I assume Apollo's velocity slowed down after it left Earth orbit, for how long was it decelerating. Did it start accelerating as it approached the Moon. What was the rate of deceleration due to Earth's gravity?
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1$\begingroup$ What do you mean? Rate of deceleration at what point? Much of the lunar transfer is unpowered. Can you clarify. $\endgroup$– Rory Alsop ♦Jan 9, 2019 at 22:35
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1$\begingroup$ @RoryAlsop I did not say powered deceleration. The velocity of Apollo changed from TLI to until Lunar orbit insertion due to the gravity of the Earth and presumably the Moon. What what that rate of change? $\endgroup$– Bob516Jan 9, 2019 at 22:56
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4$\begingroup$ While the question could have been better formulated, I don't feel it merits being down voted, as judging on the basis of OP's record here, it was asked in good faith. What seems obvious to ourselves might not to somebody else, and that is something all of us ought to keep in mind when interacting here. $\endgroup$– Happy KoalaJan 9, 2019 at 23:02
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1$\begingroup$ @uhoh That would indeed be a rather awesome program... You're full of awesome ideas, you know? If you have an official fan club, could I pretty please be its president? There are lots of charting libraries out there so I could spin one up really quickly, but first the stars orbiting around the Saggie :D (my new nick name for Sagittarius A). $\endgroup$– Happy KoalaJan 10, 2019 at 8:35
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1$\begingroup$ @RoryAlsop Ah, fair enough, after all those Karman posts I've started equating down votes to "bugger off, you sod", but of course that's not the case. I'll keep that in mind next time around my eyes start tearing up at the sight of a post getting down voted :D $\endgroup$– Happy KoalaJan 10, 2019 at 8:37
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5 Answers
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I assume Apollo's velocity slowed down after it left Earth orbit, for how long was it decelerating.
A pretty good analogy for TLI is throwing a baseball straight up into the air. The "throw" is the TLI burn; as soon as the ball leaves your hand it begins to slow down, trying to fall back towards Earth. The peak altitude of the "throw" is around where the moon will be three days later.
Did it start accelerating as it approached the Moon.
Yes, as noted in @PearsonArtPhoto's answer.
What was the rate of deceleration due to Earth's gravity?
The deceleration decreases the farther you get from Earth in an inverse-square relation:
$a=-{G M \over r^2}$
Where $GM$ (aka $\mu$) is the gravitational parameter of Earth (3.986e14) and $r$ is the distance from the center of Earth in meters. In LEO this is still 9.2 m/s2 or about 94% of Earth's surface gravity. By the time you're 3000 km up, though, it's only about 50% of surface gravity.
You can compare the results of the equations for the "downward"-pulling Earth component and the "upward"-pointing moon component, with the appropriate distances and gravitational parameters for the two bodies, to see what the crossover point is. Or use algebra if you're into that.
answered Jan 10, 2019 at 0:42
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Yes, in fact it did slow down with time, until it approached close enough that the Moon pulled it faster. That happened at a point very close to the Moon. In a diagram on this page, for Apollo 8 we can see that point was just after the second full day, and the speed was about 3578 km/hr.
Borman, Lovell and Anders were the first humans to leave the Earth’s gravity. They also never felt any physical change when the spacecraft slowed down to 3,578 kilometres per hour relative to Earth and crossed over into the Moon’s gravity field at 55:38:40 GET (0629:40 AEST). They were 326,415 kilometres from Earth and 62,598 kilometres from the Moon.
