Newton's Law of Gravity

1. Force of gravity acts on a line joining centers of masses.
2. Force of gravity is attractive.
3. Force of gravity is proportional to each mass.
4. Force of gravity is inversely proportional to the distance between the centers of the masses.

or,

$$\vec{\bf F}_{12} = - \frac{Gm_1m_2}{r^2}\hat{\bf r}$$ (4)

where $G = 6.67x10^{-11}$ N-m$^2$/kg$^2$ is the universal gravitational constant.

Kepler's first law follow from solving Newton's laws and the equations of motion for this particular force law. This is a bit difficult and beyond the scope of this course, although we will show that circular orbits are one special solution that easily satisfy Kepler's Laws.

Kepler's Second Law is proven by observing that this force is radial, and hence exerts no torque. Thus the angular momentum of a planet is constant! That is,

$$dA = \frac{1}{2}\vert\vec{\bf r}\times\vec{\bf v}dt\vert = \frac{1}{2}\vert\vec{\bf r}\vert\vert\vec{\bf v}dt\vert \sin\theta$$ (5)

$$= \frac{1}{2m}\vert\vec{\bf r}\times m\vec{\bf v}dt\vert$$ (6)

or

$$\frac{dA}{dt} = \frac{1}{2m}\vert\vec{L}\vert= {\rm constant}$$ (7)

(and Kepler's second law is proved for this force).

For a circular orbit, we can also prove Kepler's Third Law. The orbit is circular, so we have a relation between $v$ and $F_r$.

$$\frac{GM_sm_p}{r^2} = m_pa_r = m_p\frac{v^2}{r}$$ (8)

so that

$$v^2 = \frac{GM_s}{r}$$ (9)

But, $v$ is related to $r$ and the period $T$ by:

$$v = \frac{2\pi r}{T}$$ (10)

so that

$$v^2 = \frac{4\pi^2r^2}{T^2} = \frac{GM_s}{r}$$ (11)

Finally,

$$r^3 = \frac{GM_s}{4\pi^2} T^2$$ (12)

and Kepler's third law is proved for circular orbits (and the constant $C$ evaluated for the solar system!).