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This is from the Apollo 11 mission report. It says the vehicle was going at 7.6 km/s when the CSM separated from the third stage. It was not in orbit at this point, it was thousands of KM from earth.

Is this accurate? Am I missing something? That seems incredibly far below other sites saying that you need 10km/s or more to leave the earth.

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    $\begingroup$ "It was not in orbit at this point, it was thousands of KM from earth." One does not follow from the other. In this answer you can see four spacecraft that orbit as far as 20,000 to 40,000 km from the Earth, yet are still definitely in "Earth orbit". $\endgroup$
    – uhoh
    Jun 10, 2018 at 6:15
  • $\begingroup$ @JCRM No the numbers are not from that link. Those altitudes and velocities are in the 7,000's The numbers are from this link about Apollo 11. ibiblio.org/apollo/Documents/lvfea-AS506-Apollo11.pdf $\endgroup$
    – uhoh
    Jun 11, 2018 at 1:05
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    $\begingroup$ "It was not in orbit at this point, it was thousands of KM from earth." - the moon is in earth orbit. It had not reached the moon yet, let alone escaped past the moon and out of earth's gravity well. $\endgroup$ Jan 4, 2020 at 1:04

2 Answers 2

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As @Rikki-Tikki-Tavi points out escape velocity is the velocity you would need at (or near) the surface of Earth to make out out of Earth orbit. Of course, just like anything thrown up into the air, the spacecraft decelerates as it moves away from the Earth.

Having escape velocity means that your total energy (relative to the Earth in this case) is greater than or equal to zero.

Using the equations from this answer

$$\mathscr{E}_{tot} = \frac{1}{2}v^2 - \frac{GM}{r}$$

where $\mathscr{E}_{tot}$ is energy per kilogram. Since both terms use the same $m$ you can just calculate energy per kilogram by dropping it.

Let's do the calculation for the "actual" numbers in the table.

altitude       7066 km
Earth radius   6378 km
r             13444 km
GM         3.986E+5 km^3/s^2
v             7.609 km/s

$\begin{align}\frac{1}{2}v^2 & =28.948\\ -\frac{GM}{r} & = -29.649\\ \mathscr{E}_{tot} & = -0.701 \ km^2/s^2 \end{align}$

So the spacecraft has plenty of energy, but not quite enough to completely escape Earth's gravity. That means if it didn't approach the moon, it would keep going until all of the kinetic energy was used up by being stored as potential energy.

$-\frac{GM}{r_{max}} = -0.701$ gives

$r_{max} = 569,000$ km. or a little past the orbital radius of the Moon if the Moon's gravity wasn't there.

So the numbers work out just fine.

If the spacecraft returned to LEO at say 400 km altitude or $r = 6778$, the velocity would then be given by

