- Earth's gravity
**Earth's gravity**, denoted by "g", refers to thegravitational attraction that theEarth exerts on objects on or near its surface. Its strength is usually quoted in terms of the acceleration it gives to falling bodies (acceleration due to gravity), which in SI units is measured in m/s² (metres per second per second, equivalently written as m·s^{−2}or N/kg). It has an approximate value of 9.8 m/s², which means that, ignoring air resistance, the speed of an object falling freely near the Earth's surface increases by about 9.8 metres per second every second.There is a direct relationship between gravitational acceleration and the downwards

weight force experienced by objects on Earth. See also "Apparent weight ".The precise strength of the Earth's gravity varies depending on location. The nominal "average" value at the Earth's surface, known as nowrap|

is, by definition, 9.80665 m/s² (32.1740 ft/s²). This quantity is denoted variously as "g"standard gravity _{n}, "g"_{e}(though this sometimes means the normal equatorial value on Earth, 9.78033 m/s²), "g"_{0}, gee, or simply "g" (which is also used for the variable local value). The symbol "g" should not be confused with "G", thegravitational constant , or g, the abbreviation for gram (which is not italicized).**Variations on Earth**The strength (or apparent strength) of Earth's gravity varies with

latitude , altitude, and localtopography andgeology . The gravitational force is often assumed to act directly towards the centre of the Earth, but the direction varies slightly because the Earth is not a perfectly uniform sphere.**Latitude**Apparent gravity is weaker at lower latitudes (nearer the equator), for two reasons. The first is that the surface of the Earth is not an

inertial frame of reference : acentripetal force is required to keep objects on the surface moving in a circular path around the Earth's axis as the Earth rotates. In providing this centripetal force, some of the gravitational force on an object is "used up" leaving less to contribute to the object's weight. The apparent force due to rotation is called thecentrifugal force ; on the surface of the Earth, it depends on an object's mass and its latitude (it's zero at the poles and largest at the equator) soclarifyme it is customarily combined with the Newtonian gravity. Thus, the local gravity, "g", is not purely the result of gravitational force, but is actually "effective gravity", a combination of true gravity and centrifugal force. It is effective gravity that you measure when you step on a scale or hang aplumb bob . 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. [*cite conference*]

first = Richard

last = Boynton

title = "Precise Measurement of Mass"

booktitle = SAWE PAPER No. 3147

publisher = S.A.W.E., Inc.

date = 2001

location = Arlington, Texas

url = http://www.space-electronics.com/Literature/Precise_Measurement_of_Mass.PDF

accessdate = 2007-01-21The second reason is that the Earth's

equatorial bulge (itself also caused by centrifugal force) causes 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.In combination, the equatorial bulge and the effects of centrifugal force mean that sea-level gravitational acceleration increases from about 9.780 m·s

^{-2}at the equator to about 9.832 m·s^{-2}at the poles, so an object will weigh about 0.5% more at the poles than at the equator. [*[*]*http://curious.astro.cornell.edu/question.php?number=310 "Curious About Astronomy?"*] , Cornell University, retrieved June 2007The same two factors influence the direction of the effective gravity. Anywhere on Earth away from the equator or poles, effective gravity points not exactly toward the center of the Earth, but rather perpendicular to the surface of the

geoid , which, due to the flattened shape of the Earth, is somewhat toward the opposite pole. About half of the deflection is due to centrifugal force, and half because the extra mass around the equator causes a change in the direction of the true gravitational force relative to what it would be on a spherical Earth.**Altitude**Gravity decreases with altitude, since greater altitude means greater distance from the Earth's centre. All other things being equal, an increase in altitude from sea level to the top of

Mount Everest (8,850 metres) causes a weight decrease of about 0.28%. (An additional factor affecting apparent weight is the decrease in air density at altitude, which lessens an object's buoyancy. [*[*] ) It is a common misconception that astronauts in orbit are weightless because they have flown high enough to "escape" the Earth's gravity. In fact, at an altitude of 400 kilometres (250 miles), equivalent to a typical orbit of the*http://www.npl.co.uk/server.php?show=ConWebDoc.2092 "I feel 'lighter' when up a mountain but am I?"*] , National Physical Laboratory FAQSpace Shuttle , gravity is still nearly 90% as strong as at the Earth's surface, and weightlessness actually occurs because orbiting objects are infree-fall .If the Earth were of perfectly uniform composition, during a descent to the centre of the Earth, gravity would decrease linearly with distance, reaching zero at the centre. In reality, the gravitational field peaks within the Earth at the

core-mantle boundary where it has a value of 10.7 m/s², because of the marked increase in density at that boundary.Fact|date=July 2008**Local topography and geology**:"See also

physical geodesy ."Local variations in

topography (such as the presence of mountains) andgeology (such as the density of rocks in the vicinity) cause fluctuations in the Earth's gravitational field, known as gravitational anomalies. Some of these anomalies can be very extensive, resulting in bulges insea level , and throwingpendulum clocks out of synchronisation.The study of these anomalies forms the basis of gravitational

geophysics . The fluctuations are measured with highly sensitivegravimeter s, the effect of topography and other known factors is subtracted, and from the resulting data conclusions are drawn. Such techniques are now used byprospectors to findoil andmineral deposits . Denser rocks (often containing mineralore s) cause higher than normal local gravitational fields on the Earth's surface. Less densesedimentary rock s cause the opposite.Paris , France has been shown by this method to almost certainly be sitting on a huge, untouchableoilfield Fact|date=June 2007.**Other factors**In air, objects experience a supporting

buoyancy force which reduces the apparent strength of gravity (as measured by an object's weight). The magnitude of the effect depends on air density (and hence air pressure); seeApparent weight for details.The gravitational effects of the

Moon and theSun (also the cause of thetide s) have a very small effect on the apparent strength of Earth's gravity, depending on their relative positions; typical variations are 2 µm/s² (0.2 mGal) over the course of a day.**Comparative gravities in various cities around the world**The table below shows the gravitational acceleration in various cities around the world; amongst these cities, it is lowest in Mexico City (9.779 m/s²) and highest in Oslo and Helsinki (9.819 m/s²).

