Universal Gravitation
Paul explains about Universal Gravitation.

The field vector g
A field vector is a single vector that describes the strength and direction of a uniform vector field. For a gravitational field the field vector is g, which is defined in this way:
$$ g = \dfrac{F}{m} $$
\( \begin{aligned} \displaystyle \require{color}
\text{where } F &= \text{force exerted on mass } m \\
m &= \text{mass in the field} \\
g &= \text{the field vector} \\
\end{aligned} \)Vector symbols are indicated here in bold italics. The direction of the vector g is the same as the direction of the associated force of Universal Gravitation.
Note that a net force applied to a mass will cause it to accelerate. Newton’s Second Law describes this relationship:
$$ a = \dfrac{F}{m} $$
\( \begin{aligned} \displaystyle \require{color}
\text{where } a &= \text{acceleration} \\
\end{aligned} \)Hence, we can say that the field vector g also represents the acceleration due to gravity and we can calculate its value at the Earth’s surface as described below.
The Law of Universal Gravitation says that the magnitude of the force of attraction between the Earth and an object on the Earth’s surface is given by:
$$ F = G\dfrac{m_{E}m_{O}}{r^{2}_{E}} $$
\( \begin{aligned} \displaystyle \require{color}
\text{where } m_{E} &= \text{the mass of the Earth} \\
m_{O} &= \text{mass of the object in kilograms} \\
r_{E} &= \text{radius of the Earth} \\
G &= \text{the gravitational constant} \\
\end{aligned} \)From the defining equation for g, on the previous page, we can see that the force experienced by the mass can also be described by:
$$ F = m_{O}g $$
Equating these two we get \( m_{O}g = G \dfrac{m_{E}m_{O}}{r^{2}_{E}} \)
This simplifies to give: \( g = G \dfrac{m_{E}}{r^{2}_{E}} = \dfrac{6.672 \times 10^{11} \times 5.974 \times 10^{24}}{(6.378 \times 10^{6})^2} \)
Hence, \( g \approx 9.80 \text{ m s}^{2}\) 
Variations in the value of g
Variation with geographical location
The actual value of the acceleration due to gravity, g, that will apply in a given situation will depend upon geographical location. Minor variations in the value of g around the Earth’s surface occur because:
 the Earth’s crust or lithosphere shows variations in thickness and structure due to factors such as tectonic plate boundaries and dense mineral deposits. These variations can alter local values of g, Universal Gravitation.
 the Earth is not a perfect sphere, but is flattened at the poles. This means that the value of g will be greater at the poles, since they are closer to the centre of the Earth.
 the spin of the Earth creates a centrifuge effect that reduces the effective value of g. The effect is greatest at the Equator and there is no effect at the poles.
As a result of these factors, the rate of acceleration due to gravity at the surface of the Earth varies from a minimum value at the Equator of 9.782 m s^{2} to a maximum value of 9.832 m s^{2} at the poles. The usual value used in equations requiring g is 9.8 m s^{2}.
Variation with altitude
The formula for g shows that the value of g will also vary with altitude above the Earth’s surface. By using a value of r equal to the radius of the Earth plus altitude, the following values can easily be calculated. It is clear from the table below that the effect of the Earth’s gravitational field is felt quite some distance out into space.
$$ g = G \dfrac{m_{E}}{(r_{E}+\text{altitude})^2} $$
NOTE: That as altitude increases the value of g decreases, dropping to zero only when the altitude has an infinite value.
The variation of g with altitude above Earth’s surface Altitude (km) g Comment 0 9.80 Earth’s surface 8.8 9.77 Mt Everest Summit 80 9.54 Arbitrary beginning of space 200 9.21 Mercury capsule orbit altitude 250 9.07 Space shuttle minimum orbit altitude 400 8.68 Space shuttle maximum orbit altitude 1 000 7.32 Upper limit for low Earth orbit 400 000 0.19 Communications satellite orbit altitude 
Variation with planetary body
The formula for g also shows that the value of g on planetary surfaces depends upon the mass and radius of the central body which, in examples so far, has been the Earth. Other planets and natural satellites (moons) have a variety of masses and radii, so that the value of g elsewhere in our solar system can be quite different from that on Earth Universal Gravitation. The following table presents a few examples.
$$ g = G \dfrac{m_{\text{planet}}}{r_{\text{planet}}^2} $$
A comparison of g on the surface of other planetary bodies Body Radius (km) g Moon 1738 1.6 Mars 3397 3.7 Jupiter 71 492 24.8 Pluto 1 151 0.66