A 2.46*10^4Kg rocket blasts off vertically from the earth's surface with a c

Question

A 2.46*10^4Kg rocket blasts off vertically from the earth's surface with a constant acceleration. During the motion considered in the problem, assume that remains constant. Inside the rocket, a 15.1-instrument hangs from a wire that can support a maximum tension of 35.1. find the minimum time for this rocket to reach the sound barrier without breaking the inside wire. Find the maximum vertical thrust of the rocket engines under these conditions. How far is the rocket above the earth's surface when it breaks the sound barrier?

Explanation / Answer

A 2.33×10^4 kg rocket blasts off vertically from the earth's surface with a constant acceleration. During the motion considered in the problem, assume that g remains constant. Inside the rocket, a 18.5 N instrument hangs from a wire that can support a maximum tension of 37.8 N The minimum time for this rocket to reach the sound barrier 330 m/s without breaking the inside wire: 32.28sec. How do you find the maximum vertical thrust of the rocket engines under these conditions? This is an impulse problem where dP = d(M + m)V = d(M + m)V + (M + m)dV = dm V + (M + m)dV = F dt, where M = rocket mass, m = fuel mass, dt = 32.28 sec, F = average thrust over dt to yield the change in momentum dP. Thus F = (dm/dt) V + (M + m) dV/dt; where dm/dt is the fuel expenditure rate kg/sec, V is the average velocity over dt (V = a dt/2 = dV/2), dV/dt = acceleration, and m = average fuel mass over dt m = (m0 - (dm/dt)dt)/2 where m0 is blast off fuel mass. This rather complex relationship can be simplified if we assume very little fuel mass compared to the rocket mass is expended over the 32.28 sec. In which case dm/dt = 0; so that F = (M + m) dV/dt; where dV = 330 mps, dt = 32.28, and M + m ~ 2.33E4 kg and then F = 2.33E4*330/32.28 = 238,197 Newtons thrust. ANS.

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