Irodov Problem 1.226

May 19, 2026

—

dexter

in Part 1: Mechanics

Problem Statement

Find the escape velocity from the surface of the Earth.

Given

M_E, R_E, G.

Concepts & Formulas

Escape velocity: set KE = gravitational PE at surface. ½mv_e² = GMm/R → v_e = √(2GM/R) = √(2gR).

Step-by-Step Solution

Step 1: Energy conservation: ½mv_e² − GMm/R = 0 (at infinity, KE=0, PE=0).
Step 2: v_e = √(2GM/R_E).
Step 3: Using g = GM/R_E²: v_e = √(2gR_E).
Step 4: Numerically: v_e = √(2×9.8×6.37×10⁶) ≈ 11.2 km/s.

Worked Calculation

v_e = √(2GM_E/R_E) = √(2gR_E) ≈ 11.2 km/s.

Boxed Answer

v_e = √(2gR_E) ≈ 11.2 km/s

Physical Interpretation

Escape velocity is √2 times orbital velocity at the surface — launching at 45° still doesn’t help; it’s the total energy that matters, not direction.