Thank you, Glom. I greatly appreciate your bringing up topics of discussion on this point. I believe “pericynthion” is an orbital term.
Therefore, “pericynthion” does not exist unless you are first in orbit, around something.
However, I do not believe it is possible for the Moon’s gravity to slow down Apollo 11 enough to allow Apollo 11 to regain orbital velocity relative to the Earth.
And, I do not mean to argue, but I do believe that Apollo 11 needed to be in a flat elliptical orbit for the re-entry corridor to work.
If Apollo 11 were traveling escape velocity as it approached the Earth, then it would either collide at high velocity, or else it would fly right on by, but either way the command module would not have swung around to the antipode/perigee side of the Earth unless it were in elliptical orbit. But, again, thank you for responding. Your answer helps me put my own questions into perspective.
However, I do not believe . . . .
. . . But I do believe . . . .
...If Apollo 11 were traveling escape velocity as it approached the Earth, then it would either collide at high velocity, or else it would fly right on by...
But, you are getting off topic here.
First, at the end of the Trans Earth Injection Burn, Apollo 11 had three different relative velocities, which were approximately: (1) 2,624 meters per second relative to the Moon;
(2) 3,624 meters per second relative to the Earth;
First, at the end of the Trans Earth Injection Burn, Apollo 11 had three different relative velocities, which were approximately: (1) 2,624 meters per second relative to the Moon; (2) 3,624 meters per second relative to the Earth; and (3) 33,044 meters per second relative to the Sun.
...You do not get a perigee point unless you are in orbit.
Assuming that chart noted above might be correct, 1,529 meters per second at the distance of the Moon is still escape velocity with respect to the Earth, no?
However, I do not believe it is possible for the Moon’s gravity to slow down Apollo 11 enough to allow Apollo 11 to regain orbital velocity relative to the Earth.
General word is Periapsis.
But it's not really an issue for the reasons others have stated, e.g. aerobraking and vacuum perigee.
It is been said that Apollo crews never actually left earth orbit, even if beeing closed to it.Not only was it said, it was also true. At no time did an Apollo spacecraft on a lunar mission achieve earth escape velocity. The eccentricity after TLI was 0.97, which is still a closed elliptical orbit with an apogee somewhat beyond the moon's orbit.
I believe “pericynthion” is an orbital term. Therefore, “pericynthion” does not exist unless you are first in orbit, around something.
Thank you for your reply. May I get a t-shirt with "I brought up anything useful on the ApolloHoax.net-Forum before somebody else beat me to it" ? :PGeneral word is Periapsis.
Thank you, that's the word I couldn't think of earlier! :)
Not only did they all have a perijove, all four also had perikrones. And Voyager 2 even had a periuranion and a periposeidion too.
And in just a few years New Horizons will reach perihadion, after having had a perijove some years ago.
Thank you, ka9q. So, after the Trans Earth Injection Burn, Apollo 11 was indeed traveling escape velocity with respect to both the Moon and the Earth. How did it get home?
Thank you, ka9q. So, after the Trans Earth Injection Burn, Apollo 11 was indeed traveling escape velocity with respect to both the Moon and the Earth. How did it get home?
Thank you, ka9q. So, after the Trans Earth Injection Burn, Apollo 11 was indeed traveling escape velocity with respect to both the Moon and the Earth. How did it get home?
You seem to labouring under the misconception that once a satellite has escape speed it must shoot away from the attractor. Not so. If the direction is straight at the attractor, it will hit it.
In astrodynamics, direction is as important as speed.
and immune to aerobraking.You seem to labouring under the misconception that once a satellite has escape speed it must shoot away from the attractor. Not so. If the direction is straight at the attractor, it will hit it.
In astrodynamics, direction is as important as speed.
I think he is laboring under the misconception that once escape velocity has been reached the spacecraft will never slow down again; it becomes immune to gravity.
Thank you, ka9q. So, after the Trans Earth Injection Burn, Apollo 11 was indeed traveling escape velocity with respect to both the Moon and the Earth. How did it get home?No, that's not what I said (NTNWIS).
