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# Find rocket thrust, inteval between ignition and lift-off

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Please provide a detailed solution to the following problem:

A rocket stands vertically on a launching pad, the mass of the fuel and the rocket is 1.9*10^3 kg. On ignition, gas is ejected from the rocket at a speed of 2.5*10^3 m/s relative to the rocket, and fuel is consumed at a constant rate of 7.4 kg/s. Find the thrust of the rocket and hence explain why there is an interval between ignition and lift-off. (take g=10 m/s^2).

https://brainmass.com/physics/newtons-second-law/rocket-thrust-inteval-between-ignition-lift-off-287687

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A rocket stands vertically on a launching pad, the mass of the fuel and the rocket is 1.9x103 kg. On ignition, gas is ejected from the rocket at a speed of 2.5x103 m/s relative to the rocket and fuel is consumed at a constant rate of 7.4 kg/s. Find the thrust of the rocket and hence explain why there is an interval between ignition and lift-off. (take g=10 m/s^2).

Solution:

M = 1.9x103 kg
Mg=19x103 N

T

v = 2.5x103 m/s

As per Newton's second law of motion, thrust generated = T = d(mv)/dt = mdv/dt +vdm/dt ..........(1) where v = speed with which the gases are ejected, m = mass of the gases that are ejected in time dt, dv/dt = acceleration of the ejected gases = 0, dm/dt = rate ejection of gases (same as the rate at which fuel is consumed)

Substituting dv/dt = 0, we get: T = vdm/dt = 2.5x103x7.4 = 18.5x103 N

Net force on the rocket = T - Mg

For the rocket to lift off, the thrust T must be greater than the weight of the rocket. Initially, the thrust is less than rocket's weight. Hence, the rocket does not lift off immediately. As the fuel burns, rocket's weight decreases and as soon as the weight becomes less than the thrust, lift off takes place.

Mass of the rocket for its weight to become equal to the thrust = 1.85x103 kg

Reduction in mass for the rocket to lift off = 1.9x103 - 1.85x103 = 0.05x103 kg or 50 kg

Rate at which the fuel is consumed = 7.4 kg/s

Time taken to take off = 50/7.4 = 6.76 sec

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