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m74z00219
2009-Feb-13, 11:18 PM
Hi all, I'm sort of confused by orbits. I have two ideas in my head that are contradictory. Part of me knows that the further out an object is from a central body, the lesser its velocity will be to maintain a circular orbit at that particular radius.

When I think about a satellite going around (say the Earth), I imagine that it needs to speed up to get to a higher orbit, not slow down. If the satellite's orbital speed decreased to zero, it would plummet to the Earth, not rise in orbit. So...I'm a bit confused and not even sure if the phrasing of my confusion makes any sense. Thanks to anyone who can help me make sense of this.

m74

Hornblower
2009-Feb-13, 11:42 PM
Hi all, I'm sort of confused by orbits. I have two ideas in my head that are contradictory. Part of me knows that the further out an object is from a central body, the lesser its velocity will be to maintain a circular orbit at that particular radius.

When I think about a satellite going around (say the Earth), I imagine that it needs to speed up to get to a higher orbit, not slow down. If the satellite's orbital speed decreased to zero, it would plummet to the Earth, not rise in orbit. So...I'm a bit confused and not even sure if the phrasing of my confusion makes any sense. Thanks to anyone who can help me make sense of this.

m74

Let's start with a spacecraft in a low circular orbit. We speed it up with a short burn, and it starts rising. It now is moving on the outward half of an elliptical orbit, and slowing down as it goes because it is rising against Earth's gravity. Halfway around it levels off at the highest point in this new orbit, known as the apogee. At this point it is moving too slowly to stay at that elevation, and if left alone it will descend back to the same point at which we gave it the thrust. This is perigee. Now it is back to the same excessive speed it had immediately after the burn. It will continue in this eccentric orbit indefinitely.

Now let's give it another prograde burn when it reaches apogee again. If we give it just the right amount, it will speed up enough to remain in a new circular orbit. It will be faster than it was before this burn, but slower than it was in the lower circular orbit.

astromark
2009-Feb-13, 11:55 PM
m74; " The further out an object is from a central body, the lesser its velocity will be to maintain a circular orbit at that particular radius." Is what you said...

That is true. No confusion there. Radial velocity and mass are the key components to orbital stability.

" I imagine that it needs to speed up to get to a higher orbit, not slow down. "

NO. thats not the whole story.
Some body will come forward any second and share with us that magic distance a space craft needs to be to maintain a Geo stationary orbit... from a lower orbit to reach that more distant point would require a thrust or energy. Not just speeding up. Increasing the velocity alone would altar the orbital track. The trajectory is very important. Get the angle right and yes a higher orbit is possible. Get it wrong and you are on your way to some place else or coming back through the atmosphere as carbon dust. Insertion into orbit is a science. No room for guess work. hideously elliptic object orbits would be a worry. Velocity and mass balanced correctly and yes. On seeing Hornblower's post. He has cleared the 'how' to do it right part. thanks.

m74z00219
2009-Feb-14, 12:04 AM
Thanks guys, I believe you've cracked open my confusion. I think the bulk of it was me confusing velocity required to MAINTAIN a particular circular orbit with how you actually manage to reach that higher orbit in the first place. So, you may have to increase your KE to reach a higher orbit, but you will have to shed much of it if you want to stay in that higher circular orbit, rather than return on the elliptical path.

Astromark, is it typical to raise satellites from LEO to GEO by a Hohmann transfer (what I believe Horblower was describing)?

mugaliens
2009-Feb-14, 12:14 AM
Picture a satellite in a circular orbit. It gives a short burn on it's rockets in the direction of travel, which causes it's velocity to speed up, but also changes it's circular orbit into an elliptical one, with the burn position as the point of closest approach (perigee), as well as the fastest (velocity).

180 degrees around, it reaches it's point of furthest reach (apogee), which is also the slowest point. At that moment, it gives another burn, again in the direction of travel, which "inserts" it into this higher orbit, at greater altitude, but slower velocity. Still, the total energy (potential + kinetic) is greater.

astromark
2009-Feb-14, 12:20 AM
Yes. that, it is.

When you consider all of the above. The fact that the 'Casinie Prob' is orbiting Saturn as it is is... Its all done with such class.

Unrelated,. But why do we not lift the ISS to a safer altitude.?

