= Escape speed, circular speed, and propulsion
For a <circular orbit>, <gravity> supplies <centripetal acceleration>, giving $v_c^2/r=\mu/r^2$. Thus
$$
v_c=\sqrt{\frac{\mu}{r}},\qquad v_{\mathrm{esc}}=\sqrt2\,v_c,\qquad \varepsilon_c=-\frac{\mu}{2r}.
$$
A body in a <circular orbit> is bound despite its continuous <free fall>. At the same $r$, escape needs twice its <kinetic energy> per unit <mass>. An instantaneous forward tangential <velocity> increment from the <circular orbit> requires $\Delta v=(\sqrt2-1)v_c$, rather than the full <escape speed> from rest. In the synthetic example this is exactly $2(\sqrt2-1)\,\mathrm{m/s}\approx0.828427\,\mathrm{m/s}$.<University Physics Volume 1 by William Moebs and others/13.4 Satellite Orbits and Energy>[§13.4, “Circular Orbits” and “Energy in Circular Orbits”: centripetal balance and negative total energy]{ref}
<Escape speed> is a coasting threshold, not a required continuously maintained <speed> or a measure of <rocket propellant>. A powered vehicle can start at rest and acquire <energy> gradually. The <rocket equation> relates an ideal <velocity> increment to <rocket exhaust speed> and <rocket mass ratio>; <rocket thrust>, <drag>, and the timing and direction of propulsion affect a real ascent. Rotation also changes the relation between surface-relative <velocity> and the <inertial frame> used in the calculation.
Reaching another body is a finite-destination <spacecraft trajectory> problem. In the synthetic radial example, a launch at $2\,\mathrm{m/s}$ reaches $r=4\,\mathrm m$ while remaining bound. A spacecraft can similarly encounter the <Moon> on a bound geocentric transfer; actual planning must include the <Moon>'s <gravity>, target motion, arrival <velocity>, and any propulsion needed for capture. Escaping <Earth> alone does not imply escaping the <Sun>, because the relevant <reference frame> and attracting <mass> have changed. The isolated formula describes one gravitational well; a full mission requires more than its scalar threshold.
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