$-0.701 = \frac{1}{2}v^2 - \frac{GM}{6778 km}$

$v = \sqrt{2(-0.701 + GM/6778 km)} $

and that gives 10.78 km/s

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    $\begingroup$ I think the point here is that Apollo never required Earth escape velocity because it never left Earth's sphere of influence (after all, the Moon is orbiting the Earth). It would have been undesirable for Apollo to be on an Earth escape trajectory for a couple of reasons: (1) it would have involved needless expenditures of propellant acquiring and then discarding excess velocity; (2) if the burn to enter lunar orbit failed to occur, Apollo would have simply orbited back to Earth, rather than wander off into a solar orbit that may never bring it home. $\endgroup$
    – Anthony X
    Jun 10, 2018 at 23:54
  • $\begingroup$ @JCRM "not sure where 7.609 comes from" I used the values that are shown in the "actual" column of the data show right there in the OP's question. i.stack.imgur.com/3SGHo.png I've rolled back as I'm answering the question as asked. Thanks for the formula align though, that looks great! $\endgroup$
    – uhoh
    Jun 11, 2018 at 0:33
  • $\begingroup$ From the linked PDF Table 7-11 page 7-9 (117/359) I was unable to find table 4-3 in that document. $\endgroup$
    – user20636
    Jun 11, 2018 at 0:43
  • $\begingroup$ @JCRM That's an interesting, but different problem. The OP may have been reading multiple documents and linked a different pdf, but the question is about making sense of the numbers so it's best to answer using the numbers the OP gives. $\endgroup$
    – uhoh
    Jun 11, 2018 at 0:49
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    $\begingroup$ Absolutely agree you should use the OP's figures - I thought I was (reading fuzzy graphics isn't my strong point) $\endgroup$
    – user20636
    Jun 11, 2018 at 1:32
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It is normal for a spacecraft to slow down as it coasts towards the Moon. The spacecraft is still in Earth orbit and is being pulled by Earth’s gravity during its entire trip to the Moon. During the powered thrust phase known as TLI (trans lunar injection) the maximum speed reached by the Apollo 11 spacecraft was 10.8 km/sec (24,200 mph) when it was at an altitude of about 300 km (180 miles). By the time of Command Module separation from the S-IVB stage 26 minutes later it had reached an altitude of 7,000 km as you noticed in the table, and by then it had slowed to 7.6 km/sec (17,000 mph). During the approximately three day coast to the Moon it continued to slow, and by the time it reached the Moon it had slowed to 0.9 km/sec (2,000 mph). It then began to speed up again as the Moon’s gravity pulled it towards the Moon.

As explained in other answers the Apollo spacecraft did not reach escape velocity. At the 300 km altitude of TLI escape velocity was 10.9 km/sec. The velocity reached during TLI was just short of this at 10.8 km/sec. It did not have to reach escape velocity because the Apollo spacecraft never left Earth’s orbit. In fact even while orbiting the Moon the spacecraft was still in Earth orbit, as of course the Moon is.

Even if a spacecraft accelerates to escape velocity, as soon as the engines cut off and it begins coasting it will immediately begin slowing down. This slowing will occur as long as the spacecraft is close enough to Earth to be influenced by its gravity, which effectively extends several million miles out into space. After getting far away enough from Earth the spacecraft will stop slowing and remain at whatever velocity it has remaining, and will be influenced from that point forward only by the Sun’s gravity, or any planets or other objects that it comes within vicinity of.

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  • $\begingroup$ My first answer was more of a commentary about trying to understand the original question, made difficult by the fact that the OP did not engage in any further discussion, and likely never will due to the amount of time that has passed. The OP made no comments other than what was in the original question. I have now edited my answer to be a reply to the original question, based on what I think was the intent and premise of the question, as I am best able to determine it from reading between the lines of the question. $\endgroup$ Jul 16, 2023 at 11:49
  • $\begingroup$ It's probably asked by a slightly more sophisticated than average Moon-hoaxer. $\endgroup$ Jul 16, 2023 at 23:53
  • $\begingroup$ The strategy of most Moon hoaxers is to bait with an innocent sounding question, then when people make good faith responses the hoaxer goes on attack ridiculing them and of course making inane claims and arguments. In this case the OP did not engage at all after posting the question, and seems to have disappeared after their post which was apparently their only activity on this site. As I indicated in my original commentary (available in the edits) they just seem to have had the very widespread belief that spacecraft somehow "break free" of Earth's gravity when they launch into space. $\endgroup$ Jul 17, 2023 at 11:26
  • $\begingroup$ Thanks, in many years on this site, I've seen plenty of different angles, from incoherent illiterate ranting to more sophisticated flavors of mental illness. Stick around, July is the prime month. $\endgroup$ Jul 17, 2023 at 11:31
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    $\begingroup$ I know what you mean, and often my hoaxer radar gets activated by similar sounding questions. One telltale sign is the word "strange", as in "people flew to the Moon in the sixties but now it's apparently too hard, strange". What I find works is to be sincere and polite, then when they start the unprovoked personal attacks, which they can't seem to resist doing, it pretty much exposes them as a troll. They seem to be defenseless against anyone who doesn't sink to their level of personal attacks. $\endgroup$ Jul 17, 2023 at 11:45

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