{excerpt}An interaction between two massive particles resulting in an attractive force exerted on each by the other.  The force is proportional to the gravitational constant G={{{*}6.674 28(67) x 10{*}{*}^\-11{^}* *m{*}{*}{^}3{^}{*}* kg{*}{*}^\-1{^}{*}* s{*}{*}^\-2{^}{*}}}, and 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 from submillimeter distances to galactic sizes, and is the dominant force on most macroscopic objects near the earth and 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 the Universal Law of Gravity is consistent with Newton's [Third Law of Motion|Newton's Third Law]:
{latex}\begin{large}\[ \vec{F}_{21} = -G\frac{m_{2}m_{1}}{r_{21}^{2}} \hat{r}_{21}\]\end{large}{latex}
Noting that the differences of the position vectors _r{_}{~}12~ and _r{_}{~}21~ will certainly satisfy:
{latex}\begin{large}\[ \vec{r}_{12} = - \vec{r}_{21}\]\end{large}{latex}
which implies:
{latex}\begin{large}\[ \vec{F}_{12} = - \vec{F}_{21}\]\end{large}{latex}

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}