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

h3. The Force of Gravity Near Earth's Surface Near Earth's Surface

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}

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. 

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