**Mathematical models**If the terrain is at sea level, we can estimate "g"::$g\_\{phi\}=9.780\; 327\; left(\; 1+0.0053024sin^2\; phi-0.0000058sin^2\; 2phi\; ight)$

where :$g\_\{phi\}$ = acceleration in m·s

^{−2}at latitude :$phi$This is the International Gravity Formula 1967, the 1967 Geodetic Reference System Formula, Helmert's equation or Clairault's formula. [

*http://geophysics.ou.edu/solid_earth/notes/potential/igf.htm International Gravity 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 3 200 km. Thus::$g\_\{phi\}=9.780\; 327\; left(\; 1+0.0053024sin^2\; phi-0.0000058sin^2\; 2phi\; ight)\; -\; 3.086\; imes\; 10^\{-6\}h$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 2π"G" times the mass per unit area, i.e. 4.2e|-10 m

^{3}·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 1.1e|-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.

Helmert's equation may be written equivalently to the version above as either::$g\_\{phi\}=\; 9.8061999\; -\; 0.0259296cos(2phi)\; +\; 0.0000567cos^2(2phi)$or:$g\_\{phi\}=\; 9.780327\; +\; 0.0516323sin^2(phi)\; +\; 0.0002269sin^4(phi)$

An alternate formula for "g" as a function of latitude is the WGS (

World Geodetic System ) 84 Ellipsoidal Gravity Formula::$g\_\{phi\}=\; 9.7803267714\; ~\; frac\; \{1\; +\; 0.00193185138639sin^2phi\}\{sqrt\{1\; -\; 0.00669437999013sin^2phi$

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 10

^{-6}m·s^{-2}.**Estimating "g" from the law of universal gravitation**From the

law of universal gravitation , the force on a body acted upon by Earth's gravity is given by:$F\; =\; G\; frac\{m\_1\; m\_2\}\{r^2\}=left(G\; frac\{m\_1\}\{r^2\}\; ight)\; m\_2$where "r" is the distance between the centre of the Earth and the body (see below), and here we take "m"_{1}to be the mass of the Earth and "m"_{2}to be the mass of the body.Additionally,

Newton's second law , "F" = "ma", where "m" is mass and "a" is acceleration, here tells us that:$F\; =\; m\_2g,$

Comparing the two formulas it is seen that::$g=G\; frac\; \{m\_1\}\{r^2\}$So, to find the acceleration due to gravity at sea level, substitute the values of the

gravitational constant , "G", the Earth'smass (in kilograms), "m"_{1}, and the Earth'sradius (in meters), "r", to obtain the value of "g"::$g=G\; frac\; \{m\_1\}\{r^2\}=(6.6742\; imes\; 10^\{-11\})\; frac\{5.9736\; imes\; 10^\{24\{(6.37101\; imes\; 10^6)^2\}=9.822\; mbox\{\; m\}\; cdot\; mbox\{s\}^\{-2\}$Note that this formula only works because of the pleasant (but non-obvious) mathematical fact that the gravity of a uniform spherical body, as measured on or above its surface, is the same as if all its mass were concentrated at a point at its centre. This is what allows us to use the Earth's radius for "r".

The value obtained agrees approximately with the measured value of "g". The difference may be attributed to several factors, mentioned above under "Variations":

*The Earth is nothomogeneous

*The Earth is not a perfect sphere, and an average value must be used for its radius

*This calculated value of "g" does not include thecentrifugal force effects that are found in practice due to the rotation of the EarthThere are significant uncertainties in the values of "r" and "m"

_{1}as used in this calculation, and the value of "G" is also rather difficult to measure precisely.If "G", "g" and "r" are known then a reverse calculation will give an estimate of the mass of the Earth. This method was used by

Henry Cavendish .**Comparative gravities of the Earth, Sun, Moon, planets and Pluto**The table below shows comparative gravitational accelerations at the surface of the Sun, the Earth's moon, each of the planets in the solar system, and Pluto. The “surface” is taken to mean the cloud tops of the

gas giants (Jupiter, Saturn, Uranus and Neptune). For the Sun, the surface is taken to mean thephotosphere . The values in the table have not been de-rated for the centrifugal effect of planet rotation (and cloud-top wind speeds for the gas giants) and therefore, generally speaking, are similar to the actual gravity that would be experienced near the poles.For spherical bodies, surface gravity (cloud-top gravity for the gas giants) is calculated as follows: "g" = ("m"·"G")/"r"

_{m}^{2}::where…

:"g" = surface gravity, in m/s²:"m" = mass, in kilograms:"G" = 6.67428 × 10

^{–11}:"" = mean or , in metersWhen a planet’s mean radius is unavailable, it can be calculated from planetary volume as follows: "r"

_{m}= ("v"·"k")^{(1/3)}::where…

:"r"

_{m}= mean radius, in meters:"v" = volume, in km³:"k" = 238,732,415

**References****See also***

Gravitation

*Gravitational acceleration

*Gravity of the Moon

*Earth's magnetic field

*Newton's law of universal gravitation **External links*** [

*http://www.csr.utexas.edu/grace/ GRACE - Gravity Recovery and Climate Experiment*]

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