Physics, asked by sharmasuruchi160, 3 months ago

Define acceleration due to gravity​

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Answered by kimrosey1512
10

Answer:

The Acceleration of Gravity

It was learned in the previous part of this lesson that a free-falling object is an object that is falling under the sole influence of gravity. A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). This numerical value for the acceleration of a free-falling object is such an important value that it is given a special name. It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. A matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it - the symbol g. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s/s. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude. We will occasionally use the approximated value of 10 m/s/s in The Physics Classroom Tutorial in order to reduce the complexity of the many mathematical tasks that we will perform with this number. By so doing, we will be able to better focus on the conceptual nature of physics without too much of a sacrifice in numerical accuracy.

g = 9.8 m/s/s, downward

( ~ 10 m/s/s, downward)

Look It Up!

The value of the acceleration of gravity (g) is different in different gravitational environments. Use the Value of g widget below to look up the acceleration of gravity on other planets. Select a location from the pull-down menu; then click the Submit button.

Explanation:

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Answered by kimrose1512
5

Answer:

The gravity of Earth, denoted by g, is the netacceleration 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]

Earth's gravity measured by NASA GRACE mission, showing deviations from the theoretical gravity of an idealized, smooth Earth, the so-called Earth ellipsoid. Red shows the areas where gravity is stronger than the smooth, standard value, and blue reveals areas where gravity is weaker. (Animated version.)[1]

In SI units this acceleration is measured 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, gravitational acceleration is approximately 9.81 m/s2, which means that, ignoring the effects of air resistance, the speed of an object falling freely will increase by about 9.81 metres 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).

The precise strength of Earth's gravity varies depending on location. The nominal "average" value at Earth's surface, known as standard gravity is, by definition, 9.80665 m/s2.[4] 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 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 = ma (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. It is a vector (physics) quantity, and its direction coincides with a plumb bob.

Variation in magnitude

A non-rotating perfect sphere of uniform mass density, or whose density varies solely with distance from the centre (spherical symmetry), would produce a gravitational field of uniform magnitude at all points on its surface. The Earth is rotating and is also not spherically symmetric; rather, it is slightly flatter at the poles while bulging at the Equator: an oblate spheroid. There are consequently slight deviations in the magnitude of gravity across its surface.

Gravity on the Earth's surface varies by around 0.7%, from 9.7639 m/s2 on the Nevado Huascarán mountain in Peru to 9.8337 m/s2 at the surface of the Arctic Ocean.[5] In large cities, it ranges from 9.7806[6] in Kuala Lumpur, Mexico City, and Singapore to 9.825 in Oslo and Helsinki.

Conventional value

In 1901 the third General Conference on Weights and Measures defined a standard gravitational acceleration for the surface of the Earth: gn = 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.

Latitude

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]

Explanation:

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