A rocket is fired in deep space, where gravity is negligible. In the first secon

Question

A rocket is fired in deep space, where gravity is negligible. In the first second it ejects 1/160 of its mass as exhaust gas and has an acceleration of 15.3 m/s^2 . What is the speed (v_gas) of the exhaust gas relative to the rocket? Express your answer numerically in kilometers per second.

Explanation / Answer

The rocket moves by spewing of exhaust gas (ejects weight of by burnt gases). This is an “open mass system” unlike closed mass systems wherein direct momentum conservation can be applied (as done by other forum friend – see the magnitude of that v2 compared to speed of gas?) The REACTION to “momentum carried off” (dp/dt) by gas causes the rocket to accelerate. Let rocket with mass M be moving up with speed (V) [relative to earth] and gas is moving at a speed of (U) [relative to rocket]. Also, let rocket be ejecting mass at ( - dM/dt). Then, the actual speed of gasses after ejection is (V-U) [relative to earth]. It may be noted that there is absolutely no force on the rocket, and even gasses are not exerting any force on it. All that is resulting is change in the momentum of SYSTEM (variable mass) due to opposite momentum of gasses. (Gases) dp/dt = - (V - U) dM/dt and ……………… (1) (Rocket) dp/dt = d/dt (MV) = M dV/dt + V dM/dt ……..(2) (1) = (2) >>> M dV/dt = dM/dt [V - (V - U)] = U dM/dt Acceleration of rock (dV/dt) = (U/M) dM/dt Now given acceleration of rock = 14 m/s^2 , ejects (1/160) of mass in 1 second. So dM = (1/160)M OR (1/M) dM/dt=1/160 U = speed of gases (relative to rocket-as assumed) = 15.3*1 / (1/160) = 15,3* 160 (U)gas^rocket = 2448 meter/sec

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