Question

If a space vehicle consumes fuel at the rate of 1.2 Kg/s and if the velocity is maintained constant at 11.2 Km/s, the upward thrust on the vehicle is,
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Answer 1

Answer:

Details of the calculation:

(a) The rate at which the momentum of the expelled gas changes is

Δp/Δt = Δ(mv)/Δt = (1.5*104 kg/s)(2.6*103 m/s) = 3.9*107 N in the backward direction. The rate at which the momentum of the rest of the rocket changes therefore is 3.9*107 N in the forward direction, since the total momentum of the system is constant. The thrust produced is 3.9*107 N.

(b) Ftot = ma. Ftot = 3.9*107 N - mg = 3.9*107 N - 2.94*107 N = 9.6*106 N.

a = 9.6*106 N/(3*106 kg) = 3.2 m/s2 upward.