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The [gravitational force|gravitation (universal)] exerted by the Earth on an object near the Earth's surface is called gravity. 

h3. The Force of Gravity

h4. Defining "Near"

Suppose an object of mass _m_ is at a height _h_ above the surface of the earth.  Assume that the earth is spherical with radius _R{_}{~}E~.  Working in spherical coordinates with the origin at the center of the earth, the gravitational force on the object from the earth will be:
{latex}\begin{large}\[ \vec{F} = - G \frac{M_{E}m}{(R_{E}+h)^{2}} \hat{r} \]\end{large}{latex}
A Taylor expansion gives:
{latex}\begin{large}\[ \vec{F} \approx - G \frac{M_{E}m}{R_{E}^{2}}\left(1 - 2\frac{h}{R_{E}} + ...\right)\hat{r} \]\end{large}{latex}
Thus, for _h_/_R{_}{~}E~ << 1, the gravitational force from the earth on the object will be essentially independent on altitude above the earth's surface and will have a magnitude equal to:
{latex}\begin{large}\[ F_{g} = mG\frac{M_{E}}{R_{E}^{2}} \]\end{large}{latex}

The gravitational force exerted by the Earth on an object near the Earth's surface is known as the force of [gravity|gravity].

h4. Defining _g_

The above expression is of the form:
{latex}\begin{large}\[ F_{g} = mg \]\end{large}{latex}
if we take:
{latex}\begin{large}\[ g = G\frac{M_{E}}{R_{E}^{2}} = \left(6.67\times 10^{-11}\mbox{ N}\frac{\mbox{m}^{2}}{\mbox{kg}^{2}}\right)\left(\frac{5.98\times 10^{24}\mbox{ kg}}{(6.37\times 10^{6}\mbox{ m})^{2}}\right) = \mbox{9.8 m/s}^{2}\]\end{large}{latex}


h3. Gravitational Potential Energy Near Earth

Near the earth's surface, if we assume coordinates with the +{_}y_ direction pointing upward, the force of gravity can be written:

{latex}\begin{large}\[ \vec{F} = -mg \hat{y}\]\end{large}{latex}

Since the "natural" ground level varies depending upon the specific situation, it is customary to specify the coordinate system such that:

{latex}\begin{large}\[ U(0) \equiv 0\]\end{large}{latex}

The gravitational potential energy at any other height _y_ can then be found by choosing a path for the work integral that is perfectly vertical, such that:

{latex}\begin{large}\[ U(y) = U(0) - \int_{0}^{y} (-mg)\;dy = mgy\]\end{large}{latex}

For an object in vertical freefall (no horizontal motion) the associated [potential energy curve]would then be:

!nearearth.gif!

For movement under pure near-earth gravity, then, there is no equilibrium point.  At least one other force, such as a normal force, tension, etc., must be present to produce equilibrium.