A rocket mass decreases as it burns fuel. The equation of motion for a rocket in

Question

A rocket mass decreases as it burns fuel. The equation of motion for a rocket in vertical flight can be obtained from Newton's law, and it is m(t) dv/dt = T - m(t) g where T is the rocket's thrust and its mass as a function of time is given by m(t) = m_0(1 - rt/b). The rocket's initial mass is m_0, the burn time is b and r is the fraction of the total mass accounted for by the fuel. Use the values T = 48,000 N, m_0 = 2200 kg, r=0.8, g = 9.81 m/s^2, and b = 40 s. Determine the rocket's velocity at burnout.

Explanation / Answer

solution:

1) here mass of fuel at any time t is given as

m(t)=m0(1-rt/b)

where at t=b

m=m0(1-.8)=.2m0=440 kg

2) motion eqution as

mdv/dt=T-mg

on integrating we get

v=5*(T/m0)*t-g*t

at t=b=40 s

v=5*(48000/440)*40-9.81*40=21425.78 m/s=21.425 km/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.