value of g. The difference may be attributed to several factors:

  • The Earth is not homogeneous
  • The Earth is not a perfect sphere
  • The choice of a value for the radius of the Earth (an average value is used above)
  • This calculated value of g does not include the centrifugal force effects that are found in practice due to the rotation of the Earth

There are significant uncertainties in the values of r and of m1 as used in this calculation. However, the value of G can be measured precisely and in fact, Henry Cavendish performed the reverse calculation to estimate the mass of the Earth.

[edit] See also

  • g-force a measure of apparent acceleration
  • G the gravitational constant in Newtons law of gravity (not to be confused with g).
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