= Circular orbit as continuous free fall
{title2=$\mathbf r(t)=a(\cos\omega t,\sin\omega t),\quad\omega^2=\mu/a^3$}
For a fixed spherical gravitating source, place a <test particle> at $(a,0)$ with initial <velocity> $(0,\sqrt{\mu/a})$, where $a$ is outside the source. Its exact <circular orbit> is
$$
\mathbf r(t)=a(\cos\omega t,\sin\omega t),\qquad \omega=\sqrt{\frac{\mu}{a^3}}.
$$
Differentiating twice gives $\ddot{\mathbf r}=-\omega^2\mathbf r=-\mu\mathbf r/a^3$, precisely the <Newtonian gravitational field> at each point of that circle. There is no outward physical <force> to cancel gravity in this <inertial frame>. Gravity supplies the inward <centripetal acceleration> that continually changes the <velocity>'s direction.
\Image[/circular-orbit-continuous-free-fall.png]
{title=A circular free-fall orbit, inward acceleration, and the straight path the same initial velocity would produce without gravity}
Over a short interval, the displacement from the straight tangent path begins inward with magnitude $\tfrac12(\mu/a^2)t^2$. The <curvature> of the orbit matters at the same order: the particle moves sideways while continually falling away from its instantaneous tangent. The full circular solution remains at radius $a$ rather than striking the source.
Its <speed> is $\sqrt{\mu/a}$, its <oscillation period> is $2\pi\sqrt{a^3/\mu}$, and its <specific orbital energy> is $-\mu/(2a)$. Removing gravity while keeping the initial <velocity> would give the straight path $(a,a\omega t)$, rather than a stationary particle. This contrast explains why orbital <weightlessness> is compatible with a substantial gravitational <force>: a freely coasting spacecraft and its contents fall together. https://www.nasa.gov/centers-and-facilities/glenn/what-is-microgravity/[NASA Glenn: orbit as continuous free fall and the origin of weightlessness]{ref}
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