Peter B
2009-Feb-14, 12:29 AM
Picture a satellite in a circular orbit. It gives a short burn on it's rockets in the direction of travel, which causes it's velocity to speed up, but also changes it's circular orbit into an elliptical one, with the burn position as the point of closest approach (apogee), as well as the fastest (velocity).

180 degrees around, it reaches it's point of furthest reach (perigee), which is also the slowest point. At that moment, it gives another burn, again in the direction of travel, which "inserts" it into this higher orbit, at greater altitude, but slower velocity. Still, the total energy (potential + kinetic) is greater.

Uh, minor nitpick, but apogee is the high point in the orbit, and perigee is the low point.

Peter B
2009-Feb-14, 12:36 AM
Thanks guys, I believe you've cracked open my confusion. I think the bulk of it was me confusing velocity required to MAINTAIN a particular circular orbit with how you actually manage to reach that higher orbit in the first place. So, you may have to increase your KE to reach a higher orbit, but you will have to shed much of it if you want to stay in that higher circular orbit, rather than return on the elliptical path.

Er, there's no shedding of energy to transfer from an elliptical orbit to a circular orbit at the altitude of the elliptical orbit's high point (apogee). As Hornblower and Mugaliens said, the change is made by a prograde burn (nose of the spacecraft pointing in the direction of travel) at apogee. (Boy this would be so much easier with a diagram. Anyone know a good link?)


Astromark, is it typical to raise satellites from LEO to GEO by a Hohmann transfer (what I believe Horblower was describing)?

Sorry, not Astromark, but yes, that's how it's done. The satellite is launched into a circular Low Earth Orbit, say at an altitude of 400 kilometres. Then the engine is fired once to put the satellite into an elliptical orbit with its apogee at Geo Orbit altitude (I can never remember how high that is) and perigee at 400 kilometres. Then, when it reaches apogee, the engine is fired again to circularise the orbit at Geo Orbit altitude. And that's a Hohmann Transfer.

Peter B
2009-Feb-14, 12:38 AM
Unrelated,. But why do we not lift the ISS to a safer altitude.?

The higher up the ISS, the less cargo could be carried on each resupply mission. Also the Shuttle might not be able to carry up the remaining pieces of the ISS.

Presumably it's a tricky balancing act - too high, against the dangers and costs of being too low, and running into debris or experiencing atmospheric drag.

Dave J
2009-Feb-14, 12:52 AM
I have a figure here of 35,786km, give or take, for a geostationary orbit. Takes a lot of energy to get there.

Amber Robot
2009-Feb-14, 01:43 AM
(Boy this would be so much easier with a diagram. Anyone know a good link?)

There's a good one here:

http://en.wikipedia.org/wiki/Hohmann_transfer_orbit

Peter B
2009-Feb-14, 02:35 AM
Poifect! Thanks for that.

astromark
2009-Feb-14, 03:25 AM
Anoyingly... being miss understood is the normal around here... I can except that. That was posed as a question was it not... He was not quoting me.
That while I procrastinate over my bumbling pros... along comes another whom it looked like I answered... no. It was the earlier post i said " Yes. that, it is." to.
Not that I would find fault with 'mugaliens' What I was saying has now turned up. Thank you Dave J. 35,786km / hr. So I think we are clear... Get the thing into Earth orbit... 17,500 or so km/hr. we want to lift into a much higher orbit. Apply energy as in a engine burn. Spiraling up we go... on reaching the highest point of this now very elliptic path re fire that burn to establish the new velocity required to maintain this orbital path and altatude... and I was of course only mentioning the ISS because of the new field of debris... some of which could be an issue.

Fiery Phoenix
2009-Feb-14, 01:11 PM
Hi all, I'm sort of confused by orbits. I have two ideas in my head that are contradictory. Part of me knows that the further out an object is from a central body, the lesser its velocity will be to maintain a circular orbit at that particular radius.

When I think about a satellite going around (say the Earth), I imagine that it needs to speed up to get to a higher orbit, not slow down. If the satellite's orbital speed decreased to zero, it would plummet to the Earth, not rise in orbit. So...I'm a bit confused and not even sure if the phrasing of my confusion makes any sense. Thanks to anyone who can help me make sense of this.

m74

Pardon me, but I'm unsure if that's exactly true. I know that celestial objects never reach a speed of zero, due to the fact that they are constantly affected by the gravity of the other objects here and there. I'm certainly not the sharpest knife, but I'm pretty sure that's how it basically is.

