# Earth's gravity

Earth's gravity, denoted by g, refers to the attractive force that the Earth exerts on objects on or near its surface. Its strength is loosely said in terms of the acceleration produced by it (accelaration due to gravity), which in SI units is measured in m/s² (metres per second per second, equivalently written as m·s−2). 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. This is explained at weight; 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 standard gravity is, by definition, 9.80665 m/s² (32.1740 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 gravitational constant, or g, the abbreviation for gram (which is not italicized).

The change in gravitational strength per unit distance (in a given direction) is known as the gravitational gradient. In the SI system this is measured in m/s² (strength) per metre (distance), which resolves to simply s−2 (inverse seconds squared). In the cgs system, gravitational gradient is measured in eotvoses.

## Variations on Earth

The strength (or apparent strength) of Earth's gravity varies with latitude, altitude, and local topography and geology. For most purposes the gravitational force is assumed to act in a line directly towards a point at the centre of the Earth, but for very precise work the direction can also vary slightly because the Earth is not a perfectly uniform sphere.

### Latitude

Gravity is weaker at lower latitudes (nearer the equator), 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, an object's velocity vector will be perpendicular to the axis of rotation, but will be prevented from following this vector by centripetal force. The gravitational force on a body is partially offset by this resulting force (known as the pseudo-force 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.

The 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.

### 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 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.

If the Earth was of perfectly uniform composition then, 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.

### Local topography and geology

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. Paris, France has been shown by this method to almost certainly be sitting on a huge, untouchable oilfield[citation needed].

### 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); 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 world

The table below shows the gravitational acceleration in various cities around the world.

 Amsterdam 9.813 m/s² Istanbul 9.808 m/s² Paris 9.809 m/s² Athens 9.807 m/s² Havana 9.788 m/s² Rio de Janeiro 9.788 m/s² Auckland, NZ 9.799 m/s² Helsinki 9.819 m/s² Rome 9.803 m/s² Bangkok 9.783 m/s² Kuwait 9.793 m/s² San Francisco 9.800 m/s² Brussels 9.811 m/s² Lisbon 9.801 m/s² Singapore 9.781 m/s² Buenos Aires 9.797 m/s² London 9.812 m/s² Stockholm 9.818 m/s² Calcutta 9.788 m/s² Los Angeles 9.796 m/s² Sydney 9.797 m/s² Cape Town 9.796 m/s² Madrid 9.800 m/s² Taipei 9.790 m/s² Chicago 9.803 m/s² Manila 9.784 m/s² Tokyo 9.798 m/s² Copenhagen 9.815 m/s² Mexico City 9.779 m/s² Vancouver, BC 9.809 m/s² Nicosia 9.797 m/s² New York 9.802 m/s² Washington, DC 9.801 m/s² Jakarta 9.781 m/s² Oslo 9.819 m/s² Wellington, NZ 9.803 m/s² Frankfurt 9.810 m/s² Ottawa 9.806 m/s² Zurich 9.807 m/s²

### Mathematical models

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

$g_{\phi}$ = 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 3 200 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. 4.2Template:E 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 1.1Template:E 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 - 0.0259296\cos(2\phi) + 0.0000567\cos^2(2\phi) $


or

$g_{\phi}= 9.780327 + 0.0516323\sin^2(\phi) + 0.0002269\sin^4(\phi) $


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

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

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 10-6 m·s-2.

## Estimating g from the law of universal gravitation

Given the law of universal gravitation, the acceleration due to gravity, 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, 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:

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

#### By Newton's Second Law

$F = G \frac{m_1 m_2}{r^2}$
but $F = mg$ by Newton's Second Law
So, $mg = G \frac{m_1 m_2}{r^2} => g = G \frac{m_1}{r^2}$

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 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 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 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 planets

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). It is usually specified as the location where the pressure is equal to a certain value (normally 75 kPa[citation needed]). For the Sun, the "surface" is taken to mean the photosphere. 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.

Body Multiple of
Earth gravity
m/s²
Sun 27.90 274.1
Mercury 0.3770 3.703
Venus 0.9032 8.872
Earth 1 (by definition) 9.8226
Moon 0.1655 1.625
Mars 0.3895 3.728
Jupiter 2.640 25.93
Saturn 1.139 11.19
Uranus 0.917 9.01
Neptune 1.148 11.28
Pluto 0.0621 0.610

For spherical bodies, surface gravity is calculated as follows: g = (m·G)/rm2

where…
g = surface gravity, in m/s²
m = mass, in kilograms
G = 6.67428 × 10–11
rm = mean or volumetric radius, in meters

When a planet’s mean radius is unavailable, it can be calculated from planetary volume as follows: rm = (v·k)(1/3)

where…
rm = mean radius, in meters
v = volume, in km³
k = 238,732,415