GATE_AE_2027_MOCK_03_Q25 GATE 2027 Aerospace Engineering All-India Master Mock 03 (Flight Dynamics, Controls & Space Trajectories) NAT · 1 Mark
Flight Mechanics & Space Dynamics • Space Dynamics & Orbital Mechanics • Keplerian Orbits & Escape Velocity

The radius of Earth is R = 6378 km. Taking surface gravity g_0 = 9.81 m/s^2, what is the theoretical escape velocity from Earth's surface in km/s? (Round off to two decimal places).

Answer Key:
11.18

Step-by-Step Verified Solution

### Solution: v_esc = sqrt(2 * g_0 * R) = sqrt(2 * 9.81 * 6378000) = sqrt(125136360) ≈ 11186 m/s ≈ 11.18 km/s. Correct: 11.18 (range 11.15 to 11.25).
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