Rocket Fuel Consumption Rate – Rankers Physics

Definition of Force Inertia and First Law: Practice Problem & Solution

For a rocket propulsion velocity of exhaust gases relative to rocket is \( 2 \text{ km/s} \). If mass of rocket system is \( 1000 \text{ kg} \), then the rate of fuel consumption for a rocket to rise up with acceleration \( 4.9 \text{ m/s}^2 \) will be: (1998)
\( 12.25 \text{ kg/s} \)
\( 17.5 \text{ kg/s} \)
\( 7.35 \text{ kg/s} \)
\( 5.2 \text{ kg/s} \)

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Definition of Force Inertia and First Law. Understanding the underlying formula is key to arriving at the correct answer below:

Thrust force \( F_t = v_e \frac{dM}{dt} \). For upward motion with acceleration \( a \): \( F_t - Mg = Ma \). \( v_e \frac{dM}{dt} = M(g + a) \). \( \frac{dM}{dt} = \frac{M(g + a)}{v_e} = \frac{1000 \text{ kg} (9.8 + 4.9) \text{ m/s}^2}{2000 \text{ m/s}} = 7.35 \text{ kg/s} \).

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