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 m3 kg-1 s-2, and the masses of the bodies, and inversely proportional to the square of the distance between them.
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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.
Newton's Law of Universal Gravitation
Statement of the Law for Point Masses
Between any two massive bodies (masses m1 and m2, respectively) there will exist an attractive force. The force on body 1 due to body 2 will have the form:
\begin
[ \vec
_
= - G \frac{m_
m_{2}}{r_
^{2}} \hat
_
]\end
where r12 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:
\begin
[ G = \mbox
\times\mbox
^{-11}\mbox
\frac{\mbox
^{2}}{\mbox
^{2}} ]\end
Compatibility with Newton's Laws of Motion
Note that the Universal Law of Gravity is consistent with Newton's Third Law of Motion:
\begin
[ \vec
_
= -G\frac{m_
m_{1}}{r_
^{2}} \hat
_
]\end
Noting that the differences of the position vectors r12 and r21 will certainly satisfy:
\begin
[ \vec
_
= - \vec
_
]\end
which implies:
\begin
[ \vec
_
= - \vec
_
]\end
The Case of Spherical Symmetry
Although the form of the Law of Universal Gravitation is strictly valid only for point particles, 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.
Gravity Near Earth's Surface
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 RE. Working in spherical coordinates with the origin at the center of the earth, the gravitational force on the object from the earth will be:
\begin
[ \vec
= - G \frac{M_
m}{(R_
+h)^{2}} \hat
]\end
A Taylor expansion gives:
\begin
[ \vec
\approx - G \frac{M_
m}{R_
^{2}}\left(1 - 2\frac
{R_{E}} + ...\right)\hat
]\end
Thus, for h/RE << 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:
\begin
[ F_
= mG\frac{M_{E}}{R_
^{2}} ]\end
Defining g
The above expression is of the form:
\begin
[ F_
= mg ]\end
if we take:
\begin
[ g = G\frac{M_{E}}{R_
{2}} = \left(6.67\times 10{-11}\mbox
\frac{\mbox
^{2}}{\mbox
{2}}\right)\left(\frac{5.98\times 10
\mbox{ kg}}{(6.37\times 10^
\mbox
)^{2}}\right) = \mbox
^
]\end