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 the fuel) is m, the fuel is consumed at rate r, and the exhaust gasses are ejected with constant velocity v, (relative to the rocket) A model for the velocity of the rocket at time t is given by v(t)=-gt-v ln(m-rt/m) where g is the acceleration due to gravity and t is not to large. If g=9.8m/s^2 m=28,000 kg, r=190 kg/s, and v=3,100 m/s. Find the height of the rocket one minute after lift off. (round answer to nearest whole meter)

Answer 1

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 v(t)= -gt - ve(ln((m - rt)/m)) 

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 

≈ 138390 m 

= 138.39km