ESCAPE
v_esc = √(2GM/R) — the energy line that just clears the rim. Mass and angle need not apply.
Use the simulation above to change the variables and play through the guided stages. The explanation below describes the default starting values; the simulation updates its explanation as you experiment.
Setup
Point a cannon straight up and fire. Fast enough and the ball climbs, slows, stops, and falls back — the ordinary story. But is there a speed so great that it never stops, never falls, and simply leaves? The answer is one of the cleanest results in physics, and it comes entirely from bookkeeping.
Energy
Gravity is a pit. Potential energy is zero infinitely far away and negative everywhere inside — the deeper you sit, the more negative. The launch hands the ball a fixed budget of total energy, kinetic plus potential, and that sum is frozen for the whole flight. Where that budget sits relative to the rim of the pit decides everything.
Solve
At the top of this bound climb the ball is momentarily still, so setting v = 0 gives the finite apex radius.
The climb
The amber bar is live kinetic energy; the blue curve is the wall of the potential well. As the ball rises it trades kinetic for potential, and the bar shrinks. If the total-energy line sits below the rim, the bar hits zero at the apex and the ball falls back. If the line clears the rim, the bar never empties — and the ball is gone for good.
Audit
The energy sum holds constant to a part in a trillion at sixty points along the flight. The threshold is exact: escape speed squared times the radius equals twice GM, and escape speed is the square root of two times the circular-orbit speed. And the punchline the ledger insists on — escape velocity does not care about the projectile’s mass or its launch angle. Only the energy decides.