I think the name of this thread should be,
Was the SM big enough to get back to Earth? It was the service module (SM) that contained the service propulsion system (SPS), thus the SM should be the focus of attention, not the CM. The answer to this question is, of course, yes. Let’s look at Apollo 17 for an example. The Apollo 17 SPS performed six burns as follows (figures are from the NASA mission reports):
Midcourse correction #2: 9.9 ft/s
Lunar orbit insertion: 2,988 ft/s
Descent orbit insertion #1: 197 ft/s
CSM circularization: 70.5 ft/s
CSM lunar orbit plane change: 366 ft/s
Trans-Earth injection: 3,046.3 ft/s
The first three burns were performed with the LM docked and fully fueled. The last three burns were performed with the LM undocked or jettisoned. The total mass of the CM, SM, and LM were as follows:
Command module, CM: 12,874 lbm
Service Module, SM: 54,044 lbm
Lunar module, LM: 36,262 lbm
The mass of SPS propellant was 40,594 lbm, which is included in the SM mass given above.
To determine the amount of propellant used to perform the six SPS burns we will use the following equation,
Mf = Mo*e^-(dV/Ve)
Where Mf = final mass, Mo=initial mass, e=mathematical constant (2.71828), dV=change in velocity, and Ve=exhaust gas velocity. Exhaust gas velocity can be determined from the specific impulse (Isp) of the SPS, which was 314 s, therefore
Ve = g*Isp = 32.174*314 = 10,103 ft/s
We can determine the amount of propellant needed for the first three burns with a single calculation, where the total velocity change is,
dV = 9.9+2,988+197 = 3,194.9 ft/s
The initial mass is simply the combined mass of the CM, SM and LM,
Mo = 12,874+54,044+36,262 = 103,180 lbm
Therefore, the final mass is
Mf = 103,180*e^-(3,194.9/10,103) = 75,207 lbm
The amount of propellant burned is Mo-Mf,
Mp1 = 103,180-75,207 = 27,973 lbm
Now we must determine the amount of propellant burned for the final three burns, where
dV = 70.5+366+3,046.3 = 3,482.8 ft/s
The initial mass is the final mass from the previous calculation, less the LM, which is either undocked or jettisoned,
Mo = 75,207-36,262 = 38,945 lbm
Alternatively, the initial mass is the CM, plus SM, less SPS propellant already used,
Mo = 12,874+54,044-27,973 = 38,945 lbm
The final mass after the last three burns is,
Mf = 38,945*e^-(3,482.8/10,103) = 27,589 lbm
And the amount of propellant used is,
Mp2 = 38,945-27,589 = 11,356 lbm
Thus, the total SPS propellant used is,
Mp_total = 27,973+11,356 = 39,329 lbm
Since our initial propellant load was 40,594 lbm, we have remaining,
Mp_remaining = 40,594-39,329 = 1,265 lbm
According to the mission reports, the actual propellant remaining was 1,630 lbm. This is because I simplified the calculations by ignoring that the mass of the SM was further reduced by consumables used during the flight. 624.7 lbm of SM RCS propellant, 60.9 lbm of hydrogen, and 606.8 lbm of oxygen was consumed during the Apollo 17 mission. These figures are partially offset by the 243.1 lbm of lunar samples collected. There is really no need to get any more precise than my calculations above because even my somewhat conservative computations show that the SM carried enough propellant to perform the maneuvers required of it. And one more thing, the amount of volume needed to storage 40,594 lbm of aerozine 50 and nitrogen tetroxide is about 560 ft^3, which is consistent with the size of the propellant tanks housed within the SM.
All that's needed is to see if the CSM can achieve lunar escape velocity from low lunar orbit.
So why don’t we perform that calculation. Sticking with Apollo 17 as our example, the CSM was in a 61.2 X 63.9 nautical mile orbit following LM jettison and CSM separation. If we take the average and convert to kilometers, we have an orbital altitude of 116 km. The Moon’s radius is 1,738 km, thus the semi-major axis of the orbit was 1,854 km. The orbital velocity of a spacecraft is given by the equation,
V = (GM/R)^0.5
Where V is the velocity, R is the orbital radius (semi-major axis), and the value of GM for the Moon is 4.902794E+12 m^3/s^2. Therefore,
V = (4.902794E+12/1,854,000)^0.5 = 1,626 m/s
The equation for escape velocity is,
Vesc = (2*GM/R)^0.5
Thus,
Vesc = (2*4.902794E+12/1,854,000)^0.5 = 2,300 m/s
Therefore, the amount of delta-V needed to go from orbital velocity to escape velocity is,
dV = 2,300-1,626 = 674 m/s (2,211 ft/s)
On Apollo 17 TEI added 3,046.3 ft/s, which was well beyond lunar escape velocity. This was done to reduce the amount of time needed to get back home. Thus, I have shown the SPS not only had enough propellant to get Apollo 17 back to Earth, but it could get it back quickly and with propellant to spare. Similar calculations for the other Apollo missions will show similar results, that is, the SM was big enough.
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