JohnD
2009-Feb-14, 01:37 PM
Some very clear thinking masters here, so may I introduce a further factor?

Consider a stable satellite. It sends a tethered probe directly earthwards, by a small thruster on the probe, and one away from Earth. The thrusters fire radially so no orbital velocity gain or loss. They stabilise at some distance from the mother ship, perhaps by a further burn.
The inner probe is now moving too slowly for its orbit. It will 'fall' in and and the tether will be in tension.
The outer probe is travelling too fast, and its tether will also be in tension.

Right so far?

In fact you don't need a complex three part satellite, just one wide enough at right angles to its orbital path, and there will be a force in a radial direction across the satellite, stretching it apart.
Is that right?

Now the question. Is this a tide?
I had thought that this was the explanation for sea tides on Earth, in particular for the flood tide on BOTH SIDES of the Earth, in line with the Moon, as both orbit their centre of gravity. But when I offered this explanation, the BA and others countered it with an explanation involving differential gravity fields. Someone even demonstrated that a body falling directly into another body, and so not in an orbit, would suffer tides.

So, the other question. Is my understanding of tides, as above, wrong?

John

Jeff Root
2009-Feb-14, 02:18 PM
Thanks guys, I believe you've cracked open my confusion. I think
the bulk of it was me confusing velocity required to MAINTAIN a
particular circular orbit with how you actually manage to reach that
higher orbit in the first place. So, you may have to increase your
KE to reach a higher orbit, but you will have to shed much of it if
you want to stay in that higher circular orbit, rather than return
on the elliptical path.
It sounds like you may have got the idea, but one part of your
description is wrong in one way and misleading in another. You
say "you will have to shed much of the kinetic energy to stay in
the higher circular orbit". Actually you need to add more kinetic
energy when you reach apogee, in order to change your elliptical
orbit into a circular orbit. What is misleading is that you actually
do shed kinetic energy as you rise to the higher altitude. That
kinetic energy is changed into potential energy. It isn't a matter
of needing to shed kinetic energy, but of shedding it whether
you want to shed it or not.

To go from a low circular orbit to a high circular orbit, you increase
your speed. That makes you rise. As you rise toward apogee, your
speed decreases, until it is not only lower than it was to begin with,
but too low to stay in a circular orbit at such high altitude.

If you do nothing at this point, you will fall back down again.
You will gain speed as you fall. When you reach the altitude of
your original circular orbit, your speed will be the same as it was
right after you increased it for the purpose of raising your orbit.
So you are again going too fast to stay in the low circular orbit,
and start rising again. You are in an elliptical orbit.

But you want to go into a high circular orbit. So when you
reach apogee, you increase your speed a second time. Now
your speed is enough to keep you at this higher altitude. Your
speed is actually less than it was in the low circular orbit, but
you got here by increasing your speed twice.

My web page on orbital speed has a little animation 2/3 of the
way down the page, on the right side, showing a satellite in an
elliptical orbit which grazes the atmosphere each time it is at
perigee, so it loses some speed and kinetic energy each orbit.
That causes the apogee to get lower and lower until the orbit
is circular at the perigee altitude, at which point the satellite
can't stay in orbit any longer:

http://www.freemars.org/jeff/speed/index.htm

-- Jeff, in Minneapolis

Peter B
2009-Feb-14, 02:20 PM
Some very clear thinking masters here, so may I introduce a further factor?

Consider a stable satellite. It sends a tethered probe directly earthwards, by a small thruster on the probe, and one away from Earth. The thrusters fire radially so no orbital velocity gain or loss. They stabilise at some distance from the mother ship, perhaps by a further burn.
The inner probe is now moving too slowly for its orbit. It will 'fall' in and and the tether will be in tension.
The outer probe is travelling too fast, and its tether will also be in tension.

Right so far?

I think so.


In fact you don't need a complex three part satellite, just one wide enough at right angles to its orbital path, and there will be a force in a radial direction across the satellite, stretching it apart.
Is that right?

Well, it won't literally pull the spacecraft apart, if that's what you mean, unless it's a very fragile spacecraft. But yes, there's a slight difference in the Earth's gravitational pull between the part nearest the Earth and the part most distant. What I'm not sure about it what the actual effect is on real spacecraft, but it's going to be small.


Now the question. Is this a tide?

Sorry, I'm not sure about the terminology.


