5
$\begingroup$

The LM ascent engine wasn't used for direct abort. Presuming no one cares about the crew and just wanted to burn the LM to empty (both stages), what would the delta v have been? If it helps, I'm ok with it if you assume the O2 tanks were completely empty or completely full. I'm curious about ballpark ranges. How did that delta v compare to the service engine?

**EDIT: to clarify the last part: how did the delta v of the LM-CSM stack burning only the LM engines to empty compare to the delta-v of the LM-CSM stack presuming that the service engine were burned to empty using the SM fuel tanks but the LM remains unused.

$\endgroup$

1 Answer 1

6
$\begingroup$

Using the two LM stages, the delta-v should be:

$$\Delta v = v_{e_{DPS}} \cdot ln\left(\frac{m_{CSM} + m_{LM}}{m_{CSM} + m_{LM} - m_{DPS propellant}}\right) + v_{e_{APS}} \cdot ln\left(\frac{m_{CSM} + m_{LMAS}}{m_{CSM} + m_{LMAS} - m_{APS propellant}}\right)$$

Where:

  • $m_{CSM}$: the mass of the command and service modules combined. This should be slightly lower than the launch mass due to some propellant being used for a midcourse correction and some volatiles lost in the incident. (28,881 kg launch mass)
  • $m_{LM}$: total mass of the lunar module. (15,188 kg)
  • $m_{LMAS}$: mass of the lunar module ascent stage. (4,700 kg)
  • $m_{DPS propellant}$: mass of propellant in the descent stage. (8,200 kg)
  • $m_{APS propellant}$: mass of propellant in the ascent stage. (2,353 kg)
  • $v_{e_{DPS}}$: exhaust velocity of the descent propulsion system. (3,050 m/s)
  • $v_{e_{APS}}$: exhaust velocity of the ascent propulsion system. (3,050 m/s)

For these initial numbers, before diving into the subtleties of small mass differences, I get:

$$\Delta v = 850 m/s$$

With the DS contributing 628 m/s and the AS contributing 222 m/s.

This is substantially lower than the ~2100m/s the service module could apply to the full stack. (~4000m/s without the LM).

For a ballpark estimate like this, some complicating factors have not been taken into account. Tanks may have unusable slumps of leftover propellants, the RCS system may or may not be able to usefully contribute, and some of the volatiles in the SM may be possible to vent.

$\endgroup$
4
  • $\begingroup$ Most of the precious resources of oxygen and electric battery capacity were located in the LM descent stage. Using the engine of the ascent stage was possible only after separation of the descent stage. So the ascent stage engine could be used only short before the reentry. $\endgroup$
    – Uwe
    Feb 2, 2023 at 21:28
  • $\begingroup$ @Uwe well, I did say assuming we didn't care about the crew :-P $\endgroup$ Feb 2, 2023 at 21:32
  • 1
    $\begingroup$ @MichaelStachowsky you did say that, but a unmanned LM without electrical power was useless. $\endgroup$
    – Uwe
    Feb 2, 2023 at 21:36
  • $\begingroup$ @Uwe fair enough $\endgroup$ Feb 3, 2023 at 1:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.