Physics I: Mechanics › Oscillations & gravitation › Escape speed

vesc = √(2GM/R)Say it: “the escape speed equals the square root of two G M over R”

Escape speed

How fast to throw something so it never comes back — and why a pebble and a boulder need the same speed.

Notation on this page: vesc is the escape speed from the surface (m/s), R the body's radius. The shortcut form is vesc = √(2gR).

Before this lesson: Gravitational potential energy, g at a planet's surface

Where it comes from

“Escape” means reaching infinity — and at infinity, gravitational potential energy is zero. The cheapest possible escape arrives there with nothing left: K = 0 too. So the escape condition is simply total energy zero:

E = ½mv² − GMm/R = 0launch kinetic exactly pays off the negative potential — E = 0 is “just barely free”
Before reading on: a 1 kg rock and a 1000 kg boulder are thrown upward. Which needs the higher launch speed to escape — or is it a tie? Think about which energies scale with mass before reading the verdict.

A tie. Both ½mv² and GMm/R scale with m, so m cancels: vesc = √(2GM/R) has no m in it. The boulder needs 1000× the energy but exactly the same speed.

Derivation

Energy conservation does everything: write E at launch, write E at infinity for the cheapest escape, equate. The mass cancels in the third step.

E
=
K + U = ½mv² − GMm/r
Step 1 — total energy. Conserved along the flight (gravity is conservative). Write it at launch: r = R.
E∞
=
0 + 0 = 0
Step 2 — the cheapest escape. “Just barely” means arriving at infinity with zero speed: K = 0 and U = 0 there.
½mvesc²
=
GMm/R
Step 3 — equate. Elaunch = E∞: ½mvesc² − GMm/R = 0. The m cancels — both sides scale with it.
vesc
=
√(2GM/R) = √(2gR)
Step 4 — solve. vesc = √(2GM/R). Since g = GM/R², 2GM/R = 2gR — the shortcut. ∎

Why √2 × orbital speed? Circular orbit at the surface needs vorb = √(GM/R) = 7.9 km/s. Escape needs √2 × that ≈ 11.2 km/s — doubling the kinetic energy (from GMm/2R to GMm/R) is what lifts E from −GMm/2R to 0.

How to use it

The procedure, every time:

  1. Use the surface R. Escape from the surface means r = R. From altitude h, use r = R + h (smaller vesc).
  2. Prefer √(2gR). If you know g and R, skip G and M entirely.
  3. Ignore the projectile's mass. It's not in the formula — any m given for the projectile is decoration.
  4. Compare with √2 × orbital speed. vesc = √2 · vorb — a built-in sanity check.
  5. Remember it's energy, not a speed limit. Rockets escape without ever hitting vesc at once — they accumulate the energy gradually.

The two faces

vesc = √(2GM/R)from mass and radius — the fundamental form
vesc = √(2gR)from surface gravity — the shortcut; needs no GSay it: “escape speed equals root two g R”
Common mistake: “rockets must reach 11.2 km/s or they fall back.” No — 11.2 km/s is for an unpowered projectile. A rocket climbing at 1 km/s with engines burning is gaining the same energy ledger gradually; it escapes once K + U ≥ 0.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — Earth: the famous 11.2 km/s

  1. Shortcut form. vesc = √(2gR) = √(2 × 9.82 × 6.371×106).
  2. Evaluate. 2 × 9.82 × 6.371×106 = 1.251×108; √(1.251×108) ≈ 11,186 m/s ≈ 11.2 km/s.
  3. Cross-check. √(2GM/R) = √(2 × 3.986×1014/6.371×106) = √(1.251×108) ✓ — both faces agree.
Common mistake: v = √(gR) ≈ 7.9 km/s — that's the orbital speed, not escape. Forgetting the 2 underestimates by √2 ≈ 1.41×. Say “two g R” every time.
Your turn — Mars: g = 3.73 m/s², R = 3.390×106 m. vesc?

Answer: ≈ 5.03 km/s. v = √(2 × 3.73 × 3.390×106) = √(2.529×107) ≈ 5029 m/s. Cross-check via 2GM/R: √(2 × 4.283×1013/3.390×106) ≈ 5027 m/s ✓.

Example 2 — the Moon: M = 7.348×1022 kg, R = 1.737×106 m

  1. Fundamental form. vesc = √(2GM/R) = √(2 × 6.674×10−11 × 7.348×1022/1.737×106).
  2. Evaluate. = √(5.647×106) ≈ 2376 m/s ≈ 2.38 km/s.
  3. Sanity check. The Moon's gravity is ~1/6 of Earth's, so escape should be much cheaper — 2.38 vs 11.2 km/s ✓. (This is why the Apollo ascent stage could be tiny.)
Common mistake: scaling Earth's 11.2 km/s by 1/6 (the g ratio) → 1.87 km/s. Escape goes as √(gR), not g — the Moon's small R partly offsets its small g. Always compute, never scale by g alone.
Your turn — ratio check: vesc,Earth/vesc,Moon?