I had thought that this was the explanation for sea tides on Earth, in particular for the flood tide on BOTH SIDES of the Earth, in line with the Moon, as both orbit their centre of gravity. But when I offered this explanation, the BA and others countered it with an explanation involving differential gravity fields. Someone even demonstrated that a body falling directly into another body, and so not in an orbit, would suffer tides.

So, the other question. Is my understanding of tides, as above, wrong?

John

In the case of ocean tides, the cause is the difference in the gravitational pull of the Moon on the oceans on the near side of the Earth, the Earth itself, and the oceans on the far side. That is, the oceans nearest the Moon experience a greater gravitational pull from the Moon than the Earth does, because it's more distant. And the oceans on the far side are still further from the Moon, so they experience even less lunar gravity.

Over the diameter of the Earth, the decrease in the Moon's gravity is significant enough to raise visible tides in the ocean. There's a similar sort of gravity gradient in the example you give above, but because we're talking about a couple of metres, rather than 13,000 kilometres, the effect is minuscule in comparison.

Jeff Root
2009-Feb-14, 02:38 PM
If the satellite's orbital speed decreased to zero, it would
plummet to the Earth, not rise in orbit.
Pardon me, but I'm unsure if that's exactly true. I know that celestial
objects never reach a speed of zero, due to the fact that they are
constantly affected by the gravity of the other objects here and there.
No, the statement was fine. If a satellite's orbital speed was
reduced to zero, it would fall straight down. Zero orbital speed
means that the satellite stays on a line drawn between the center
of the Earth and a distant star. The Earth will be rotating under
the satellite, so it would appear to be moving across the sky in
step with the distant stars. But it would also be falling straight
toward the center of the Earth.

-- Jeff, in Minneapolis

Jeff Root
2009-Feb-14, 03:25 PM
Consider a stable satellite. It sends a tethered probe directly
earthwards, by a small thruster on the probe, and one away from
Earth. The thrusters fire radially so no orbital velocity gain or loss.
They stabilise at some distance from the mother ship, perhaps by
a further burn.
The inner probe is now moving too slowly for its orbit. It will 'fall'
in and and the tether will be in tension.
The outer probe is travelling too fast, and its tether will also be
in tension.

Right so far?
The details are wrong, I think, but the bottom line is right: The
tether will be in tension.

Thrusting radially is more complicated than thrusting along the
orbit. The exact result isn't obvious to me.



In fact you don't need a complex three part satellite, just one
wide enough at right angles to its orbital path, and there will be
a force in a radial direction across the satellite, stretching it apart.
Is that right?
Sure. Gravity-gradient stabilization. The International Space
Station uses it, and many smaller satellites in LEO.



Now the question. Is this a tide?
I had thought that this was the explanation for sea tides on Earth, in
particular for the flood tide on BOTH SIDES of the Earth, in line with
the Moon, as both orbit their centre of gravity. But when I offered
this explanation, the BA and others countered it with an explanation
involving differential gravity fields. Someone even demonstrated that
a body falling directly into another body, and so not in an orbit, would
suffer tides.

So, the other question. Is my understanding of tides, as above, wrong?
Tides are very complex and unintuitive. We would need a more
detailed description from you to determine whether it is correct.
The differences you have encountered could merely be a matter
of terminology, rather than an incorrect description. There could
also be different but equivalent ways of correctly describing the
phenomenon. Or you might be describing the phenomenon correctly
but describing its causes incorrectly.

A tether is under tension because of the gravity gradient across
its length. I would call that a tidal force. A ball of liquid in orbit--
say, a glob of liquid mercury-- would be slightly elongated by that
tidal force into a prolate spheroid. It would have tidal bulges.

The tides on Earth are not aligned with the Moon, mainly because
water flows when it piles up, so the tidal bulges move a little bit
ahead of where you might expect them to be.

-- Jeff, in Minneapolis

Fiery Phoenix
2009-Feb-14, 03:40 PM
No, the statement was fine. If a satellite's orbital speed was
reduced to zero, it would fall straight down. Zero orbital speed
means that the satellite stays on a line drawn between the center
of the Earth and a distant star. The Earth will be rotating under
the satellite, so it would appear to be moving across the sky in
step with the distant stars. But it would also be falling straight
toward the center of the Earth.

-- Jeff, in Minneapolis

That doesn't apply to planets, though, does it? I mean, the case maybe different with satellites around planets, but not with planets around stars, right?