answered Jan 9, 2019 at 23:26
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$\begingroup$ from your page: "At 2:50:37.79 GET (0141:37 AEST), the S-IVB stage burned for 5 minutes 17.7 seconds to boost the spacecraft’s velocity by 7,451.2 kilometres per hour, and Apollo 8 left Earth orbit and headed for the Moon at 38,959.4 kilometres per hour." $\endgroup$ Jan 10, 2019 at 1:39
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1$\begingroup$ "Borman, Lovell and Anders were the first humans to leave the Earth’s gravity. They also never felt any physical change when the spacecraft slowed down to 3,578 kilometres per hour relative to Earth and crossed over into the Moon’s gravity field at 55:38:40 GET (0629:40 AEST). They were 326,415 kilometres from Earth and 62,598 kilometres from the Moon." $\endgroup$ Jan 10, 2019 at 1:52
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1$\begingroup$ @bitchaser What exactly turned on the light indicating Apollo 8 was starting to fall towards the Moon? $\endgroup$– uhohFeb 17, 2019 at 21:48
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2$\begingroup$ Just to nitpick, the statement "Borman, Lovell and Anders were the first humans to leave the Earth’s gravity" is incorrect. They never left earth's gravity. No one can, everything that exists and will ever exist is within earth's gravity (as long as earth exists). What the author probably meant is that they were the first humans to be subject to a gravitational field that was stronger than that of earth. $\endgroup$ Mar 16, 2019 at 18:41
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2$\begingroup$ Yup, I understand it's a direct quote. The reason why I brought it up is that the answer could be improved by pointing out that this direct quote is in fact incorrect (the mistake is very common but fundamental and very much relates to the subject of the question). $\endgroup$ Mar 18, 2019 at 11:44
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I think you are all referring to the "Lagrange Point" between the Earth and the moon; this is a physical location where the gravitational forces of the Earth and moon are equal on a spacecraft. As soon as TLI is completed, maximum speed is achieved for the burn period. The spacecraft slows gradually, and at a less rate with more distance from the Earth, all the way to Lagrange Point. At which the spacecraft starts falling to the moon and speeds up gradually, at an increasing rate, while getting closer to the moon.
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1$\begingroup$ "...this is a physical location where the gravitational forces of the Earth and moon are equal on a spacecraft." I don't think this is the correct way to explain how the point is defined and calculated unless you mention that this is in the rotating frame of the Earth-Moon system. In reality it is the point where an object would remain in orbit around the center of mass of the Earth-Moon system with the same period as that of the Moon. If you looked at it in a rotating frame, it would look like the forces balanced, but that's not what's really happening. $\endgroup$– uhohAug 26, 2020 at 0:47
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1$\begingroup$ It would be great if you'd update your answer a bit to reflect this difference, and perhaps add a link to Wikipedia or your favorite site about Lagrange points. Welcome to Stack Exchange! $\endgroup$– uhohAug 26, 2020 at 0:48
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$\begingroup$ OK, thank you for reading my input. I mentioned the Lagrange point concept as a top of the hill sort of idea. After a TLI acceleration burn is completed, the spacecraft will slightly decelerate, ie traversing uphill due to the earth gravity field, along its orbital path to the moon. Eventually, the gravitational influence of the earth and moon are equal; ie the top of the hill (similar to a Lagrange point). Then the moon gravity field accelerates the spacecraft, ie a downhill orbital path with respect to velocity, until a burn to decelerate to a lunar orbit is achieved. $\endgroup$ Aug 26, 2020 at 21:01
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One thing to remember is that the Apollo spacecraft was not launched towards the moon. It was launched towards where the moon was going to be three days later. Just as it was about to start falling back towards the earth, the moon came in from the side and captured it (with a little help from the service module engine).
If you leave the moon out of the picture, Apollo would have never left earth's orbit at all. It just changed from a circular orbit to a very tall, skinny elliptical orbit.
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$\begingroup$ I want to say "but it never did leave Earth orbit..." but I wont. :-) $\endgroup$– uhohAug 26, 2020 at 4:34
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Typically the closer you get to a larger celestial body, you speed up. that's why space probes use Jupiter for gravity assists
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$\begingroup$ Not an answer to the question, which is velocity change while moving away. $\endgroup$ Aug 26, 2020 at 8:05