A rocket starting from rest ejects gas at exhaust speed

A rocket starting from rest ejects gas at exhaust speed $v_{1} = 300\ \text{m/s}$, at a rate of $m_{1} = 90\ \text{g/s}$, and its initial mass is $m_{2} = 990\ \text{g}$. After how many seconds does its acceleration reach $w = 30\ \text{m/s}^{2}$?

Step 1: Mass and thrust

In SI units $m_{1} = 0.090\ \text{kg/s}$ and $m_{2} = 0.990\ \text{kg}$. The mass remaining at time $t$ is

\[ m(t) = m_{2}-m_{1}t = 0.990-0.090\,t, \]

while the thrust, the reaction of the ejected gas, is constant: \[F = m_{1}v_{1} = 0.090\times300 = 27\ \text{N}.\]

Step 2: Acceleration at time $t$

The acceleration grows as the rocket lightens: \[w(t) = \frac{F}{m(t)} = \frac{27}{0.990-0.090\,t}. \tag{1}\]

Step 3: Solve for the required moment

Setting $w = 30$ and inverting: \[m(t^{*}) = \frac{F}{w} = \frac{27}{30} = 0.900\ \text{kg},\]

\[ 0.990-0.090\,t^{*} = 0.900 \;\Longrightarrow\; 0.090\,t^{*} = 0.090 \;\Longrightarrow\; t^{*} = 1\ \text{s}. \]

Step 4: The same answer through Tsiolkovsky's formula

Tsiolkovsky's velocity law gives the speed acquired while the mass falls from $m_{2}$ to $m$: \[v = v_{1}\ln\frac{m_{2}}{m} = 300\ln\frac{0.990}{0.900} = 300\ln1.1 \approx 28.593\ \frac{\text{m}}{\text{s}}.\]

Inverting it, $m = m_{2}e^{-v/v_{1}}$, and since $m = m_{2}-m_{1}t$, \[t = \frac{m_{2}}{m_{1}}\left(1-e^{-v/v_{1}}\right) = 11\left(1-e^{-0.09531}\right) = 11\left(1-\frac{10}{11}\right) = 1\ \text{s} \;\;\checkmark\]

the exponential reducing exactly to $\tfrac{10}{11}$ because $e^{0.09531} = 1.1$.

Step 5: Verification

\[m(1) = 0.990-0.090 = 0.900\ \text{kg}, \qquad w = \frac{27}{0.900} = 30\ \frac{\text{m}}{\text{s}^{2}} \;\;\checkmark\]

\[w(0) = \frac{27}{0.990} \approx 27.27\ \frac{\text{m}}{\text{s}^{2}}, \text{ so the acceleration rises by about } 10\% \text{ in the first second} \;\;\checkmark\]

\[\text{The fuel runs out at } t = \frac{m_{2}}{m_{1}} = 11\ \text{s}, \text{ where (1) would diverge} \;\;\checkmark\]