But, in your calculations, wherein you came up with your 16,000 km figure, did you assume that the Moon was just making Apollo 11 slow down, or did you assume that the Moon would be pulling Apollo 11 along on the Moon's orbital trajectory?For purposes of illustration it is possible to approximate a lunar mission as two patched conics, i.e., an ordinary 2-body orbit around the earth connected to one around the moon. But that's only a rough approximation, and what actually happens is much messier. During critical periods the spacecraft is significantly affected by both the earth and the moon and that means solving the three-body problem, made famous by the fact that several brilliant mathematicians looked for but failed to find any general analytic solutions, unlike the 2-body problem that does have such solutions.
Chew, yes I read your posts. Thank you for taking the time. I would like to know whether your calculation to come up with the 16,000 km distance from the Moon assumed the Moon was stationary, or did you account for the fact the Moon would be moving away at the rate of approximately 1,000 meters per second?
So, somewhere in the calculations, NASA had to account for the velocity of 1,000 m/s, which is the approximate velocity of the Moon in orbit around the Earth. That is what I am looking for: Where and how did NASA account for this velocity?
So, somewhere in the calculations, NASA had to account for the velocity of 1,000 m/s, which is the approximate velocity of the Moon in orbit around the Earth. That is what I am looking for: Where and how did NASA account for this velocity?
Yes, I realize it is much more fun beating up lunar landing doubters on issues like lunar photographs and radiation belts, but the velocity numbers are much more important.
As soon as the Apollo 11 direction changed, and it was no longer traveling retrograde relative to the Moon's orbital trajectory, then Apollo 11's velocity relative to the Earth should have been recalculated.
The retrograde motion on the back side of the Moon was not a braking force, and Apollo 11 would have retained its momentum.
Back in 1969, did NASA forget to include the Moon’s orbital velocity when it computed Apollo 11’s return velocity to the Earth?
Telling me that the orbital velocity was included in some type of computer program does not answer the question, because the velocity numbers that NASA did provide back in 1969 do not include the Moon's orbital velocity.
So, without insulting me by calling me naive, could someone please explain to me why the Moon's orbital velocity was not included.
It seems to me that an elliptical orbit is necessary, else Apollo 11 could never get all the way around to the antipode.
So, somewhere in the calculations, NASA had to account for the velocity of 1,000 m/s, which is the approximate velocity of the Moon in orbit around the Earth. That is what I am looking for: Where and how did NASA account for this velocity?
The retrograde motion on the back side of the Moon was not a braking force, and Apollo 11 would have retained its momentum. So, I am back to the same question: Back in 1969, did NASA forget to include the Moon’s orbital velocity when it computed Apollo 11’s return velocity to the Earth?
Back in 1969, did NASA forget to include the Moon’s orbital velocity when it computed Apollo 11’s return velocity to the Earth?No.
the velocity numbers that NASA did provide back in 1969 do not include the Moon's orbital velocity.
It seems to me that an elliptical orbit is necessary
the velocity numbers are much more important.
(http://2.bp.blogspot.com/_UuybM7Z4NjI/SAYMfEsRWSI/AAAAAAAACI8/t-NcL2OP8gg/s400/Poop.jpg)
The retrograde motion on the back side of the Moon was not a braking force, and Apollo 11 would have retained its momentum.How can you possibly say this?
It seems to me that an elliptical orbit is necessary
It's more correct to say that a conic section is necessary. Ellipses and circles are conic sections, but so is the parabola, the hyperbola and even the straight line. These last three correspond to open trajectories, i.e., the object is on an escape trajectory.
circle: eccentricity=0
ellipse: 0 < e < 1
parabola: e = 1
hyperbola: e > 1
straight line: e = infinity
It's more correct to say that a conic section is necessary. Ellipses and circles are conic sections, but so is the parabola, the hyperbola and even the straight line. These last three correspond to open trajectories, i.e., the object is on an escape trajectory.
circle: eccentricity=0
ellipse: 0 < e < 1
parabola: e = 1
hyperbola: e > 1
straight line: e = infinity
With respect, parabolas and hyperbolas cannot, by definition, be "orbits"
In celestial mechanics, a Kepler orbit (or Keplerian orbit) describes the motion of an orbiting body as an ellipse, parabola, or hyperbola, which forms a two-dimensional orbital plane in three-dimensional space.
No matter your terminology, I'm not sure how a hyperbola can be an orbit, open or otherwise.Technically true, I think... the terminology I usually see is a parabolic or hyperbolic trajectory.