Jeff Root
2009-Feb-14, 03:58 PM
No, the statement was fine. If a satellite's orbital speed was
reduced to zero, it would fall straight down. Zero orbital speed
means that the satellite stays on a line drawn between the center
of the Earth and a distant star. The Earth will be rotating under
the satellite, so it would appear to be moving across the sky in
step with the distant stars. But it would also be falling straight
toward the center of the Earth.
That doesn't apply to planets, though, does it? I mean, the case
maybe different with satellites around planets, but not with planets
around stars, right?
Sure it would apply. If a planet were somehow to lose all of its
speed relative to the star it orbits, so that its orbital speed were
zero, it would begin to fall straight down toward the star,
accelerating all the way.

Maybe you have a different situation in mind than I have, and I
just haven't imagined the situation you are assuming?

-- Jeff, in Minneapolis

Fiery Phoenix
2009-Feb-14, 04:06 PM
Sure it would apply. If a planet were somehow to lose all of its
speed relative to the star it orbits, so that its orbital speed were
zero, it would begin to fall straight down toward the star,
accelerating all the way.

Maybe you have a different situation in mind than I have, and I
just haven't imagined the situation you are assuming?

-- Jeff, in Minneapolis

I'm actually just going by what people pointed out in this thread of mine, which I made less than a month ago:

http://www.bautforum.com/space-astronomy-questions-answers/84122-orbit-calculation.html

From what I understood from the majority of posts in that thread, a planet's orbital velocity never, ever reaches zero. No matter what.

I'm sorry if I'm making too much out of this, Jeff. I am trying here. Thank you. :)

Jeff Root
2009-Feb-14, 06:09 PM
I'm actually just going by what people pointed out in this thread of
mine, which I made less than a month ago:

http://www.bautforum.com/space-astronomy-questions-answers/84122-orbit-calculation.html

From what I understood from the majority of posts in that thread,
a planet's orbital velocity never, ever reaches zero. No matter what.

I'm sorry if I'm making too much out of this, Jeff. I am trying here.
Thank you. :)
Oh my! I read halfway down through that thread and I think I
see now what you mean. Oh my! If I have got this right, you
are making a silly little mistake. Your words just above are more
accurate than you might expect: You are making too much out
of what you were told in that thread.

Centaur told you in post #8 that "the planet’s orbital velocity
would slow down with greater distance but never reach zero."

Is that what you are referring to? If so...

Centaur was talking about the orbital speed of a body that is
not affected by outside forces. An example of an outside force
would be a rocket on the body, changing its speed. Another
example would be gravitational interactions with other bodies.

The speed of a body in orbit around another body depends on
the the total mass of the system and the distance between the
two bodies. The farther apart the bodies are, the slower the
orbital speed. But if you give one of the bodies a shove (let's
say one body is much less massive than the other, and that is
the one you shove), its velocity will change. That is, its speed
and/or its direction of motion will change. The speed it has after
the shove depends on the shove. If you give it the right shove,
you will stop it in its orbit. Then its orbital velocity will be zero,
and it will begin to fall straight toward the body it is orbiting.

Does that do it for ya? Lemme know if it doesn't. I might have
jumped to the wrong conclusion about what you meant.

-- Jeff, in Minneapolis

Fiery Phoenix
2009-Feb-14, 08:32 PM
That's exactly it, Jeff. So, normally, the orbital velocity of an object (or a planet, in this case) does not reach reach zero unless it is affected by an outside force, which is Newton's first law of motion. I get it now.

I'm so sorry for the confusion, Jeff. I take it back.

astromark
2009-Feb-14, 08:49 PM
Looking back through this I can only offer this; Yes tidal forces are gravity at work. Yes if you have a two part satellite, tethered. Slow one and yes the tether would be under tension.. but no it would be unlikely to tear a satellite apart. The strength of materials would be greater than the weak gravity force.

JohnD
2009-Feb-15, 10:29 AM
Thanks.

Astro and others,
My "pulling the satellite apart" words were merely to indicate the tension at right angles to the orbit. Let's leave tensile strength aside, especially as I want to know about tides in the sea.

Jeff, you said,
"Tides are very complex and unintuitive. We would need a more
detailed description from you to determine whether it is correct.
The differences you have encountered could merely be a matter
of terminology, rather than an incorrect description. There could
also be different but equivalent ways of correctly describing the
phenomenon. Or you might be describing the phenomenon correctly
but describing its causes incorrectly."

