SpaceX launches a rocket m = 1 X 10^5 kg. b) Ignoring gravity, estimate the time

Question

SpaceX launches a rocket m = 1 X 10^5 kg.

b) Ignoring gravity, estimate the time it will it take to travel 50 miles (vertically up) to reach the upper atmosphere? Tip: after solving the equation of motion to get position as a function of time, you might want to plot the mass as a function of time and the position as a function of time. How much fuel is burnt to reach the upper atmosphere?

c) Now including the effect of gravity, how long will it take to travel 50 miles vertically up to reach the upper atmosphere? How much fuel is burnt to reach this height?

Explanation / Answer

spaceX launches rocekt

mo = 10^5 kg

a. Vex = 2000 m/s

m' = 12,000 kg/s

h = 300 mil = 482.803 km

delV = 8000 m/s

fuel mass = m

now from rocket equation

dV = Vex*ln(1 + m/mo)

m = 53.59815*10^5 kg = mass of fuel

b. ignoring gravity

dv/dt = Vex*m'/(mo + m) = 24000000/(10^5 + m(t))

also

m(t) = m - m'(t) = 53.59815*10^5 - 12000*t

dv = -2000*dt/(t - 454.98458)

integrating

8000 = -2000*ln(t/454.98 - 1)

t = 463.317916 s

c. if we include effect of gravity

dh = [2000*ln(1 + 53.5981 - 0.12t) - gt]dt

integrating

80467.2 = 2000*(( t - 454.984166)*ln(54.5981 - 0.12t) + 1819.936249 + 0.12t)

t = 462.855 s

fuel burnt = m'*t = 5554261.2 kg = 55.542612 kg ( more than fuel on board)

d. to use air drag

the equation becomes

dv/dt = 2000*(-0.12)/(1 + 53.5981 - 0.12t) - g - c(y)v^2/m(t)

c(y) = 0.25*030^2*exp(-y/10,000) = 225*exp(-y/10,000)

dv/dt = 2000*(-0.12)/(1 + 53.5981 - 0.12t) - g - 225*exp(-y/10,000)v^2/m(t)

the above equation can be used to find y as a function of t

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