## FANDOM

332 Pages

 This page uses content from the English Wikipedia. The original article was at Gravity of Earth. The list of authors can be seen in the page history. As with the Units of Measurement Wiki, the text of Wikipedia is available under Creative Commons License see Wikia:Licensing.

The gravity of Earth, denoted g, refers to the acceleration that the Earth imparts to objects on or near its surface. In SI units this acceleration is measured in meters per second per second (in symbols, m/s2hi

or m·s-2) or equivalently in newtons per kilogram (N/kg or N·kg-1). It has an approximate value of 9.81 m/s2, which means that, ignoring the effects of air resistance, the speed of an object falling freely near the Earth's surface will increase by about 9.81 meters (about 32.2 ft) per second every second. This quantity is sometimes referred to informally as little g (in contrast, the gravitational constant G is referred to as big G).

There is a direct relationship between gravitational acceleration and the downwards weight force experienced by objects on Earth, given by the equation F = ma (force = mass × acceleration). However, other factors such as the rotation of the Earth also contribute to the net acceleration.

Although the precise strength of Earth's gravity varies depending on location, the nominal "average" value at the Earth's surface, known as standard gravity is, by definition, 9.80665 m/s2 (32.1737 ft/s2). This quantity is denoted variously as gn, ge (though this sometimes means the normal equatorial value on Earth, 9.78033 m/s2), g0, gee, or simply g (which is also used for the variable local value). The symbol g should not be confused with g, the abbreviation for gram (which is not italicized).[1][2]

## Variations in gravity and apparent gravityEdit

A perfect sphere of spherically uniform density (density varies solely with distance from center) would produce a gravitational field of uniform magnitude at all points on its surface, always pointing directly towards the sphere's center. However, the Earth deviates slightly from this ideal, and there are consequently slight deviations in both the magnitude and direction of gravity across its surface. Furthermore, the net force exerted on an object due to the Earth, called "effective gravity" or "apparent gravity", varies due to the presence of other forces, such as inertia caused by the Earth's rotation. A scale or plumb bob measures only this effective gravity.

Parameters affecting the apparent or actual strength of Earth's gravity i

At latitudes nearer the Equator, the inertia produced by Earth's rotation is stronger than at polar latitudes. This counteracts the Earth's gravity to a small degree – up to a maximum of 0.3% at the Equator – reducing the downward acceleration of falling objects.

The second major cause for the difference in gravity at different latitudes is that the Earth's equatorial bulge (itself also caused by inertia) causes objects at the Equator to be farther from the planet's center 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, an object at the Equator experiences a weaker gravitational pull than an object at the poles.

In combination, the equatorial bulge and the effects of the Earth's inertia mean that sea-level gravitational acceleration increases from about 9.70999 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.[3][4]

The 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 inertia, 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 ...the value of gravitional constant is not correct value i have a proofEdit

Gravity decreases with altitude, since greater altitude means greater distance from the Earth's center. All other things being equal, an increase in altitude from sea level to the top of Mount Everest (8,850 meters) 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.[5]) 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 kilometers (250 miles), equivalent to a typical orbit of the Space Shuttle, gravity is still nearly 90% as strong as at the Earth's surface, and weightlessness actually occurs because orbiting objects are in free-fall.[6]

The following formula approximates the Earth's gravity variation with altitude:

$g_h=g_0\left(\frac{r_e}{r_e+h}\right)^2$

Where

This formula treats the Earth as a perfect sphere with a radially symmetric distribution of mass; a more accurate mathematical treatment is discussed below.

