◄ WORLD V · SONNY 5DART 550 · a helldive down the well

THE ESCAPE VELOCITY the speed to climb out of the well

How fast must you leave to never fall back? Set kinetic energy equal to the depth of the well: ½v² = GM/r, so v = √(2GM/r) — and note it does not depend on your mass, only the well’s. From Earth’s surface that is 11,186 m/s (11.2 km/s). Beyond it you are unbound; below it the void keeps you.

THE TECHNIQUE v = √(2GM/r)

The demo computes √(2·μ/R) with μ=3.986×10¹⁴ (Earth); i13 has no sqrt, so it runs Newton’s method: live demo







  
HISTORY & CREDIT Newton · escape energy

  
“A heavier rocket needs a higher escape velocity.” — escape velocity is independent of the escaping mass; it depends only on the body you leave. cited

  
the balance · kinetic energy ½v² = potential depth GM/r.
the speed · v = √(2GM/r) — mass-independent.
Newton · the well is the 1/r potential of universal gravitation (Principia, 1687).

  
One speed divides bound from free. physics






  
RECOMMEND FOR I-13 11.2 km/s, on the compiler

  
On i-13 (Newton sqrt, 40 iterations), Earth’s escape velocity is 11,186 m/s:

  
$ i13 run gw_escape.i13   # v = sqrt(2*mu/R), Newton's method
RUN OK · 796 step(s) · call depth 42
v2    = 125129493.0   -- 2*mu/R
v_esc = 11186.13       -- m/s (11.2 km/s)

  
Recommend as a NULL — a theorem (B39) + a resource workaround (B40). The escape speed is a forall-pinned function of the body’s M and r; i13 lacks sqrt so it computes one via Newton iteration — a resource cost, not a keeper. NULL — the speed to leave the well.