= Escape-energy balance diagram
Use the synthetic <standard gravitational parameter> $\mu=8\,\mathrm{m^3/s^2}$ and launch distance $r_0=2\,\mathrm m$. Then <escape speed> is $2\sqrt2\,\mathrm{m/s}$. Three outward radial launches give exact comparisons: $v_0=2\,\mathrm{m/s}$ has <specific orbital energy> $-2\,\mathrm{J/kg}$ and stops outward at $r=4\,\mathrm m$; $v_0=2\sqrt2\,\mathrm{m/s}$ has zero <specific orbital energy>; $v_0=2\sqrt3\,\mathrm{m/s}$ has <specific orbital energy> $+2\,\mathrm{J/kg}$ and asymptotic <speed> $2\,\mathrm{m/s}$. The second case has <speed> $2\,\mathrm{m/s}$ at $r=4\,\mathrm m$. Doubling $r_0$ divides <escape speed> by $\sqrt2$, whereas doubling $M$ at fixed $r_0$ multiplies it by $\sqrt2$. Changing the <test particle>'s <mass> changes $E$ but cancels from the <escape speed>.
\Image[/inverse-square-escape-energy-balance.png]
{title=The <inverse-square potential> and <kinetic energy> budget for three radial launch <speeds>}
{description=Computed curves use $x=r/r_0$ and $\mathcal E=\varepsilon/(\mu/r_0)$. The left panel compares $U/m=-\mu/r$ with constant <specific orbital energies> $\mathcal E=-1/2,0,+1/2$. The right panel shows <kinetic energy> per unit <mass>, divided by $\mu/r_0$, equal to $\mathcal E+1/x$. Only the physically allowed outward branch is drawn for negative <energy>, ending at $x=2$. These are synthetic predictions, not measurements.}
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