Acceleration due to gravity

For other uses, see g force.

The acceleration due to gravity denoted g (also gee, g-force or g-load) is a non-SI unit of acceleration defined as exactly 9.80665 m·s⁻², which is approximately equal to the acceleration due to gravity on the Earth's surface at sea level.

The symbol g is properly written in lowercase and italic, to distinguish it from the symbol G, the gravitational constant, which is always written in uppercase; and from g, the abbreviation for gram, which is not italicized.

This conventional value was established by the 3rd CGPM (1901, CR 70). The total acceleration is found by vector addition of the opposite of the actual acceleration (in the sense of rate of change of velocity) and a vector of 1 g downward for the ordinary gravity (or in space, the gravity there). For example, being accelerated upward with an acceleration of 1 g doubles the experienced gravity. Conversely, weightlessness means an acceleration of 1 g downward in an inertial reference frame. Therefore, the term μg-force is a comparative measure of acceleration applied to a body with respect to sea-level gravity on earth. It is a normalized force vector since dividing the resultant force vector applied to a body by the body's weight (magnitude at sea level) cancels the mass, resulting in a "fractional-g" -magnitude vector, e.g. a person sitting on a chair at sea level is experiencing "1g," due to their weight.

The value of g defined above is an arbitrary midrange value on the Earth, approximately equal to the sea level acceleration of free fall at a geodetic latitude of about 45.5°; it is larger in magnitude than the average sea level acceleration on Earth, which is about 9.797 645 m·s⁻². The standard acceleration of free fall is properly written as g_n (sometimes g₀) to distinguish it from the local value of g that varies with position.

The units of acceleration due to gravity, meters per second squared, are interchangeable with newtons per kilogram. The quantity, 9.806 65, stays the same. These alternate units may be more helpful when considering problems involving pressure due to gravity, or weight.

Variations of Earth's gravity

The actual acceleration of a body at the Earth's surface depends on the location at which it is measured, smaller at lower latitudes, for two reasons.

The first is that the rotation of the Earth imposes an additional acceleration on the body that opposes gravitational acceleration. The net downward force on the body is therefore offset by a centrifugal force that acts upwards, reducing its weight. This effect on its own would result in a range of values of g from 9.789 m·s⁻² at the equator to 9.823 m·s⁻² at the poles.

The second reason is the Earth's equatorial bulge (itself also caused by centrifugal force), which causes objects at the equator to be further 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.

The combined result of these two effects is that g is 0.052 m·s⁻² more, hence the force due to gravity of an object is 0.5 % more, at the poles than at the equator.

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 3200 km. Thus:

g_{\phi}=9.780 318 \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. 4.2 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 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.

Local variations in both the terrain and the subsurface cause further variations; the gravitational geophysical methods are based on these: the small variations are measured, the effect of the topography and other known factors is subtracted, and from the resulting variations conclusions are drawn. See also physical geodesy and gravity anomaly.

Calculated value of g

Given the law of universal gravitation, g is merely a collection of factors in that equation:

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

where g is the bracketed factor and thus:

g=G \frac {m_1}{r^2}

We can plug in values of

G

and the mass (in kilograms) and radius (in meters) of the Earth to obtain the calculated 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{ ms}^{-2}

This agrees approximately with the measured value of g. The difference may be attributed to several factors:

The Earth is not homogeneous
The Earth is not a perfect sphere
The choice of a value for the radius of the Earth (an average value is used above)
This calculated value of g does not include the centrifugal force effects that are found in practise due to the rotation of the Earth

There are significant uncertainties in the values of G and of m₁ as used in this calculation. However, the value of g can be measured precisely and in fact, Henry Cavendish performed the reverse calculation to estimate the mass of the Earth.

Usage of the unit

The g is used primarily in aerospace fields, where it is a convenient magnitude when discussing the loads on aircraft and spacecraft (and their pilots or passengers). For instance, most civilian aircraft are capable of being stressed to 4.33 g (42.5 m·s⁻², 139 ft/s²), which is considered a safe value. The g is also used in automotive engineering, mainly in relation to cornering forces and collision analysis.

One often hears the term being applied to the limits that the human body can withstand without losing conciousness, sometimes referred to as "blacking out", or g-loc (loc stands for loss of consciousness). A typical person can handle about 5 g (50 m·s⁻²) before this occurs, but through the combination of special g-suits and efforts to strain muscles —both of which act to force blood back into the brain— modern pilots can typically handle 9 g (90 m·s⁻²) sustained (for a period of time) or more. Resistance to "negative" or upward gees which drive blood to the head, is much less. This limit is typically in the -2 to -3 g (-20 m·s⁻² to -30 m·s⁻²) range. The vision goes red and is also referred to as a "red-out". This is probably due to capillaries in the eyes bursting under the increased blood pressure. Humans can survive about 20 to 40 g instantaneously (for a very short period of time). Any exposure to around 100 g or more, even if momentary, is likely to be lethal.

Human g-force experience

Amusement park rides such as roller coasters typically do not expose the occupants to much more than about 3 g (a notable exception is the Sheikra Rollercoaster at Tampa wich pulls 4 g and the Titan Rollercoaster in Texas with a maximum of 4.5 g)^*

^*^*^*)

A sky-diver in a stable free-fall experiences his full weight of 1 g (see terminal velocity)
A scuba diver or swimmer experiences his full weight of 1 g, but buoyancy largely cancels the weight of their body. However, density differences do create forces, noteably the lungs are significantly buoyant.
Astronauts in earth orbit experience 0 g (Weightlessness.)
Passengers on planes on a parabolic trajectory experience 0 g (as in the Vomit Comet)
Aerobatic and fighter pilots may sometimes experience a greyout between 6 and 8 g. This is not a total loss of consciousness but is characterised by temporary loss of colour vision, tunnel vision, or an inability to interpret verbal commands. They also experience a 'redout' at negative g. These effects are mostly caused by blood pressure differences between the heart and the brain.

Everyday g-forces

10.4 g when plopping down into a chair.
8.1 g when hopping off a step.
3.5 g during a cough.
2.9 g during a sneeze. ^*

Strongest g-forces survived by humans

Voluntarily: Colonel John Stapp in 1954 sustained 46.2 g in a rocket sled, while conducting research on the effects of human deceleration. See Martin Voshell (2004), 'High Acceleration and the Human Body'.

Involuntarily: Formula One race car driver David Purley survived an estimated 179.8 g in 1977 when he decelerated from 172 km·h⁻¹ (107 mph) to 0 in a distance of 26 inches (66 cm) after his throttle got stuck wide open and he hit a wall.

External links

Wired article about enduring a human centrifuge at the NASA Ames Research Center ^*

Units of acceleration | Gravimetry

Gourt Home::Gourt :: Acceleration due to gravity