Rocket performance: The propulsive (thrust) power of a rocket equals thrust multiplied by flight speed. With mass flow rate m_dot, jet speed v1, and aircraft speed v2, the useful propulsive power is P = m_dot * (v1 - v2) * v2.

Introduction / Context:
Propulsive (useful) power in flight is the rate at which the propulsion system does work on the vehicle to increase or maintain its kinetic energy against drag. For rockets or pure jets, this equals thrust times the flight speed.

Given Data / Assumptions:

• m_dot is the exhaust mass flow rate (propellant for rockets).
• v1 is the jet exhaust speed relative to the vehicle.
• v2 is the aircraft (vehicle) speed relative to the air/space reference.
• Ideal momentum theory; pressure thrust term neglected for simplicity.

Concept / Approach:
Thrust T from momentum balance in 1-D is T = m_dot * (v1 - v2) for a generic momentum jet (with v2 representing incoming stream speed). The useful propulsive power is the power delivered to the vehicle, P = T * v2. Combining gives P = m_dot * (v1 - v2) * v2.

Step-by-Step Solution:

Verification / Alternative check:
Propulsive efficiency η_p = useful power / jet power = [m_dot * (v1 - v2) * v2] / [0.5 * m_dot * (v1 - v2)^2] = 2 v2 / (v1 - v2). This standard relationship is consistent with the derived power expression (for idealized cases).

Why Other Options Are Wrong:

• P = m_dot * v1^2 / 2: This is the exhaust kinetic power, not the useful propulsive power to the vehicle.
• P = m_dot * v2^2 / 2: Relates to vehicle kinetic energy rate only in a special acceleration case, not general propulsion power expression.
• P = (v1 + v2) / 2: Dimensionally incorrect and not a power.

Common Pitfalls:
Confusing jet power with propulsive power; only the portion of jet energy that goes into pushing the vehicle forward at speed v2 counts as useful propulsive power.

Final Answer:
P = m_dot * (v1 - v2) * v2