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<h1>Geopotential and Geometric Altitude</h1> 
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<p>
If the mathematical expressions do not display properly on your browser,
you may <a href="geopot.pdf">view the page in PDF</a>. 
I have a page to help you
<a href="mathml.htm">get started with MathML</a>.</p>


<p>
The standard atmosphere is defined in terms of geopotential altitude.
The idea behind this concept
is that a small change in geometric altitude will make a change in gravitational
potential energy per unit mass.
To make this same change in potential energy at sea level requires a
equivalent change in geopotential altitude.
Mathematically, this is expressed as g dZ = G dH 
where H stands for geopotential altitude and Z stands for
geometric altitude, g is the acceleration
of gravity and G is the value of g at sea level.
The value of g varies with altitude and is shown in
elementary physics texts to vary as
</p>


<table class="eq"><tr><td>
<math mode="display" xmlns="&mathml;">
<mrow>
<mfrac><mrow><mi>g</mi></mrow>
<mrow><mi>G</mi></mrow>
</mfrac>
<mo>=</mo>
<msup><mrow><mo>(</mo>
<mfrac><mrow><mi>E</mi></mrow>
<mrow><mi>Z</mi><mo>+</mo><mi>E</mi></mrow>
</mfrac>
<mo>)</mo></mrow><mrow><mn>2</mn></mrow>
</msup>
</mrow>
</math>
</td><td class="eqno">(1)</td></tr></table>

<p>where E is the radius of the earth. So,</p>

<table class="eq"><tr><td>
<math mode="display" xmlns="&mathml;">
<mrow><mi>dH</mi><mo>=</mo>
<mfrac><mrow><mi>g</mi></mrow>
<mrow><mi>G</mi></mrow>
</mfrac>
<mi>dZ</mi><mo>=</mo>
<msup><mrow><mo>(</mo>
<mfrac><mrow><mi>E</mi></mrow>
<mrow><mi>Z</mi><mo>+</mo><mi>E</mi></mrow>
</mfrac>
<mo>)</mo></mrow><mrow><mn>2</mn></mrow>
</msup>
<mi>dZ</mi></mrow>
</math></td>
<td class="eqno">(2)</td></tr></table>

 
<p>and integrating yields</p>


<table class="eq"><tr><td>
<math mode="display" xmlns="&mathml;">
<mrow>
<msubsup><mo>&int;</mo><mrow><mn>0</mn> </mrow>
<mrow><mi>H</mi></mrow></msubsup>
<mi>dH</mi><mo>=</mo>
<msubsup><mo>&int;</mo><mrow><mn>0</mn></mrow>
<mrow><mi>Z</mi></mrow></msubsup>
<msup><mrow><mo>(</mo>
<mfrac><mrow><mi>E</mi></mrow>
<mrow><mi>Z</mi><mo>+</mo><mi>E</mi></mrow>
</mfrac>
<mo>)</mo></mrow><mrow><mn>2</mn></mrow>
</msup>
<mi>dZ</mi></mrow>
</math></td>
<td class="eqno">(3)</td></tr></table>


<table class="eq"><tr><td>
<math mode="display" xmlns="&mathml;">
<mrow><mi>H</mi><mo>=</mo>
<mfrac><mrow><mi>E</mi><mi>Z</mi></mrow>
<mrow><mi>E</mi><mo>+</mo><mi>Z</mi></mrow>
</mfrac>
</mrow>
</math></td>
<td class="eqno">(4)</td>
</tr></table>


<table class="eq"><tr><td>
<math mode="display" xmlns="&mathml;">
<mrow><mi>Z</mi><mo>=</mo>
<mfrac><mrow><mi>E</mi><mi>H</mi></mrow>
<mrow><mi>E</mi><mo>-</mo><mi>H</mi></mrow>
</mfrac>
</mrow>
</math></td>
<td class="eqno">(5)</td>
</tr></table>

<p>
While Z and H are virtually identical at low altitudes, you can calculate
that Z = 86 km corresponds to H=84.852 km. 
(Use 6356 km for the radius of the earth).
At this altitude, g is 0.9735 times the value at sea level.
If you don't like differentials,
you can regard H=EZ / (E+Z) as the definition
of H and then derive dH/dZ=g/G.
</p>


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