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 point masses (masses m1 and m2, respectively) there will exist an attractive force along the line joining the masses. 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, G is the gravitational constant equal to:
\begin
[ G = \mbox
\times\mbox
^{-11}\mbox
\frac{\mbox
^{2}}{\mbox
^{2}} ]\end
r12 is the distance between the two objects and is obtained as the magnitude of the vector difference between the position vector of object 1 and the position vector of object 2. The vector difference is expressed as:
\begin
[\vec
_
= \vec
_
-\vec
_
]\end
Compatibility with Newton's Laws of Motion
Note that the Universal Law of Gravition 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.
Gravitational Potential Energy
Form of the Potential Energy
For two spherically symmetric objects (objects 1 and 2), it is customary to analyze the energy of the gravitational interaction by constructing spherical coordinates with one of the objects at the origin (if one of the objects dominates the mass of the system, its position is typically used as the origin). Newton's Law of Universal Gravitation then takes the form:
\begin
[ \vec
= - G\frac{m_
m_{2}}{r^{2}} \hat
]\end
where r is the position of the object that is not placed at the origin.
It is also customary to make the assignment that the potential energy of the system goes to zero as the separation goes to infinty:
\begin
[ \lim_
U(r) = 0 ]\end
Thus, we can define the potential for any separation r as:
\begin
[ U(r) = U(\infty) - \lim_{r_
\rightarrow \infty}\int_{r_{0}}^
\left(-G\frac{m_
m_{2}}{r^{2}}\right) \;dr
= - Gm_
m_
\left(\frac
-\lim_{r_
\rightarrow \infty}\frac
{r_{0}}\right)]
[U(r) = -G\frac{m_
m_{2}}
]\end
Potential Energy Curve
If the two objects are isolated from other influences, their [potential energy curve] is then:
This potential energy curve is somewhat misleading, since the potential is spherically symmetric. Thus, although in spherical coordinates, r cannot go negative, if we define a one-dimensional coordinate system by following a radial line through the origin (suppose, for instance, we chose to follow the z axis where z = rcosθ) we would generate a curve:
which indicates the possibility of stable equilibrium when the objects' separation goes to zero. Of course, this is technically impossible for objects of finite size.
Gravitational Potential Energy of a System
In a system composed of many spherically symmetric objects, the total gravitational potential energy can be found by adding up the contribution from each distinct interaction.
It is very important to note that any pair of the bodies experiences only one interaction between them. Take, for example, a system composed of four objects labeled 1, 2, 3 and 4. There are six distinct interactions among these bodies, each of which has an associated potential energy:
\begin
[ 1 \leftrightarrow 2 \mbox
U_
]
[ 1 \leftrightarrow 3 \mbox
U_
]
[ 1 \leftrightarrow 4 \mbox
U_
]
[ 2 \leftrightarrow 3 \mbox
U_
]
[ 2 \leftrightarrow 4 \mbox
U_
]
[ 3 \leftrightarrow 4 \mbox
U_
]\end
The total potential energy would then be given by:
\begin
[ U_
= U_
+U_
+U_
+U_
+U_
+U_
]\end
It is important to beware of the temptation to double-count. The potential energy U12 is associated with the interaction between objects 1 and 2, it is not associated with both object 1 and object 2.
Gravitation 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 on altitude above the earth's surface and will have a magnitude equal to:
\begin
[ F_
= mG\frac{M_{E}}{R_
^{2}} ]\end
The gravitational force exerted by the Earth on an object near the Earth's surface is known as the force of [gravity].
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
Gravitational Potential Energy Near Earth
Near the earth's surface, if we assume coordinates with the +y direction pointing upward, the force of gravity can be written:
\begin
[ \vec
= -mg \hat
]\end
Since the "natural" ground level varies depending upon the specific situation, it is customary to specify the coordinate system such that:
\begin
[ U(0) \equiv 0]\end
The gravitational potential energy at any other height y can then be found by choosing a path for the work integral that is perfectly vertical, such that:
\begin
[ U
= U(0) - \int_
^
(-mg)\;dy = mgy]\end
For an object in vertical freefall (no horizontal motion) the associated [potential energy curve]would then be:
Unable to render embedded object: File (nearearth.gif) not found.
For movement under pure near-earth gravity, then, there is no equilibrium point. At least one other force, such as a normal force, tension, etc., must be present to produce equilibrium.