Question:

A rocket is launched in a constant gravity, 9.80 m/s\(^2\). The initial velocity is 400 m/s, and the burn time is 100 seconds. If the exhaust velocity is 2000 m/s, and the mass decreases by a factor of three; namely, the current mass divided by the original mass is equal to \(\frac{1}{3}\), find the final velocity of the rocket.

Answer:

The equation of motion is given as

\[  m\frac{dv}{dt}=-mg-u \frac{dm}{dt}  \]

where \(u\) is the exhaust velocity. Divide both sides

by the mass, \(m\).

\[  \frac{dv}{dt}=-g-u \frac{dm}{dt}\frac{1}{m}  \]

Multiply \(dt\) by both sides. (This manipulation is not for mathematicians.)

\[  dv=-gdt – u \frac{dm}{m}  \]

Integrate this.

\[  \int^{v}_{v_0} dv=\int^{t}_0 -gdt – u \int^{m}_{m_0}\frac{dm}{m}  \]

Therefore,

\[  v-v_0 = -gt – u \ln \frac{m}{m_0}  \]

Plug in the numbers.

\[  v = 400 – 9.80 \cdot 100 – 2000 \ln \frac{1}{3}=1620 \ \mathrm{m/s}  \]