Deriving v_esc = √(2GM/r) = √(2gr)

Open with the energy statement, not the formula: KE_i + U_i = KE_∞ + U_∞ = 0 + 0. Both terms at infinity are zero, KE because 'minimum' means arriving with nothing left, U because infinity is where we set the zero. That sentence is usually worth a mark on its own.

Then substitute: ½mv² + (−GMm/r) = 0, so ½mv² = GMm/r. All the launch kinetic energy is spent paying off the depth of the potential well. Cancel m, multiply by 2, take the root.

Say out loud that m cancels, and say why it matters: escape speed is a property of the PLANET, not of the thing escaping. A pebble and a rocket need the same 11 km s⁻¹.

The √(2gR) form comes from GM = gR² at the surface. Use it when the question gives you g and R but not M, which it usually does. For Earth, √(2 × 9.81 × 6.4 × 10⁶) = 1.1 × 10⁴ m s⁻¹.

Compare with orbit: v_orbit = √(GM/r), so v_esc = √2 × v_orbit at the same radius. Handy sanity check, and it also answers the 'how much extra KE to escape from orbit' question, since KE scales as v² so you need exactly double the kinetic energy.

Two facts the mark scheme wants stated explicitly: independent of the object's mass, and independent of launch direction (energy is a scalar). Air resistance and the planet's own rotation are ignored in this treatment, which is worth flagging if asked for assumptions.