I agree that I am pursuing an intuitive model of tides, rather than an arithmetical one. What details of the concept do you need to tell if I am wrong or not expressing it clearly enough?

John

Jeff Root
2009-Feb-15, 02:51 PM
John,

First of all, I'm not the person to help with tides. They are beyond my
abilities. I understand several of the basic concepts, probably well
enough to explain them, but I do not have a comprehensive, deep
understanding of the subject, and I cannot do much of the math.

The situation here is a classic one. You want to know whether your
understanding is correct. I tell you that we would need to know more
about your understanding in order to determine whether it is correct,
so naturally you ask exactly what it is we need to know. That reminds
me of a scene in 2001: A Space Odyssey (either in the book or in an
unused draft, not in the film) in which HAL is asked whether there is
something he knows that the crew doesn't know, and HAL replies that
he's sorry, but in order to answer that question he would have to know
everything that the astronauts know.

I hope you see the problem there. :)

The usual way to do things is for the correct information to be given
to the student, and then the student is quizzed to determine whether
he understands it correctly.

My best advice at the moment is for you to find web pages or books
on whatever aspects of tides interest you, study those, and come back
here with whatever questions you have. If you have specific questions
right now, that's fine too. If your questions are very basic, I may be
able to answer them or to help you find answers. If they are more
advanced, then someone else may be able to help you.

If you want to start over again and attempt to explain what question
or disagreement you were having, I can try to sort it out. But I can't
tell you what more I need to know until I know more. :)

-- Jeff, in Minneapolis

hhEb09'1
2009-Feb-15, 03:22 PM
You want to know whether your
understanding is correct. I tell you that we would need to know more
about your understanding in order to determine whether it is correct,
so naturally you ask exactly what it is we need to know. JohnD went into more depth about his explanation in this thread, Why is there a tidal bulge on the far side of the Earth? (http://www.bautforum.com/bad-astronomy-books/7530-why-there-tidal-bulge-far-side-earth-5.html#post305318), a few years ago.

mugaliens
2009-Feb-15, 09:30 PM
Uh, minor nitpick, but apogee is the high point in the orbit, and perigee is the low point.

:doh: DOAH!!! :doh:

Thanks!

Fixed.

mugaliens
2009-Feb-15, 09:52 PM
Consider a stable satellite. It sends a tethered probe directly earthwards, by a small thruster on the probe, and one away from Earth. The thrusters fire radially so no orbital velocity gain or loss. They stabilise at some distance from the mother ship, perhaps by a further burn.

You could accomplish this merely by reeling them out, the upper one upward and forward of the satellite, and the lower one downward and beyond. They'd actually form an S until you stopped paying out the tether, at which point both ends would wiggle around for a long time!


Now the question. Is this a tide?

Yep! Well, to be more precise, it's the same differential in gravitational attraction that results in a tide, which is the rising and ebbing of Earth's oceans. Same thing in principle, though.


I had thought that this was the explanation for sea tides on Earth, in particular for the flood tide on BOTH SIDES of the Earth, in line with the Moon, as both orbit their centre of gravity. But when I offered this explanation, the BA and others countered it with an explanation involving differential gravity fields.

Same thing.


Someone even demonstrated that a body falling directly into another body, and so not in an orbit, would suffer tides.

Same thing. Only the differential between your toes and your head isn't enough for you to feel it (a few thousandths, if not merely a few millionths of a g). Try a close approach to a black hole, however, and you'd feel it. Too close and it'll rip you apart. Even closer and it's enough to pull apart solid matter (spaghettification (http://en.wikipedia.org/wiki/Spaghettification)).


So, the other question. Is my understanding of tides, as above, wrong?

Nope. :)

hhEb09'1
2009-Feb-16, 02:53 AM
I had thought that this was the explanation for sea tides on Earth, in particular for the flood tide on BOTH SIDES of the Earth, in line with the Moon, as both orbit their centre of gravity. But when I offered this explanation, the BA and others countered it with an explanation involving differential gravity fields.
Same thing.Almost the same thing.

A tether that points to the center of the earth through its orbit will also rotate once per orbit, and the result of that rotation is a stretching that is not a tidal force, in the classical definition. That difference is pretty much the bone of contention in those old threads.