### DepthEdit

If the Earth were a sphere of uniform density then gravity would decrease linearly to zero as one travelled in a straight line from the Earth's surface to its center. This is a consequence of Gauss' law for gravity. Because of the spherical symmetry, gravity is radially downward and equal in magnitude at all points at a given radius r. The surface area of a sphere of radius r being 4πr2, Gauss's law gives

$\displaystyle 4\pi r^2 g = 4 \pi GM,$

where G is the gravitational constant and M is the total mass enclosed within the surface. Since, for r less than the Earth's radius and a constant density ρ, M = ρ(4/3)πr3, the dependence of gravity on depth is

$g(r) = \frac{4\pi}{3} G \rho r.$

If the density decreases linearly with increasing radius from a density ρ0 at the center to ρ1 at the surface, then ρ(r) = ρ0 − (ρ0 − ρ1) r / re, and

$g(r) = \frac{4\pi}{3} G \rho_0 r - \frac{4\pi}{3} G \left(\rho_0-\rho_1\right) \frac{r^2}{r_e}.$

The actual depth-dependence of density and gravity, inferred from seismic travel times (see Adams–Williamson equation), is shown in the graphs below.

 Earth's radial density distribution according to the Preliminary Reference Earth Model (PREM).[7] Earth's gravity according to the Preliminary Reference Earth Model (PREM).[7] Two models for a spherically symmetric Earth are included for comparison. The straight dashed line is for a constant density equal to the Earth's average density. The curved dotted line is for a density that decreases linearly from center to surface. The density at the center is the same as in the PREM, but the surface density is chosen so that the mass of the sphere equals the mass of the real Earth.

### Local topography and geologyEdit

Template:See also Local variations in topography (such as the presence of mountains) and geology (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 in sea level, and throwing pendulum clocks out of synchronisation.

The study of these anomalies forms the basis of gravitational geophysics. The fluctuations are measured with highly sensitive gravimeters, the effect of topography and other known factors is subtracted, and from the resulting data conclusions are drawn. Such techniques are now used by prospectors to find oil and mineral deposits. Denser rocks (often containing mineral ores) cause higher than normal local gravitational fields on the Earth's surface. Less dense sedimentary rocks cause the opposite.

### Other factorsEdit

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); see Apparent weight for details.

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/s2 (0.2 mGal) over the course of a day.

### Comparative gravities in various cities around the worldEdit

The table below shows the gravitational acceleration in various cities around the world;[8] amongst these cities, it is lowest in Mexico City (9.779 m/s2) and highest in Oslo (Norway) and Helsinki (Finland) (9.819 m/s2).

### Mathematical modelsEdit

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

$g_{\phi}=9.780327 \left(1+0.0053024\sin^2 \phi-0.0000058\sin^2 2\phi \right) \frac{\mathrm{m}}{\mathrm{s}^2}$

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 Clairaut's formula.[9]

The first correction to be applied 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 is 9.8 m·s−2 per 3,200 km.

Using the mass and radius of the Earth:

$r_\mathrm{Earth}= 6.371 \times 10^{6}\,\mathrm{m}$
$m_\mathrm{Earth}= 5.9736 \times 10^{24}\,\mathrm{kg}$

The FAC correction factor (Δg) can be derived from the definition of the acceleration due to gravity in terms of G, the Gravitational Constant (see Estimating g from the law of universal gravitation, below):

$g_0 = G \, m_\mathrm{Earth} / r_\mathrm{Earth}^2 = 9.8331\,\frac{\mathrm{m}}{\mathrm{s}^2}$

where:

$G = 6.67428 \times 10^{-11}\,\frac{\mathrm{m}^3}{\mathrm{kg}\cdot\mathrm{s}^2}.$

At a height h above the nominal surface of the earth gh is given by:

$g_h = G \, m_\mathrm{Earth} / \left( r_\mathrm{Earth} + h \right) ^2$

So the FAC for a height h above the nominal earth radius can be expressed:

$\Delta g = \left [ G \, m_\mathrm{Earth} / \left( r_\mathrm{Earth} + h \right) ^2 \right ] - \left[G \, m_\mathrm{Earth} / r_\mathrm{Earth}^2 \right]$

This expression can be readily used for programming or inclusion in a spreadsheet. Collecting terms, simplifying and neglecting small terms (h<<rEarth), however yields the good approximation:

$\Delta g = - \, \dfrac{ G \, m_\mathrm{Earth}}{ r_\mathrm{Earth} ^2} \times \dfrac{ 2 \,h}{r_\mathrm{Earth}}$

Using the numerical values above and for a height h in meters:

$\Delta g = - 3.084 \times 10^{-6}\, h$

Grouping the latitude and FAC altitude factors the expression most commonly found in the literature[10] is:

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

where $\ g_{\phi}$ = acceleration in m·s−2 at latitude $\ \phi$ and altitude h in meters. Alternatively (with the same units for h) the expression can be grouped as follows:

$g_{\phi}=9.780327 \left[ \left( 1+0.0053024\sin^2 \phi-0.0000058\sin^2 2\phi \right) - 3.155 \times 10^{-7}h \right] \,\frac{\mathrm{m}}{\mathrm{s}^2}$

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.

