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`g = 9.80665 m/s^2`

9.80665

**Acceleration due to Gravity (g)** at sea level on Earth is 9.80665 m/s^{2}.

The mass of the Earth causes a force of gravity to be exerted on objects. On the surface of the Earth, the value most often used is 9.80665 m/s^{2 }which can be computed using the force of gravity equation, the mass of the Earth and the mean radius to the Earth's surface from the center:

`g = G*m_e*m/R_E^2`

The earth is not a point mass, nor is it a perfect sphere. Because of the effect of the Earth's rotation, a resulting centrifugal force has caused the Earth to have a bulge around the equator. The Earth's rotation and the resultant centrifugal force (heading outward) counteracts the effect of gravity (downward). This has a measurable effect in the apparent acceleration due to gravity at different latitudes. A good approximation of the total effect is modeled in the International Gravity Formula below. To indicate the ascension or decline from the equator, latitude (φ) can be used.

The International Gravity Formula, `g(phi) = 9.7803267714*( (1+ 0.00193185138639*sin^2(phi))/sqrt(1- 0.00669437999013* sin^2(phi)))`

Altitude also has an effect on the apparent acceleration due to gravity because of the increased distance from the center of mass. The following equation approximates the acceleration due to gravity as affected by altitude (**h**):

Acceleration due to gravity at different altitudes: `g(h) = g(r_e/(r_e + h))`