How does the lunar module of Apollo 11 reached an orbit speed of 1600 meter/second in order to rejoin the CSM (Command/Service Module)?

Additional question, Where are the Oxygen & Fuel tanks of the module and what kind of such small engine was used to reach that speed? is there a schema / specification of the lunar module we can look?

Lunar Module

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    $\begingroup$ The speed required to be in a low circular orbit around the Moon isn't 1600 miles per second. It's about 1600 meters per second. 1600 miles per second is more than four times escape velocity from the surface of the Sun. $\endgroup$
    – notovny
    Sep 7, 2022 at 19:23
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    $\begingroup$ @notovny thanks, fixed $\endgroup$ Sep 7, 2022 at 19:30
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    $\begingroup$ Apollo Lunar Module didn't use oxygen for propulsion. It used hydrazine mix and dinitrogen tetroxide. They have significant advantages if you need 1) to store fuel in tanks for many days (or even years!) 2) to start engines many times. Both the substances are liquid at ambient temperature (liquid oxygen requires very low temperature). Also they ignite just in contact - so the engine construction is simple. Downsides - they are toxic, and have some lower exaust velocity. $\endgroup$
    – Heopps
    Sep 8, 2022 at 9:41

2 Answers 2



How does the lunar module of Apollo 11 reached an orbit speed of 1600 meter/second in order to rejoin the CSM (Command/Service Module)?

This table from APOLLO EXPERIENCE REPORT - MISSION PLANNING FOR LUNAR MODULE DESCENT AND ASCENT by my former colleague Floyd Bennett shows the delta-v requirements for the Apollo 11 ascent to be 6055.7 feet per second and that the Lunar Module Ascent Engine had plenty of performance to make it with the propellant load on board.

enter image description here

For detailed information on the ascent propulsion system, please consult the Experience Report here.


Where are the Oxygen & Fuel tanks of the module? is there a schema / specification of the lunar module we can look?

enter image description here

From the Air and Space Museum website.

  • $\begingroup$ Thanks! looks super small though, still wonder how it got into orbit speed with that amount of fuel. what type of engine was it so it can reach that speed? (even with no air fraction). $\endgroup$ Sep 7, 2022 at 19:28
  • $\begingroup$ @RoeeYossef see this answer space.stackexchange.com/a/29862/6944 $\endgroup$ Sep 7, 2022 at 19:32
  • $\begingroup$ In addition to the excellent link in the comment above, weight was ruthlessly shaved from the ascent module. The thinnest possible wires were used; so thin that breaking wires were a concern during development. Paneling is so thin that it appears flimsy in photos... because it is just enough to get the job done. And so on. $\endgroup$ Sep 7, 2022 at 21:22
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    $\begingroup$ @RoeeYossef The ascent engine was liquid fueled, bipropellant, hypergolic, non-cryogenic, fixed thrust, non-gimballed, and pressure fed. Other types of engine perform better, but it was optimized for reliability first and performance second. $\endgroup$ Sep 7, 2022 at 22:38
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    $\begingroup$ Beyond just that the lunar module was very lightly built; a pair of big tanks full of fuel are pretty heavy compared to that fairly small, entirely hollow lunar module with no heavy cargo other than the astronauts themselves. The engine is an ordinary hypergolic powered engine, it is nothing special. $\endgroup$
    – ikrase
    Sep 10, 2022 at 8:58

How did they reach 1600 m/s:

  1. the dry weight of the LM ascent module was only 2150 kg. The propellant mass was 2350 kg, so slightly over 50% of the ascent module weight was propellant.
  2. this weight was launched from the moon, in 1/6 g gravity.

The Tsiolkovsly rocket equation allows us to calculate which speed we can achieve with these numbers.

$$\Delta v = v_\text{e} \ln \frac{m_0}{m_f} = I_\text{sp} g_0 \ln \frac{m_0}{m_f}$$

  • $v_\text{e}$ is the effective exhaust velocity;
  • $I_\text{sp}$ is the specific impulse in dimension of time: 311 s
  • $g_0$ is the gravitational constant: 9.8 m/s2
  • $\ln$ is the natural logarithm
  • $m_0$ is the initial total mass, including propellant and payload, a.k.a. wet mass; 4700 kg
  • $m_f$ is the final total mass without propellant,a.k.a. dry mass; 2150 kg

So 311 * 9.8 * ln(4500/2150) is 2383 m/s of delta-V is available, far more than the 1600 m/s we need.

This ignores gravity losses, which will reduce the available delta-v.

Approximation of gravity losses: burn time (465 s) * gravity (1.6 m/s2) is 744 m/s, 2383-744 is 1640 m/s.

  • $\begingroup$ not so much an approximation of gravity losses, as a worst case $\endgroup$
    – JCRM
    Feb 15 at 13:02

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