Acceleration of gravity as a function of the height above the surface of the Earth

Factors:
g(h) is the acceleration of gravity - the dependant variable
G = universal gravitational constant
M = mass of the Earth
R = radius of the Earth
h = height above the Earth - the independant variable

Formula:

\( g(h) = \dfrac{G \cdot M}{(R + h)^2} \)

Values for the Earth (Metric):
\( G = 6.67430 \times 10^{-11} \text { m}^3/\text{kg} \cdot\text {s}^2 \)
\( M = 5.972 \times 10^{24} \text{ kg} \)
\( R = 6,371 \text{ km} \)

Values for the Earth (English):
\( G = 1.068846 \times 10^{-9} \text { ft}^3/\text{lb} \cdot\text {s}^2 \)
\( M = 4.092 \times 10^{23} \text{ slugs} = 1.3 \times 10^{25} \text{ lbs equivalent} \)
\( R = 3,948.8 \text{ mi} \)

Derivation of the above formula
F is the force exerted by the Earth on an object.
g is the acceleration of gravity exerted on the object by the Earth.
G = universal gravitational constant
M = mass of the Earth
m = mass of the Object
R = radius of the Earth

\( F = \dfrac{GMm}{R^2} \)

\( F = ma = mg \)    Force of the Earths gravity on the object

\( mg = \dfrac{GMm}{R^2} \)

\( \color{red}{\cancel{m}}\color{black}g = \dfrac{GM\color{red}{\cancel{m}}}{R^2} \)

g in various units
\( g = 32 \text { feet}/\text {second}^2 = \dfrac {21.9 \text { miles}/\text {hour}} {second} = 9.8 \text { meters}/\text {second}^2 = \dfrac {35.3 \text { km}/\text {hour}} {second} \)

vertical coordinate due to acceleration of gravity with MathJax
\(y = h - {v_yt / 2} \text { = } h - { ( gt)t / 2} = h - {gt^2 / 2} \)

vertical coordinate due to acceleration of gravity without MathJax
y = h - v_yt/2    =    h - (gt)t / 2    =    h - gt²/2

MathJax sample Quadratic Equation
When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are

\( x = {-b \pm \sqrt{b^2-4ac} \over 2a}.\)

Unit Examples
\(feet/sec^2\)