Page History

\begin{large}\[ \vec{F}_{12} = - G \frac{m_{1}m_{2}}{r_{12}^{2}} \hat{r}_{12} \]\end{large}

where, G is the gravitational constant equal to:

\begin{large}\[ G = \mbox{6.67}

\times\mbox{10}^{-11}\mbox{ N}\frac{\mbox{m}^{2}}{\mbox{kg}^{2}} \]\end{large}

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{large}\[\vec{r}_{12} = \vec{r}_{1}-\vec{r}_{2}\]\end{large}

Compatibility with Newton's Laws of Motion

Note that the Universal Law of Gravition is consistent with Newton's Third Law of Motion:

\begin{large}\[ \vec{F}_{21} = -G\frac{m_{2}m_{1}}{r_{21}^{2}} \hat{r}_{21}\]\end{large}

Noting

that

the differences of the

position

vectors

r

₁₂ and

r

₂₁ will

certainly

satisfy:

which implies:

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{large}\[ \vec{F} = - G\frac{m_{1}m_{2}}{r^{2}} \hat{r}\]\end{large}

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{large}\[ \lim_{r \rightarrow \infty} U(r) = 0 \]\end{large}

Thus, we can define the potential for any separation r as:

\begin{large}\[ U(r) = U(\infty) - \lim_{r_{0}\rightarrow \infty}\int_{r_{0}}^{r} \left(-G\frac{m_{1}m_{2}}{r^{2}}\right) \;dr 
= - Gm_{1}m_{2} \left(\frac{1}{r}-\lim_{r_{0}\rightarrow \infty}\frac{1}{r_{0}}\right)\]
\[U(r) = -G\frac{m_{1}m_{2}}{r} \]\end{large}

Potential Energy Curve

If the two objects are isolated from other influences, their potential energy curve is then:

Image Added

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 = r cosθ) we would generate a curve:

Image Added

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{large}\[ 1 \leftrightarrow 2 \mbox{ gives rise to }U_{12}\]
\[ 1 \leftrightarrow 3 \mbox{ gives rise to }U_{13}\]
\[ 1 \leftrightarrow 4 \mbox{ gives rise to }U_{14}\]
\[ 2 \leftrightarrow 3 \mbox{ gives rise to }U_{23}\]
\[ 2 \leftrightarrow 4 \mbox{ gives rise to }U_{24}\]
\[ 3 \leftrightarrow 4 \mbox{ gives rise to }U_{34}\]\end{large}

The total potential energy would then be given by:

\begin{large}\[ U_{\rm sys} = U_{12}+U_{13}+U_{14}+U_{23}+U_{24}+U_{34}\]\end{large}