{excerpt}A force exerted by any massive body on any other massive body, regardless of the distance separating them, that is proportional to the masses of the bodies and inversely proportional to the square of the distance between them.{excerpt}
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h2. Motivation for Concept
Newton's Law of Universal Gravitation provides an effective description of the movement of objects near the earth and also of the orbits of the bodies (planets, moons, comets, asteroids, etc.) in the solar system.
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h2. Newton's Law of Universal Gravitation
h4. Statement of the Law for Point Masses
Between any two massive bodies (masses _m_~1~ and _m_~2~, respectively) there will exist an attractive force. The force on body 1 due to body 2 will have the form:
{latex}\begin{large}\[ \vec{F}_{12} = - G \frac{m_{1}m_{2}}{r_{12}^{2}} \hat{r}_{12} \]\end{large}{latex}
where _r_~12~ is the position vector of object 1 in a coordinate system with object 2 located at the origin and _G_ is a constant of proportionality equal to:
{latex}\begin{large}\[ G = \mbox{6.67}\times\mbox{10}^{-11}\mbox{ N}\frac{\mbox{m}^{2}}{\mbox{kg}^{2}} \]\end{large}{latex}
h4. Compatibility with Newton's Laws of Motion
Note that this Law implies there is also a force on body 2 due to body 1:
{latex}\begin{large}\[ \vec{F}_{21} = -G\frac{m_{2}m_{1}}{r_{21}^{2}} \hat{r}_{21}\]\end{large}{latex}
Noting that the position vectors _r_~12~ and _r_~21~ will certainly satisfy:
{latex}\begin{large}\[ \vec{r}_{12} = - \vec{r}_{21}\]\end{large}{latex}
implies:
{latex}\begin{large}\[ \vec{F}_{12} = - \vec{F}_{21}\]\end{large}{latex}
so that the Law of Universal Gravitation is perfectly compatible with the [Third Law of Motion|Newton's Third Law].
h4. The Case of Spherical Symmetry
Although the form of the Law of Universal Gravitation is strictly valid only for [point particles|point particle], it is possible to show that for extended objects with a spherically symmetric mass distribution, the Law will hold in the form stated above *provided that the positions of the spherical objects are specified by their centers*.
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h2. 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}\begin{large}\[ \vec{F} = - G \frac{M_{E}m}{(R_{E}+h)^{2}} \hat{r} \]\end{large}{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 of 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}\begin{large}\[ F_{g} = mg \]\end{large}{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}
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