A rocket accelerates by burning its onboard fuel, so its mass decreases with tim

Question

A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at liftoff (including its fuel) is m, the fuel is consumed at rate r, and the exhaust gases are ejected with constant velocity ve (relative to the rocket). A model for the velocity of the rocket at time t is given by the equation v(t) = gt ve ln m rt m where g is the acceleration due to gravity and t is not too large. If g = 9.8 m/s2, m = 31,000 kg, r = 170 kg/s, and ve = 2,800 m/s, find the height of the rocket one minute after liftoff. (Round your answer to the nearest whole meter.)

Explanation / Answer

v(t) =gt ve ln((m rt)/m)

g = 9.8 m/s2, m = 31,000 kg, r = 170 kg/s, and ve = 2,800 m/s

m-rt=0, m=rt

t (final) = As you after all the fuel consumption rocket can't move further, and there is no information regarding mass of fuel so final time can't be decided but to have the vague idea about the final lets calculate fuel mass at t=60s. which will be l = rt, l is the mass of fuel.

l= 170*60 =10200 kg , as you know weight of rocket is 31000 should be technically heavier than fuel. It shows that rocket probably can't go any further after 1 minute. So height of the rocket after 1 minute probably be ZERO.

x(t) = v(t) dt

= [-gt - ve(ln((m - rt)/m))] dt (Note lnu du = ulnu - u + k)

Let u = (m - rt)/m Thus du = -r/m dt ie dt = -m/r du

So x(t) = -½gt² - ve*-m/r (u(lnu - 1) + c
= mve/r [ (m - rt)/m * (ln((m - rt)/m) - 1] - ½gt²
= ve(m - rt)/r * (ln((m - rt)/m) - 1) - ½gt²

If g = 9.8 m/s^2, m = 30000 kg, r = 155 kg/s, and ve = 3000 m/s and t = 1 minute = 60 s

x(60) = 3000 * (30000 - 155 * 60)/155 * (ln (30000 - 155 * 60)/155 - 1) - ½ * 9.8 * 60² = 138.39km

( above gives the height from t=0 to t=60)

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