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

$\ g_{\phi}= \left(9.8061999 - 0.0259296\cos(2\phi) + 0.0000567\cos^2(2\phi)\right)\,\frac{\mathrm{m}}{\mathrm{s}^2}$

or

$\ g_{\phi}= \left( 9.780327 + 0.0516323\sin^2(\phi) + 0.0002269\sin^4(\phi) \right)\,\frac{\mathrm{m}}{\mathrm{s}^2}$

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

$\ g_{\phi}= \left(9.7803267714 ~ \frac {1 + 0.00193185138639\sin^2\phi}{\sqrt{1 - 0.00669437999013\sin^2\phi}} \right)\,\frac{\mathrm{m}}{\mathrm{s}^2}$

The difference between the WGS-84 formula and Helmert's equation is less than 0.68·10−6 m·s−2.

## Estimating g from the law of universal gravitationEdit

From the law of universal gravitation, the force on a body acted upon by Earth's gravity is given by

$F = G \frac{m_1 m_2}{r^2}=\left(G \frac{m_1}{r^2}\right) m_2$

where r is the distance between the center of the Earth and the body (see below), and here we take m1 to be the mass of the Earth and m2 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's mass (in kilograms), m1, and the Earth's radius (in meters), r, to obtain the value of g:

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

Note that this formula only works because of the 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 center. 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 not homogeneous
• The Earth is not a perfect sphere, and an average value must be used for its radius
• This calculated value of g only includes true gravity. It does not include the reduction of constraint force that we perceive as a reduction of gravity due to the rotation of Earth, and some of gravity being "used up" in providing the centripetal acceleration

There are significant uncertainties in the values of r and m1 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, and planetsEdit

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 their major moons, Pluto and Eris. 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 the photosphere. The values in the table have not been de-rated for the inertia 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.

Body Multiple of
Earth gravity
m/s2
Sun 27.90 274.1
Mercury 0.3770 3.703
Venus 0.9032 8.872
Earth 1 9.8067
Moon 0.1655 1.625
Mars 0.3895 3.728
Jupiter 2.640 25.93
Io 0.182 1.789
Europa 0.134 1.314
Ganymede 0.145 1.426
Callisto 0.126 1.24
Saturn 1.139 11.19
Titan 0.138 1.3455
Uranus 0.917 9.01
Titania 0.039 0.379
Oberon 0.035 0.347
Neptune 1.148 11.28
Triton 0.079 0.779
Pluto 0.0621 0.610
Eris 0.0814 (approx.) 0.8 (approx.)

1. Template:Cite paper
2. "SI Unit rules and style conventions". National Institute For Standards and Technology (USA). September 2004. Retrieved 2009-11-25. "Variables and quantity symbols are in italic type. Unit symbols are in roman type."
3. Template:Cite conference
4. "Curious About Astronomy?", Cornell University, retrieved June 2007
5. "I feel 'lighter' when up a mountain but am I?", National Physical Laboratory FAQ
6. "The G's in the Machine", NASA, see "Editor's note #2"
7. 7.0 7.1 Dziewonski, A.M.; Anderson, D.L.. "Preliminary reference Earth model". Physics of the Earth and Planetary Interiors 25: 297–356.
8. International Gravity formula
9. Chsfootball.net
10. This value excludes the adjustment for centrifugal force due to Earth's rotation and is therefore greater than the 9.80665 m/s2 value of standard gravity.

## Edit

Community content is available under CC-BY-SA unless otherwise noted.