{excerpt}The [gravitational force|gravitation (universal)] exerted by the earth on an object near the earth's surface.{excerpt}
h3. The Force of Gravity 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}
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A Taylor expansion gives:
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{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}
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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:
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{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}
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if we take:
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{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's Surface
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.
h3. Example Problems involving Near-earth Gravity
h4. {toggle-cloak:id=force} Example Problems involving Gravitational Force
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h4. {toggle-cloak:id=en} Example Problems involving Gravitational Potential Energy
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{contentbylabel:gravitational_potential_energy,example_problem|maxResults=50|operator=AND|showSpace=false|excerpt=true}
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