Universal Law of Gravitation

Galileo had already shown that all objects, heavy or light, fall towards the Earth with the same acceleration when released from the same height near the surface. Newton went a decisive step further: he proposed that the force of attraction between masses is UNIVERSAL, applying not just to falling objects near the Earth but to any two masses anywhere in the universe -- explaining both terrestrial gravity and, simultaneously, Kepler's empirical laws of planetary motion.

Newton's own reasoning (1665) started with the Moon. The Moon orbits the Earth roughly circularly, completing one revolution in 27.3 days at a mean distance of $3.85\times10^5$ km, with (nearly) constant angular speed $\omega$. Since this is circular motion, the Moon must be under a constant centripetal force directed towards the Earth, $F=mr\omega^2$, giving a centripetal acceleration $a=r\omega^2=r\left(\frac{2\pi}{T}\right)^2$. Substituting the Moon's numbers gives $a\approx2.7\times10^{-3}$ m/s^2 -- a value about 3600 times SMALLER than the $g=9.8$ m/s^2 felt by objects at the Earth's surface. Comparing this ratio to the SQUARE of the ratio of distances (distance of Moon from Earth's centre, about 60 Earth radii, squared is also about 3600) shows the two ratios match almost exactly: $a_{object}/a_{moon}\approx(\text{distance of moon}/\text{distance of object})^2$. Newton concluded that the acceleration due to Earth's gravity falls off as the INVERSE SQUARE of distance from the Earth's centre, $a\propto1/r^2$, and since $F=ma$, the force itself is $F\propto m/r^2$ for a fixed object of mass m at distance r.

By Newton's third law, the object also pulls back on the Earth with an equal and opposite force, so this same force must also be proportional to the Earth's mass M, giving $F\propto Mm/r^2$. Newton then generalised this beyond the Earth-object pair to state the Universal Law of Gravitation: every particle of matter in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them. For two point masses $m_1$ and $m_2$ separated by distance r, $F=\dfrac{Gm_1m_2}{r^2}$, where G is the universal gravitational constant, $G=6.67\times10^{-11}$ N m^2/kg^2, with dimensional formula $[G]=[L^3M^{-1}T^{-2}]$. This force is always attractive, acts along the line joining the two masses (Fig. 5.3), and the two mutual forces form an action-reaction pair of equal magnitude and opposite direction.

In vector form, if $m_1$ and $m_2$ have position vectors $\vec{r}_1$ and $\vec{r}_2$ from some origin, and $\vec{r}_{21}=\vec{r}_2-\vec{r}_1$ is the vector from $m_1$ to $m_2$ (with unit vector $\hat{r}_{21}$), the force on $m_2$ due to $m_1$ is $\vec{F}_{21}=-\dfrac{Gm_1m_2}{r^2}\hat{r}_{21}$ (directed from $m_2$ back towards $m_1$), and by Newton's third law $\vec{F}_{12}=-\vec{F}_{21}$. For a COLLECTION of point masses, the law applies pairwise, and the total force on any one mass is the VECTOR SUM of the individual pairwise forces from every other mass in the system (Fig. 5.4); for n particles, the force on the i-th mass is $\vec{F}_i=\sum_{j\neq i}\vec{F}_{ij}$. …