A spacecraft is in a circular orbit at radius r = 2R_E from Earth's centre. Use GM_E = 3.986 × 10¹⁴ m³ s⁻² and R_E = 6.37 × 10⁶ m. Find its circular speed, orbital period and specific mechanical energy. Determine the local escape speed and the minimum instantaneous tangential speed increase that places it on a zero-energy escape trajectory. Explain why the engine energy per kilogram is not simply ½(Δv)².

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Circular speed = 5.59 km s⁻¹, period = 1.43 × 10⁴ s, specific energy = −1.56 × 10⁷ J kg⁻¹, escape speed = 7.91 km s⁻¹, and minimum Δv = 2.32 km s⁻¹. Circular gravity supplies centripetal acceleration, so v_c = √(GM/r) = √[3.986 × 10¹⁴/(1.274 × 10⁷)] = 5.59 × 10³ m s⁻¹. The period is T = 2πr/v_c = 2π√(r³/GM) = 1.43 × 10⁴ s, about 3.97 h. Specific orbital energy is ε = v_c²/2 − GM/r = −GM/(2r) = −GM/(4R_E) = −1.56 × 10⁷ J kg⁻¹. Zero total energy at the burn point requires v_esc = √(2GM/r) = √2v_c = 7.91 km s⁻¹. A tangential prograde burn therefore needs Δv = 7.91−5.59 = 2.32 km s⁻¹. Kinetic-energy change is ½(v_esc²−v_c²) = v_cΔv + ½(Δv)². The cross term matters because the spacecraft already has substantial orbital velocity before the burn.

Escape from a circular orbit

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