Answer: ≈ 4.7. 11186/2376 ≈ 4.71. Equivalently √[(ME/MM)(RM/RE)] = √[81.3 × 0.2726] = √22.2 ≈ 4.71 ✓.

Example 3 — from altitude: escape from the ISS orbit (r = 6.771×106 m)

  1. Use r, not R. vesc = √(2GM/r) = √(2 × 3.986×1014/6.771×106).
  2. Evaluate. = √(1.177×108) ≈ 10,850 m/s ≈ 10.9 km/s.
  3. Compare. Only 3% less than the surface 11.2 km/s — altitude barely helps escape, because gravity falls slowly. (What altitude buys you is orbital energy, not escape.)
Common mistake: “from orbit you're halfway to escape.” Energy-wise you're not even close: circular orbit has E = −GMm/2r, and escape needs E = 0 — you must double the kinetic energy (hence the √2).
Your turn — escape speed from r = 2REarth?

Answer: ≈ 7.91 km/s. v = 11.2/√2 ≈ 7.91 km/s (since v ∝ 1/√r, doubling r divides by √2). Amusingly, that equals Earth's surface orbital speed — no coincidence: √(2GM/2R) = √(GM/R).

Before reading on: a 1 kg rock and a 1000 kg boulder are each thrown upward at exactly 11.2 km/s from Earth's surface (no air). Which escapes — and how do their required launch energies compare?

Example 4 — judgment call: mass independence

  1. Both escape. vesc = 11.2 km/s has no m — at exactly escape speed, E = 0 for any mass.
  2. Energies differ hugely. ½mv²: rock → 0.5 × 1 × (11186)² ≈ 6.26×107 J; boulder → 6.26×1010 J — 1000× more.
  3. The lesson. Speed is democratic; energy is not. The formula promises equal speeds, and the energy bill scales with who's paying.
Common mistake: “the boulder needs a higher speed because it's heavier.” Heavier needs more force and more energy — but the speed that zeroes the energy ledger is mass-free.
Your turn — launch energy for 1 kg at the Moon's 2.38 km/s escape speed?

Answer: ≈ 2.82×106 J. ½ × 1 × 2376² ≈ 2.82×106 J — about 22× less than Earth's 6.26×107 J per kilogram. Escaping the Moon is cheap.

Memorization tips

  • Chant it: “escape speed equals root two G M over R.” The 2 is the whole difference from orbital speed.
  • Earth anchor: 11.2 km/s. Orbital 7.9 km/s × √2 = 11.2 km/s. If your answer isn't near these, recheck.
  • The shortcut: v = √(2gR) — no G, no M. Given g and R, it's one line.
  • E = 0 is the definition. “Just barely escapes” = arrives at infinity at rest. Every escape problem starts here.
  • Mass-free speed, mass-ful energy. The projectile's m cancels for v but not for the energy bill — say which one the question asks for.
  • Derive in 20 seconds: ½mv² = GMm/R → v = √(2GM/R). Energy in, energy out.

Final challenge

Five mixed questions — computations, the mass trap, and the rocket subtlety, all in one. Score 5/5 and escape speed is yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

Why doesn't the projectile's mass matter for escape speed?

Both the kinetic energy ½mv² and the potential energy GMm/r scale with m, so m cancels in ½mv² = GMm/R. A heavier projectile needs more energy but the same speed — energy scales, speed doesn't.

Do rockets actually reach 11.2 km/s?

Not all at once. Escape speed is about total energy, not instantaneous speed: rockets climb slowly, trading thrust for altitude, and escape once their total energy K + U reaches zero. The 11.2 km/s applies to an unpowered projectile fired from the surface.

What's the difference between escape speed and orbital speed?

Circular orbital speed at the surface is √(GM/R) = 7.9 km/s; escape speed is √(2GM/R) = 11.2 km/s — exactly √2 times larger. Orbiting means staying bound; escaping means reaching infinity.

Does launch direction matter?

For the speed value, no (ignoring air and rotation): energy is a scalar. In practice launching eastward from the equator gets a free ~0.46 km/s boost from Earth's rotation.

Can anything escape a black hole?

Setting vesc = c gives the Schwarzschild radius Rs = 2GM/c² — the Newtonian hint of the event horizon. Inside it, the Newtonian formula breaks down, but the energy intuition (E ≥ 0 to escape) survives in relativity.

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