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{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}
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 Excerpt
The gravitational force exerted by the earth on an object near the earth's surface.
The Force of Gravity Near Earth's Surface 
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 RE. 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}

 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}
RE << 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}
Defining g
The above expression is of the form: 
 Latex
\begin{large}\[ F_{g} = mg \]\end{large}
{latex}
\\
if we take:
\\
{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 
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 
Image Added
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 
Example Problems involving Near-earth 
 Gravity

h4. {
Gravity
 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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