Gravitation
Class 11 · Gravitation

Escape Velocity

v_esc = √(2GM/R) — launch a rocket with various v₀ and see if it escapes.

Key Notes

01

Escape velocity v_esc is the MINIMUM speed needed to escape a gravitational field, going to infinity with zero residual KE.

02

Derivation: ½mv_esc² = GMm/R ⇒ v_esc = √(2GM/R) = √(2gR).

03

Earth: v_esc ≈ 11.2 km/s. Moon: ~2.4 km/s. Sun (at its surface): ~617 km/s.

04

Relation to orbital velocity: v_esc = √2 · v_orb. So escape = ~1.41× orbital.

05

Independent of the direction of launch (assumes no atmosphere, no rotation).

06

Escape velocity does NOT depend on the escaping body's mass. Same v for an atom or a rocket.

07

If launch speed > v_esc: body follows a hyperbolic trajectory. v = v_esc: parabola. v < v_esc: ellipse (bound).

08

Above v_esc, body has POSITIVE total energy ⇒ unbound, leaves to infinity.

Formulas

Escape velocity (surface)

Standard formula. R = radius of central body.

From height h

Less than surface escape velocity; ratio √(R/(R+h)).

Relation to orbital velocity

Same r; orbiting ↔ escaping differ by factor √2.

Energy condition

E ≥ 0 ⇒ unbound (escape). E < 0 ⇒ bound.

Important Points

v_esc = √2 × v_orb at the SAME radius. Always factor of √2 (≈ 1.41) between them.

Earth's escape velocity (11.2 km/s) is achievable — that's why we have rockets to Mars.

If a planet has v_esc > c (speed of light), it's a BLACK HOLE — light can't escape (Schwarzschild radius).

Moon's low v_esc is why it lost its atmosphere — gas molecules with v > 2.4 km/s eventually leak away.

v_esc independent of direction (in ideal case). In practice, launching east takes advantage of Earth's rotation.

Common mistake: forgetting the factor of 2 vs orbital velocity formula.

Escape Velocity notes from sciphylab (also known as SciPhy, SciPhy Lab, SciPhy Labs, Physics Lab). Class 11 physics revision for JEE Mains, JEE Advanced, NEET UG, AP Physics 1/2/C, SAT, and CUET-UG.