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The Universal Law of Gravitation

By a process of logic that I will not attempt to describe, Newton deduced that the force  F  between two objects with masses  m  and  M  separated by a distance  r  was given by

 \begin{displaymath}F = {GmM \over r^2}
\end{displaymath} (10.3)

where   $G = (6.67259 \pm 0.00085) \times
10^{-11}$ m$^3\cdot$kg $^{-1}\cdot$s-2  is the Universal Gravitational Constant. Actually, Newton didn't know the value of G; he only postulated that it was universal - i.e. that it was the same constant of proportionality for every pair of masses in this universe. The actual determination of the value of G was first done by Cavendish in an experiment to be described below.

We should also express this equation in vector form to emphasize that the force on either mass acts in the direction of the other mass: if   $\vec{\mbox{\boldmath$\space F $\unboldmath }}_{12}$  denotes the force acting on mass  m2  due to its gravitational attraction by mass  m1  then

 \begin{displaymath}\vec{\mbox{\boldmath$ F $\unboldmath }}_{12} =
- \, {G \, m . . . 
 . . . er r_{12}^2} \;
\hat{\mbox{\boldmath$ r $\unboldmath }}_{12}
\end{displaymath} (10.4)

where   $\hat{\mbox{\boldmath$\space r $\unboldmath }}_{12}$  is a unit vector in the direction of   $\vec{\mbox{\boldmath$\space r $\unboldmath }}_{12}$,  the vector distance from m1 to m2, and  r12  is the scalar magnitude of $\vec{\mbox{\boldmath$\space r $\unboldmath }}_{12}$. Note that the reaction force   $\vec{\mbox{\boldmath$\space F $\unboldmath }}_{21}$  on m1 due to m2 is obtained by interchanging the labels "1" and "2" which ensures that it is equal and opposite because   $\vec{\mbox{\boldmath$\space r $\unboldmath }}_{21} \equiv
- \vec{\mbox{\boldmath$\space r $\unboldmath }}_{12}$  by definition.



 
next up previous
Next: Weighing the Earth Up: Celestial Mechanics Previous: Kepler's Laws of Planetary Motion
Jess H. Brewer - Last modified: Sat Nov 14 12:27:25 PST 2015