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Mass ratio

Rockets and landing · level 6, lesson 4 of 6

A question to start

A drinks can is about 95% drink and 5% can by mass. A good rocket stage is about the same: 95% propellant, 5% tanks and engines. Why do rockets need to be built that thin?

The idea

The mass ratio of a rocket is

\[R = \frac{m_0}{m_1} = \frac{\text{full mass}}{\text{empty mass}},\]

where \(m_0\) is the mass with all its propellant and \(m_1\) the mass after burning it. The propellant fraction is

\[f = \frac{m_0 - m_1}{m_0} = 1 - \frac{1}{R}.\]
Propellant fraction \(f\) Mass ratio \(R\)
50% 2
67% 3
80% 5
90% 10
95% 20

The empty mass \(m_1\) includes everything that is not propellant: tanks, engines, pipes, computers, and the payload. The mass ratio says how much of the rocket can be "thrown away" as exhaust.

What limits it. Tanks must hold the propellant (often under pressure and at very low temperature), and the engines must push the whole stack. Even the best stages manage about 5% of their full mass in structure. Add a payload, and \(R\) falls further. As lesson 5 shows, the speed a rocket can reach grows only as the logarithm of \(R\), so each extra km/s needs a much bigger \(R\).

Worked examples

1. A stage. Full: 400 t. Empty (with payload): 40 t. \(R = 10\), \(f = 90\%\).

2. Adding payload. The same stage carries an extra 10 t of payload, so full is 410 t and empty is 50 t: \(R = 8.2\). The payload is only 2.4% of the full mass, but it cuts \(R\) by almost a fifth.

3. A Mars ascent rocket. It needs \(R = 3\) (lesson 5). If its tanks and engines are 10% of its full mass and the crew capsule is \(m_p\): \(m_1 = 0.1\,m_0 + m_p\) and \(m_0/m_1 = 3\), so \(m_1 = m_0/3\), \(m_p = m_0/3 - 0.1\,m_0 = 0.233\,m_0\). Nearly a quarter of the rocket can be the crew capsule.

Try it

  1. A rocket is 85% propellant. Find its mass ratio.
  2. A stage has \(R = 12\). What is its propellant fraction?
  3. A rocket of 100 t full has 70 t of propellant. Find \(R\) and \(f\).
  4. Why can't a rocket be 99% propellant?
  5. In example 3, what if the tanks and engines were 15% of the full mass?
Answers
  1. \(R = 1/(1 - 0.85) = 6.7\).
  2. \(f = 1 - 1/12 = 91.7\%\).
  3. \(R = 100/30 = 3.33\); \(f = 70\%\).
  4. The tanks, engines and structure needed to hold and burn the propellant weigh more than 1% of it.
  5. \(m_p = m_0/3 - 0.15\,m_0 = 0.183\,m_0\).

Common mistake

Leaving the payload out of \(m_1\). The empty mass is everything still there after the burn, payload included.

Where it leads