Halc

Tidal disruption (the breakup of moons that fall below the Roche limit) is essentially Newtonian physics and has been fairly well understood since the 19th century. It has little if anything to do with gravitational waves and more with physical stress on orbiting objects. Earth for instance puts out a total energy of about 200 watts in the form of gravitational waves. It has plenty of tidal gradient to say destroy the moon if it ever gets close enough (it will be destroyed before this happens), but that 200 watts will not register on any detector we make. Yes, an orbiting GW detector will presumably be more sensitive than LIGO, but would lack the redundancy of the multiple GW detectors on Earth unless they orbit several of them. Not sure how much redundancy is needed in space where trucks driving nearby are not going to trigger false positives.

OK, it does seem to be a measurement topic, and not one of expressability or predictability Measurement of planets is hard due to the long delay between where it is and where you see it. The OP didn't seem to reference prediction Your OP seemed to have little to do with entropy. Your card shuffle example was one of a chaotic function, but the deck seems no more entropic before or after the shuffle, as opposed to if you play 52 pickup. This seem to have nothing to do with relativity theory. Choice of coordinate system was necessary even in Newton's physics. The sun's frame is an accelerating one, more complicated. I chose an inertial frame, but I didn't compute many numbers in it. Just >180 and 'faces the other way'. This is true, and you need two of them, not just one. Given distant stars not in the system, we have that reference. The OP said to ignore the gravitational influence of the rest of the galaxy, but that doesn't mean we don't have external references. Without it, picking a stable reference is possible (since rotation is absolute), but not as easy. But I chose the CoM as the reference. That's not an absolute reference, sure, but the relationships between the planets can be derived from each planet's coordinate relative to that CoM. Measuring it all is another problem since there's nowhere to be that sees where everything is at a given time since the distances are so large.

Well, you posted this in relativity, so it needs to be stated that you seem to be referencing states at times relative to some inertial frame, say the frame of the center of mass of the collection of objects, which is stable in isolation. The question seems to be how to express the states 1,2,3 Sure, in state 2 the Earth object has rotated just over 180 degrees and is facing the other way. It has also moved around the solar system just like all the other objects. I referenced the solar system center of mass, so given that reference, each of the objects has a position relative to that at each of the times 1,2,3. It's a stable point of reference, so nobody is worrying about saying where Earth is relative to Jupiter since both have moved. Measuring it is another thing, but measuring doesn't seem to be your question.

This two-year old topic started with the below comment. I've not read almost any of it, but I see a lot of word salad that seems unrelated to this original question. 1) No rigid object can instantly change its velocity without breaking, per Bell's spaceship scenario. You can accelerate it over time (finite proper acceleration), and how much time that takes depends on what clock is used to do it. There are limits. I worked out the minimum time it takes to move a 100 LY rigid train a distance of one light day, being stopped at beginning and end of trip. It takes almost 2 months and cannot be done faster without violating rigidity. It cannot be done at all without applying force to all parts of the object instead of say pulling it with an engine up front. Now regarding this latest post, forgive me if I am unaware of any context that might help make sense of any of it. How can a velocity be treated as a distance? This seems meaningless. How does a simultaneity or a nonsimultaneity have a size? Two events not simultaneous in some frame would have a time difference in time, but that difference would be a time, not a speed. You are seemingly comparing a time to a speed, which is total nonsense. How is any of the posts (in any of 2024) relevant to the topic? Is this just one of those blogs left open to keep the forum from filling up with dozens of crazy topics from one user?

No, not at all. As swansont points out, the term is used to refer to mass issues involving frames of reference. Mass of any kind is a resistance to acceleration. Coordinate mass is simply a coordinate dependent mass that you've been speaking of: Say your proper (physical) mass is 80 kg and so is your coordinate mass relative to the frame of your shoe. Relative to the frame of some muon, your mass is still 80 kg but your coordinate mass (some sites call it relativistic mass) would be say 500 kg. Notice that all we did was an abstract change of reference frames (coordinate systems) and there was no requirement for energy or acceleration. Mass is physical and frame independent: it is 80 kg in any frame. But coordinate mass is a frame dependent abstraction, ranging from 80 on up to any arbitrary value depending on the coordinate system chosen. Notice also that no mention of location was made in any of that.

