User talk:Millsng/Sandbox

Mathematical models


If the terrain is at sea level, we can estimate g:
 * $$g_{\phi}=9.780 327 \left( 1+0.0053024\sin^2 \phi-0.0000058\sin^2 2\phi \right) \ m/s^2  $$

where
 * $$ \ g_{\phi}$$ = acceleration in m·s−2 at latitude :$$ \ \phi$$

This is the International Gravity Formula 1967, the 1967 Geodetic Reference System Formula, Helmert's equation or Clairault's formula.

The first correction to this formula is the free air correction (FAC), which accounts for heights above sea level. Gravity decreases with height, at a rate which near the surface of the Earth is such that linear extrapolation would give zero gravity at a height of one half the radius is 9.8 m·s−2 per 3 200 km.

Using the mass and radius of the Earth:


 * $$ r_\mathrm{Earth}= 6.371 \times 10^{6} \ m $$


 * $$ m_\mathrm{Earth}= 5.9736 \times 10^{24} \ kg $$

The FAC correction factor (&Delta;g) can be derived from the definition of the acceleration due to gravity in terms of G, the Gravitational Constant (see Estimating g from the law of universal gravitation, below):


 * $$ g_0 = G \, m_\mathrm{Earth} / r_\mathrm{Earth}^2 = 9.8331 \ m/s^2$$

where:


 * $$G = 6.67428 \times 10^{-11} \ m^3  \ kg^{-1}  \ s^{-2}.$$

At a height h above the nominal surface of the earth gh is given by:


 * $$ g_h = G \, m_\mathrm{Earth} / \left( r_\mathrm{Earth} + h \right) ^2 $$

So the FAC for a height h above the nominal earth radius can be expressed:


 * $$ \Delta g =   \left  [ G \, m_\mathrm{Earth} / \left( r_\mathrm{Earth} + h \right) ^2 \right ] - \left[G \, m_\mathrm{Earth} / r_\mathrm{Earth}^2 \right] $$

This expression can be readily used for programming or inclusion in a spreadsheet. Collecting terms, simplifying and neglecting small terms (h<<rEarth), however yields the good approximation:


 * $$ \Delta g = - \, \dfrac{ G \, m_\mathrm{Earth}}{ r_\mathrm{Earth} ^2} \times  \dfrac{  2 \,h}{r_\mathrm{Earth}}$$

Using the numerical values above and for a height h in metres:


 * $$ \Delta g = - 3.084 . 10^{-6}\, h $$

Grouping the latitude and FAC altitude factors the expression most commonly found in the literature is:


 * $$g_{\phi}=9.780 327 \left( 1+0.0053024\sin^2 \phi-0.0000058\sin^2 2\phi \right) - 3.086  .  10^{-6}h $$

where $$ \ g_{\phi}$$ = acceleration in m·s−2 at latitude :$$ \ \phi $$ and altitude h in metres. Alternatively (with the same units for h) the expression can be grouped as follows:


 * $$g_{\phi}=9.780 327 \left[ \left( 1+0.0053024\sin^2 \phi-0.0000058\sin^2 2\phi \right) - 3.155 \times 10^{-7}h \right] \ m/s^2 $$

For flat terrain above sea level a second term is added, for the gravity due to the extra mass; for this purpose the extra mass can be approximated by an infinite horizontal slab, and we get 2πG times the mass per unit area, i.e. 4.2 m3·s−2·kg−1 (0.042 μGal·kg−1·m²)) (the Bouguer correction). For a mean rock density of 2.67 g·cm−3 this gives 1.1 s−2 (0.11 mGal·m−1). Combined with the free-air correction this means a reduction of gravity at the surface of ca. 2 µm·s−2 (0.20 mGal) for every metre of elevation of the terrain. (The two effects would cancel at a surface rock density of 4/3 times the average density of the whole Earth.)

For the gravity below the surface we have to apply the free-air correction as well as a double Bouguer correction. With the infinite slab model this is because moving the point of observation below the slab changes the gravity due to it to its opposite. Alternatively, we can consider a spherically symmetrical Earth and subtract from the mass of the Earth that of the shell outside the point of observation, because that does not cause gravity inside. This gives the same result.

Helmert's equation may be written equivalently to the version above as either:
 * $$ \ g_{\phi}= \left(9.8061999 - 0.0259296\cos(2\phi) + 0.0000567\cos^2(2\phi)\right) \ m/s^2

$$ or
 * $$ \ g_{\phi}= \left( 9.780327 + 0.0516323\sin^2(\phi) + 0.0002269\sin^4(\phi) \right) \ m/s^2

$$

An alternate formula for g as a function of latitude is the WGS (World Geodetic System) 84 Ellipsoidal Gravity Formula:


 * $$ \ g_{\phi}= \left(9.7803267714 ~ \frac {1 + 0.00193185138639\sin^2\phi}{\sqrt{1 - 0.00669437999013\sin^2\phi}} \right) \ m/s^2

$$

A spot check comparing results from the WGS-84 formula with those from Helmert's equation (using increments 10 degrees of latitude starting with zero) indicated that they produce values which differ by less than 10−6 m·s−2.

my appology you may have an error, or your chart may be unclear but according to the text, four times your chart height at 8850m there is only a minor 0.28% reduction in weight, Altitude Gravity decreases with altitude, since greater altitude means greater distance from the Earth's centre. All other things being equal, an increase in altitude from sea level to the top of Mount Everest (8,850 metres) causes a weight decrease of about 0.28%. (An additional factor affecting apparent weight is the decrease in air density at altitude, which lessens an object's buoyancy.[5]) It is a common misconception that astronauts in orbit are weightless because they have flown high enough to "escape" the Earth's gravity. In fact, at an altitude of 400 kilometres (250 miles), equivalent to a typical orbit of the Space Shuttle, gravity is still nearly 90% as strong as at the Earth's surface, and weightlessness actually occurs because orbiting objects are in free-fall.[6] http://en.wikipedia.org/wiki/Gravity_of_Earth#Altitude