A rocket with initial mass m_0 = 600 t burns fuel at a rate…

A rocket with initial mass $m_0 = 600\ \mathrm{t}$ burns fuel at a rate $k = 3\ \mathrm{t/s}$; the exhaust speed relative to the rocket is $u = 2000\ \mathrm{m/s}$. (a) What is its initial acceleration if it is launched vertically? (b) Can it reach the first cosmic velocity if the mass of the shell and equipment is $m_0' = 60\ \mathrm{t}$?

Explanation of Method

Can the rocket reach the first cosmic velocity?

Step 1: (a) Initial acceleration

Thrust $uk$ upward, weight $m_0 g$ downward: $uk = 2000 \cdot 3000 = 6.0 \cdot 10^6\ \mathrm{N}$.

\[ m_0 g = 6 \cdot 10^5 \cdot 9.81 = 5.886 \cdot 10^6\ \mathrm{N}. \]

\[ a_0 = \frac{uk - m_0 g}{m_0} = \frac{1.14 \cdot 10^5}{6 \cdot 10^5}. \]

\[a_0 \approx 0.19\ \mathrm{m/s^2}\]

The thrust only just exceeds the weight, so the rocket lifts off very slowly. As fuel burns, the acceleration grows; just before burnout it is $uk/m_0' - g \approx 90\ \mathrm{m/s^2}$.

Step 2: (b) Final speed, ideal case

The Tsiolkovsky equation, which neglects gravity and air resistance, gives an upper limit: $v_{\text{ideal}} = u\ln\frac{m_0}{m_0'} = 2000\ln 10 \approx 4605\ \mathrm{m/s}$.

Step 3: Including gravity

The 540 t of fuel burn in $t_b = 540/3 = 180\ \mathrm{s}$. Launched vertically, gravity takes away $g t_b$ during that time: $v = u\ln\frac{m_0}{m_0'} - g t_b = 4605 - 1766 \approx 2840\ \mathrm{m/s}$.

\[\begin{gathered} v \le 4.6\ \mathrm{km/s} \ (\approx 2.8 \text{ with gravity}) \\ \lt 7.9\ \mathrm{km/s} \end{gathered}\]

So the rocket cannot reach the first cosmic velocity.

$0$ $2$ $4$ $6$ $8$ $v\ (\mathrm{km/s})$ $\text{needed (1st cosmic)}$ $7.91$ $\text{ideal, no gravity}$ $4.61$ $\text{with gravity loss}$ $2.84$

Speeds the rocket could reach after burning its 540 t of fuel (180 s), compared with the 7.91 km/s needed. Even the ideal value falls well short.

Result

\[\boxed{\,a_0 \approx 0.19\ \mathrm{m/s^2}\,}\]

\[\boxed{\,\begin{gathered} v \le 4.6\ \mathrm{km/s} \ (\approx 2.8 \text{ with gravity}) \\ < 7.9\ \mathrm{km/s} \end{gathered}\,}\]

Remark

Check. The original solution is correct: $a_0 \approx 0.19\ \mathrm{m/s^2}$, and $4.61\ \mathrm{km/s} \lt 7.91\ \mathrm{km/s}$, so the answer to (b) is no. Its value 4.61 km/s is the ideal (gravity-free) speed; taking gravity into account during the 180 s burn makes it even lower (about 2.84 km/s), so the conclusion only becomes stronger. To reach 7.9 km/s even ideally, the rocket would need $m_0/m_0' = e^{7.91/2} \approx 52$, i.e. a shell of only about 11.5 t, which is why real rockets use several stages.