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Tsiolkovsky's rocket equation

Rockets and landing · level 6, lesson 5 of 6

A question to start

A rocket leaving the surface of Mars must gain about 4.1 km/s to reach orbit. Its methane engine has an exhaust speed of 3.7 km/s. What fraction of the rocket must be propellant? And what about leaving Earth?

The idea

Derivation. In space, far from any planet, a rocket of mass \(m\) moving at \(v\) throws out a small mass \(dm\) of exhaust at speed \(v_e\) backwards, relative to itself. In the frame moving with the rocket at that instant, momentum before is 0; after, the rocket of mass \(m - dm\) moves at \(dv\) and the exhaust at \(-v_e\):

\[0 = (m - dm)\,dv - dm\,v_e \quad\Rightarrow\quad m\,dv = v_e\,dm\]

(dropping the tiny \(dm\,dv\)). Here \(dm\) is the mass thrown out, so the rocket's mass falls by \(dm\). Writing the change in the rocket's mass as \(dM = -dm\): \(dv = -v_e\,dM/M\). Integrate from the full mass \(m_0\) to the empty mass \(m_1\):

\[\Delta v = v_e \ln\frac{m_0}{m_1}.\]

This is Tsiolkovsky's rocket equation, published by Konstantin Tsiolkovsky in 1903. It says:

  • The speed gained depends only on the exhaust speed and the mass ratio, not on how fast the propellant is burned.
  • It grows only as the logarithm of the mass ratio. Doubling \(R\) adds a fixed \(v_e \ln 2 = 0.69\,v_e\).
  • Turned round, \(m_0/m_1 = e^{\Delta v / v_e}\): the mass ratio grows exponentially with the speed needed.
\(\Delta v\) needed (\(v_e\) = 3.7 km/s) Mass ratio \(e^{\Delta v/v_e}\) Propellant fraction
2.0 km/s 1.72 42%
3.7 km/s 2.72 63%
4.1 km/s (Mars surface to orbit) 3.03 67%
7.4 km/s 7.39 86%
9.4 km/s (Earth surface to orbit) 12.7 92%
11.1 km/s 20.1 95%

The answer to the question. From Mars: a mass ratio of about 3, two thirds propellant. From Earth with the same engines: about 12.7, so 92% propellant, which leaves almost nothing for tanks, engines and payload in one stage. That is the "tyranny of the rocket equation", and why Earth launchers are staged (lesson 6) while a single stage can get off Mars.

With gravity. Near a planet, the rocket also fights gravity while it burns. The 9.4 and 4.1 km/s above already include those gravity losses (and drag). Level 8 shows how to estimate them.

Worked examples

1. Mars ascent. \(R = e^{4.1/3.7} = e^{1.108} = 3.03\). A 30 t rocket (fuelled) ends at 9.9 t.

2. A lander's last burn. A lander must remove 600 m/s with engines of \(v_e = 2{,}200\) m/s. \(R = e^{0.273} = 1.31\): 24% of its mass at the start of the burn is propellant.

3. Leaving Earth orbit for Mars. 3.6 km/s with hydrogen engines (\(v_e = 4.4\) km/s): \(R = e^{0.818} = 2.27\). With methane (\(3.7\)): \(R = e^{0.973} = 2.65\).

Try it

  1. A rocket with \(v_e = 3\) km/s has \(R = 5\). Find \(\Delta v\).
  2. What mass ratio gives \(\Delta v = 6\) km/s with \(v_e = 4.5\) km/s?
  3. Use the rocket equation simulator: with \(v_e\) = 3.7 km/s and an empty mass of 100 t, how much propellant reaches 4.1 km/s?
  4. Why does doubling the mass ratio not double \(\Delta v\)?
  5. Show that a rocket can go faster than its own exhaust speed. What mass ratio does it need to reach \(2v_e\)?
Answers
  1. \(3 \ln 5 = 3 \times 1.609 = 4.83\) km/s.
  2. \(e^{6/4.5} = e^{1.333} = 3.79\).
  3. \(R = 3.03\), so \(m_0 = 303\) t: about 203 t of propellant.
  4. \(\Delta v\) grows with \(\ln R\), and \(\ln 2R = \ln R + \ln 2\): it only adds \(0.69\,v_e\).
  5. \(\Delta v > v_e\) when \(R > e = 2.72\). For \(2v_e\), \(R = e^2 = 7.39\).

Common mistake

Using common logarithms. The equation uses the natural logarithm, \(\ln\). Using \(\log_{10}\) gives answers 2.3 times too small.

Where it leads