Gravity of Earth

In 1901 the third General Conference on Weights and Measures defined a standard gravitational acceleration for the surface of the Earth: g_n = 9.80665 m/s². It was based on measurements done at the Pavillon de Breteuil near Paris in 1888, with a theoretical correction applied in order to convert to a latitude of 45° at sea level.[7] This definition is thus not a value of any particular place or carefully worked out average, but an agreement for a value to use if a better actual local value is not known or not important.[8] It is also used to define the units kilogram force and pound force.

The differences of Earth's gravity around the Antarctic continent.

The surface of the Earth is rotating, so it is not an inertial frame of reference. At latitudes nearer the Equator, the outward centrifugal force produced by Earth's rotation is larger than at polar latitudes. This counteracts the Earth's gravity to a small degree – up to a maximum of 0.3% at the Equator – and reduces the apparent downward acceleration of falling objects.

The second major reason for the difference in gravity at different latitudes is that the Earth's equatorial bulge (itself also caused by centrifugal force from rotation) 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, 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 surface centrifugal force due to rotation mean that sea-level gravity increases from about 9.780 m/s² at the Equator to about 9.832 m/s² at the poles, so an object will weigh approximately 0.5% more at the poles than at the Equator.[2][9]

The graph shows the variation in gravity relative to the height of an object above the surface

Gravity decreases with altitude as one rises above the Earth's surface because greater altitude means greater distance from the Earth's centre. All other things being equal, an increase in altitude from sea level to 9,000 metres (30,000 ft) causes a weight decrease of about 0.29%. (An additional factor affecting apparent weight is the decrease in air density at altitude, which lessens an object's buoyancy.[10] This would increase a person's apparent weight at an altitude of 9,000 metres by about 0.08%)

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 mi), equivalent to a typical orbit of the ISS, gravity is still nearly 90% as strong as at the Earth's surface. Weightlessness actually occurs because orbiting objects are in free-fall.[11]

The effect of ground elevation depends on the density of the ground (see Slab correction section). A person flying at 9,100 m (30,000 ft) above sea level over mountains will feel more gravity than someone at the same elevation but over the sea. However, a person standing on the Earth's surface feels less gravity when the elevation is higher.

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

${\displaystyle g_{h}=g_{0}\left({\frac {R_{\mathrm {e} }}{R_{\mathrm {e} }+h}}\right)^{2}}$

Where

• g_h is the gravitational acceleration at height h above sea level.
• R_e is the Earth's mean radius.
• g₀ is the standard gravitational acceleration.

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

Earth's radial density distribution according to the Preliminary Reference Earth Model (PREM).[12]

Earth's gravity according to the Preliminary Reference Earth Model (PREM).[12] Two models for a spherically symmetric Earth are included for comparison. The dark green straight line is for a constant density equal to the Earth's average density. The light green curved 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.

An approximate value for gravity at a distance r from the center of the Earth can be obtained by assuming that the Earth's density is spherically symmetric. The gravity depends only on the mass inside the sphere of radius r. All the contributions from outside cancel out as a consequence of the inverse-square law of gravitation. Another consequence is that the gravity is the same as if all the mass were concentrated at the center. Thus, the gravitational acceleration at this radius is[13]

${\displaystyle g(r)=-{\frac {GM(r)}{r^{2}}}.}$

where G is the gravitational constant and M(r) is the total mass enclosed within radius r. If the Earth had a constant density ρ, the mass would be M(r) = (4/3)πρr³ and the dependence of gravity on depth would be

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

The gravity g' at depth d is given by g'=g(1-d/R) where g is acceleration due to gravity on the surface of the Earth, d is depth and R is the radius of the Earth. If the density decreased linearly with increasing radius from a density ρ₀ at the center to ρ₁ at the surface, then ρ(r) = ρ₀ − (ρ₀ − ρ₁) r / r_e, and the dependence would be

${\displaystyle g(r)={\frac {4\pi }{3}}G\rho _{0}r-\pi G\left(\rho _{0}-\rho _{1}\right){\frac {r^{2}}{r_{\mathrm {e} }}}.}$

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

Local differences in topography (such as the presence of mountains), geology (such as the density of rocks in the vicinity), and deeper tectonic structure cause local and regional differences in the Earth's gravitational field, known as gravitational anomalies.[14] 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.

In air or water, 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 the air density (and hence air pressure) or the water density respectively; 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.

If the terrain is at sea level, we can estimate ${\displaystyle g\{\phi \}}$, the acceleration at latitude ${\displaystyle \phi }$:

${\displaystyle {\begin{aligned}g\{\phi \}&=9.780327\,\,\mathrm {m} \cdot \mathrm {s} ^{-2}\,\,\left(1+0.0053024\,\sin ^{2}\phi -0.0000058\,\sin ^{2}2\phi \right),\\&=9.780327\,\,\mathrm {m} \cdot \mathrm {s} ^{-2}\,\,\left(1+0.0052792\,\sin ^{2}\phi +0.0000232\,\sin ^{4}\phi \right),\\&=9.780327\,\,\mathrm {m} \cdot \mathrm {s} ^{-2}\,\,\left(1.0053024-0.0053256\,\cos ^{2}\phi +0.0000232\,\cos ^{4}\phi \right),\\&=9.780327\,\,\mathrm {m} \cdot \mathrm {s} ^{-2}\,\,\left(1.0026454-0.0026512\,\cos 2\phi +0.0000058\,\cos ^{2}2\phi \right)\end{aligned}}}$.

