g (acceleration due to gravity at sea level)

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g_{earth} = 9.80665 m/ s^{2}

$g_"earth" = 9.80665" m/"s^2$

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The effect of location (Latitude)

The earth is not a perfect sphere, because of the effect of the Earth's rotation and the 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, $g (ϕ) = 9.7803267714 \cdot ⎛ ⎜ ⎜ ⎝ \frac{1 + 0.00193185138639 \cdot {sin}^{2} (ϕ)}{\sqrt{1 - 0.00669437999013 \cdot {sin}^{2} (ϕ)}} ⎞ ⎟ ⎟ ⎠$ $g(phi) = 9.7803267714*( (1+ 0.00193185138639*sin^2(phi))/sqrt(1- 0.00669437999013* sin^2(phi)))$

The effect of location (Altitude)

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: $g (h) = g (\frac{r_{e}}{r_{e} + h})$ $g(h) = g(r_e/(r_e + h))$

Approximation of g

Most students studying physics are allowed to use an approximation of g of value 9.8 m/s².

g (acceleration due to gravity at sea level)

The effect of location (Latitude)

The effect of location (Altitude)

Approximation of g

Gravity Calculators: