= Solution
For a light <test particle> of <mass> $m$ outside a spherical source of <mass> $M$, <Newton's law of universal gravitation> gives $\mathbf F=-GMm\widehat{\mathbf r}/r^2$. It is a <central force>, so its <torque> about the source vanishes and <angular momentum> $\mathbf L=m\mathbf r\times\mathbf v$ is conserved. The motion therefore lies in a fixed plane when $L\ne0$. Its <gravitational potential energy> is $-GMm/r$, and <conservation of mechanical energy> gives
$$
\epsilon=\frac{v^2}{2}-\frac{GM}{r}=\text{constant},\qquad
h=\frac Lm=|\mathbf r\times\mathbf v|=\text{constant}.
$$
Here $\epsilon$ is <specific orbital energy> and $h$ is <specific angular momentum>. A <Kepler orbit> has polar equation
$$
\boxed{r(\theta)=\frac{p}{1+e\cos\theta},\qquad
p=\frac{h^2}{GM},\qquad
e^2=1+\frac{2\epsilon h^2}{(GM)^2}.}
$$
The angle is measured from periapsis. The <conic section> is an <ellipse> for $\epsilon<0$ and $0\le e<1$, a <parabola> for $\epsilon=0$ and $e=1$, and a <hyperbola> for $\epsilon>0$ and $e>1$. The circle is the $e=0$ elliptical case. The special case $h=0$ is radial motion and must be handled as a degenerate limit. Bound ellipses have the source at one focus, which is <Kepler's first law>.
For a bound orbit of <semi-major axis> $a$, $\epsilon=-GM/(2a)$. Combining this with the energy equation gives the <vis-viva equation>:
$$
\boxed{v^2=GM\left(\frac2r-\frac1a\right).}
$$
The particle therefore moves faster near periapsis and slower near apoapsis. Conservation of <angular momentum> also gives constant <areal velocity>, $dA/dt=h/2$, expressing <Kepler's second law>. Using the ellipse's area and $h^2=GMa(1-e^2)$ gives the <orbital period>
$$
\boxed{T^2=\frac{4\pi^2a^3}{GM},}
$$
which is <Kepler's third law>.
A <circular orbit> follows from <centripetal acceleration> $v_c^2/r=GM/r^2$, giving $v_c=\sqrt{GM/r}$ and $T=2\pi\sqrt{r^3/(GM)}$. The minimum <escape velocity> from radius $r$ follows from the marginally unbound condition $\epsilon=0$:
$$
\boxed{v_c=\sqrt{\frac{GM}{r}},\qquad v_{\rm esc}=\sqrt{\frac{2GM}{r}}=\sqrt2\,v_c.}
$$
These ideal orbits neglect <drag force>, other gravitating bodies and changes in the central source. For two comparable <masses>, the relative <two-body problem> uses $G(M+m)$ instead of $GM$. An orbiting object is continuously in <free fall>; the inward gravitational <force> changes its direction even when its <speed> is constant. These inverse-square results assume the physical variation of gravity with radius, rather than extending a near-surface constant-$g$ approximation to an entire orbit.
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