This is the International Gravity Formula 1967, the 1967 Geodetic Reference System Formula, Helmert's equation or Clairaut's formula.[17]

An alternative formula for g as a function of latitude is the WGS (World Geodetic System) 84 Ellipsoidal Gravity Formula:[18]

${\displaystyle g\{\phi \}=\mathbb {G} _{e}\left[{\frac {1+k\sin ^{2}\phi }{\sqrt {1-e^{2}\sin ^{2}\phi }}}\right],\,\!}$

where,

• ${\displaystyle a,\,b}$ are the equatorial and polar semi-axes, respectively;
• ${\displaystyle e^{2}=1-(b/a)^{2}}$ is the spheroid's eccentricity, squared;
• ${\displaystyle \mathbb {G} _{e},\,\mathbb {G} _{p}\,}$ is the defined gravity at the equator and poles, respectively;
• ${\displaystyle k={\frac {b\,\mathbb {G} _{p}-a\,\mathbb {G} _{e}}{a\,\mathbb {G} _{e}}}}$ (formula constant);

then, where ${\displaystyle \mathbb {G} _{p}=9.8321849378\,\,\mathrm {m} \cdot \mathrm {s} ^{-2}}$,[18]

${\displaystyle g\{\phi \}=9.7803253359\,\,\mathrm {m} \cdot \mathrm {s} ^{-2}\left[{\frac {1+0.001931850400\,\sin ^{2}\phi }{\sqrt {1-0.006694384442\,\sin ^{2}\phi }}}\right]}$.

where the semi-axes of the earth are:

${\displaystyle a=6378137.0\,\,{\mbox{m}}}$

${\displaystyle b=6356752.3\,\,{\mbox{m}}}$

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

The first correction to be applied to the model is the free air correction (FAC) that accounts for heights above sea level. Near the surface of the Earth (sea level), gravity decreases with height such that linear extrapolation would give zero gravity at a height of one half of the Earth's radius - (9.8 m·s⁻² per 3,200 km.)[19]

Using the mass and radius of the Earth:

${\displaystyle {\begin{aligned}r_{\mathrm {Earth} }&=6.371\cdot 10^{6}\,\mathrm {m} \\m_{\mathrm {Earth} }&=5.9722\cdot 10^{24}\,\mathrm {kg} \end{aligned}}}$

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):

${\displaystyle {\begin{aligned}g_{0}&={\frac {G\,M_{\mathrm {e} }}{R_{\mathrm {e} }^{2}}}=9.81998\,{\frac {\mathrm {m} }{\mathrm {s} ^{2}}}\\G&=6.67408\cdot 10^{-11}\,{\frac {\mathrm {m} ^{3}}{\mathrm {kg} \cdot \mathrm {s} ^{2}}}.\end{aligned}}}$

At a height h above the nominal surface of the Earth g_h is given by:

${\displaystyle g_{h}={\frac {G\,M_{\mathrm {e} }}{\left(R_{\mathrm {e} }+h\right)^{2}}}}$

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

${\displaystyle \Delta g_{h}=\left[{\frac {G\,M_{\mathrm {e} }}{\left(R_{\mathrm {e} }+h\right)^{2}}}\right]-\left[{\frac {G\,M_{\mathrm {e} }}{R_{\mathrm {e} }^{2}}}\right]}$

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

${\displaystyle \Delta g_{h}\approx -\,{\frac {G\,M_{\mathrm {e} }}{R_{\mathrm {e} }^{2}}}\cdot {\frac {2\,h}{R_{\mathrm {e} }}}}$

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

${\displaystyle \Delta g_{h}\approx -3.086\cdot 10^{-6}\,h}$

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

${\displaystyle g\{\phi ,h\}=g\{\phi \}-3.086\cdot 10^{-6}h}$

where ${\displaystyle g\{\phi ,h\}}$ is acceleration in m·s⁻² at latitude ${\displaystyle \ \phi }$ and altitude h in metres.

Note: The section uses the galileo (symbol: "Gal"), which is a cgs unit for acceleration of 1 centimetre/second².

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 metre 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. The density of the whole Earth is 5.515 g·cm⁻³, so standing on a slab of something like iron whose density is over 7.35 g·cm⁻³ would increase one's weight.)

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.

Currently, the static and time-variable Earth's gravity field parameters are being determined using modern satellite missions, such as GOCE, CHAMP, Swarm, GRACE and GRACE-FO.[20][21] The lowest-degree parameters, including the Earth's oblateness and geocenter motion are best determined from Satellite laser ranging.[22]