Applying Newton's Second (F_{net} = ma) to the small object we can determine its acceleration if no other force are acting on the small mass m but gravity (F_{net} = F_{G}).
Acceleration of Gravity Near the Surface of a Large Body
Assume that a large body of mass M and radius R has a radically uniform mass distribution. The gravitational force on a small mass m at height h above the surface of the large body can be expressed as,

If h << R, then
If we neglect air resistance, then this is the net force acting on the mass m. Applying Newton's Second Law, F_{net} = ma,
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Because of its importance we usually give the acceleration of gravity due to the Earth a separate name,
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* This equation does not depend upon the mass of the small object  the mass canceled out on both sides of the force equation. This proves that the acceleration due to gravity alone is the same on all objects independent of their mass, their size, their density, or even their speed.
* Local concentrations of mass in the Earth can change the magnitude of the acceleration of gravity in the third significant figure. In fact, it is often true that as you go up a mountain the vale of g gets larger not smaller as you would expect because mountains contain a larger concentration of denser rocks.
* Since the value of g is not a fixed constant, in this course we will normally use 9.80 m/s^{2} when solving problems.
Acceleration of Gravity at the North Pole
The Earth's polar radius is 6356 km and its mass is 5.9736x10^{24} kg.
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Acceleration of Gravity at the Equator
The Earth's equatorial radius is 6378.1 km. Neglecting the Earth's rotation,
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The Earth is also rotating so that the acceleration of gravity at the equator will be reduced by the centripetal acceleration. Since these are both along the same direction (radially) at the equator we don't have to subtract them vectorially  just numerically,
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Since the Earth makes one revolution per day, w = 2p rev/day
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Weight as Function of Latitude:
If you are at some latitude F above the equator then your period of rotation will still be one day, but your distance from the axis of rotation will be reduced to R cos(F). This reduces the effect of the centripetal acceleration. Moreover the Earth is not a perfect sphere, so R will also change.
Assuming a spherical Earth we can determine the weight on a mass m at a latitude F. By applying the law of cosines to the force vectors we get
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Where
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For Bellingham WA, F = 48.78
A value that the student's here a Western Washington University have consistently found when measuring the acceleration of gravity over the pasted 30 years. There are other factors that could effect the local value of g such as elevation above sea level and the local concentrations of denser rock.
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