Gravity of Earth
The gravity of Earth, denoted by g, is the net acceleration that is imparted to objects due to the combined effect of gravitation (from mass distribution within Earth) and the centrifugal force (from the Earth's rotation).[2][3] It is a vector quantity, whose direction coincides with a plumb bob and strength or magnitude is given by the norm .
In SI units, this acceleration is expressed in metres per second squared (in symbols, m/s2 or m·s−2) or equivalently in newtons per kilogram (N/kg or N·kg−1). Near Earth's surface, the acceleration due to gravity, accurate to 2 significant figures, is 9.8 m/s2 (32 ft/s2). This means that, ignoring the effects of air resistance, the speed of an object falling freely will increase by about 9.8 metres per second (32 ft/s) every second. This quantity is sometimes referred to informally as little g (in contrast, the gravitational constant G is referred to as big G).
The precise strength of Earth's gravity varies with location. The agreed upon value for standard gravity is 9.80665 m/s2 (32.1740 ft/s2) by definition.[4] This quantity is denoted variously as gn, ge (though this sometimes means the normal gravity at the equator, 9.7803267715 m/s2 (32.087686258 ft/s2)),[5] g0, or simply g (which is also used for the variable local value).
The weight of an object on Earth's surface is the downwards force on that object, given by Newton's second law of motion, or F = m a (force = mass × acceleration). Gravitational acceleration contributes to the total gravity acceleration, but other factors, such as the rotation of Earth, also contribute, and, therefore, affect the weight of the object. Gravity does not normally include the gravitational pull of the Moon and Sun, which are accounted for in terms of tidal effects.
are the equatorial and polar semi-axes, respectively;
is the spheroid's , squared;
eccentricity
is the defined gravity at the equator and poles, respectively;
(formula constant);
If the terrain is at sea level, we can estimate, for the Geodetic Reference System 1980, , the acceleration at latitude :
This is the International Gravity Formula 1967, the 1967 Geodetic Reference System Formula, Helmert's equation or Clairaut's formula.[18]
An alternative formula for g as a function of latitude is the WGS (World Geodetic System) 84 Ellipsoidal Gravity Formula:[19]
where,
then, where ,[19]
where the semi-axes of the earth are:
The difference between the WGS-84 formula and Helmert's equation is less than 0.68 μm·s−2.
Further reductions are applied to obtain gravity anomalies (see: Gravity anomaly#Computation).
From the law of universal gravitation, the force on a body acted upon by Earth's gravitational force is given by
where r is the distance between the centre of the Earth and the body (see below), and here we take to be the mass of the Earth and m 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
Comparing the two formulas it is seen that:
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 metres), r, to obtain the value of g:[20]
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 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 "Variation in magnitude":
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.
