# About Earth Gravity and its characteristics in different scenarios

1. Earth's gravitational field gives rise to a typical acceleration of 9.80 m/s². If you say gravity changes with altitude, then what will be the gravity at per say 8900 meters. Is there any specific formula?

2. Does the moon have any noticeable effects on the gravitational acceleration? If so, what will be the change?

3. If gravity changes during the above 2 scenarios, then is it possible for an aircraft to save fuel during take-off at high altitude and when flying relatively close to the moon?

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Related (essentially duplicates?): physics.stackexchange.com/q/6074 and physics.stackexchange.com/q/49870 – dmckee Sep 25 '13 at 14:26
The field we actually measure at the earth's surface includes fictitious forces due to the earth's rotation. For this reason, there is a latitude dependence. Also, the geoid is (approximately) an equipotential, not a surface of constant field strength. – Ben Crowell Sep 25 '13 at 15:52

1. The value you quote, $g\approx 9.80\ m/s^2$, is obtained from simply plugging in numbers into the formula for Newtonian gravitation: $$F_g=\frac{GMm}{r^2}$$ where $G\approx 6.67*10^{-11}$ is the gravitational constant, $M$ and $m$ are the masses of the two gravitationally interacting bodies, and $r$ is the distance between their centers of mass. If you assume that our movements on Earth are confined to a distance from the ground which is negligible in comparison with the radius of Earth itself, $r_E\approx6.37*10^6\ m$, then we may simplify by using $$F=mg$$where $$g=\frac{GM_E}{r_E^2}\approx\frac{6.67*10^{-11}*5.97*10^{24}}{(6.37*10^6)^2}\approx9.81\ m/s^2$$ Now, you can see that this $g$ is really nothing fundamental, but rather something to make it easier for high schoolers to calculate. So if you want to calculate the gravitational force at 8900m above the Earth, simply plug in $r=r_E+9800$ in the formula for $g$. This will give you $g\approx 9.78$, so it does have a slight impact.