Time lords live in a universe with significantly different laws of physics. Same for Star Wars/Trek. The problem is that there are two kinds of mass: proper mass (frame invariant) and coordinate mass (frame dependent). Use of the term 'mass' was sort of ambiguous until around 1950 when the term was formally assigned to mean proper mass. But using it to mean coordinate mass (as Einstein did) persists in pop sources to this day, and chatbots will likely still use it since those pop sources are the larger percentage of its training data. The article is nice, but makes a lot of errors, some of which are just in clarity, but some actually wrong. The wonderland example for instance treats a bicycle as a brick instead of a system with moving parts. The cycle wheels contract if they rotate, so the spokes would become slack and the bicycle would fail (collapse) if a plodding walking pace was approached. The article treats speed as absolute, which violates special relativity. You're moving at 0.99c right now relative to some muon. No fancy acceleration or Time Lord engine required. Do you mass more because of that? No. Hence their choice of what 'mass' means. It starts with: "nothing can move faster than [the speed of] light", which should read that it's only true relative to an inertial frame. For instance, if I shine a light to a reflector on Mars and measure the round trip, it will go faster than c. Not much, but it will, and it is due to spacetime not being flat between here and there, so inertial frames are not applicable. Reading more, I see SR, 1905 "begins with the astonishing experimental fact that c never changes". This is wrong. It is a premise, not a fact, and to date it has never been experimentally verified. It says that an object's mass increases with speed, but that's a coordinate effect, not a physical one, so only the coordinate mass thus increases, and they don't correctly say that. It later says 'ship time is flowing at less than 7/8 the normal rate' which is confusing. Ship time is at 100% per the first postulate of SR. The wording makes it sound like time dilation is a physical effect instead of a coordinate one. Again, does your clock run slow because it is moving at .99c relative to some muon? No, it doesn't. Discussion of a more general version of E=mc2 is given by other posters. The bottom left of the 2nd page says something crazy about a journey to 283 light years away taking just over a year from Earth's point of view. That's totally wrong. It might take that much proper time on the ship, but that's not an Earth observer.

The Lorentz factor is almost exactly 7. Some call the 1.4 million kg the relativistic mass but technically the mass of the ship is the frame invariant proper mass of 200,000 kg. That's just a terminology thing. The ship relativistic momentum goes up to γvm, so you can compute its kinetic energy by multiplying that by v. Also note that once you've given your ship a specific mass, one has to wonder how it accelerates without losing any of it, which is why most hypothetical scenarios avoid any mention of masses since it introduces so many problems that are not illustrative of the point of the exercise. Not sure how your fusion engine produces thrust. Most fusion reactors only produce heat, and maybe electricity from that, neither of which immediately translates to thrust. So anyway, unless you are firing this object out of an insanely long rail gun, any calculation of energy usage needs to factor in where the energy is produced and where it is going, and what is left at each point. Don't know what that phrase means. You talk about energy, but then follow it up with a specification of power (so many joules per sec). If the ship is accelerating at 1g, then it's power consumption is presumably proportional to its current mass, so integration is needed. One also has to factor in energy imparted to the reaction mass, and all that scaled by some kind of efficiency rating. The most efficient engine might be something like an ion drive, but those don't put out enough force for 1g of acceleration. 200 tons is seemingly not enough to provide life support for Joanne for all those years, let alone any left over for trivialities like propulsion. Perhaps it is really good at recycling.

Acceleration in physics is a vector change in velocity over time (and is not a scalar increase in speed, the dictionary definition). That change can only be in one direction, so if you're using lateral jets and still are maintaining 1G of proper acceleration then you're just finding an inefficient way of accelerating at some angle. Anyway, if you're getting up to .99c in the solar system frame, your circle at 1G would be at minimum far larger in diameter than the distance to AC, and you'd be using your main engines pointed sideways to maintain the turn radius, with no increase in speed. This would take at least 7 years to get up to this speed linearly and far more if you maintained a circle centered on something like Earth. It would take decades to do one 'orbit', one lap of this huge circle. Once you 'cruise' and go straight, you'd be heading away from your destination since at no point in that circle are you heading toward it. Fastest way to get there at fixed acceleration is straight out and back, but getting up to a speed higher than your ability to stop in time is just wasted effort. It would take at least 3g to get up to .99c and still stop 4 light years away

You can't orbit anything much faster than its escape velocity at 1G, so no, dropping to the sun requires one to lose energy rather than gain it, which is what you want to get to AC. OK, technically a ship always loses energy as its fuel drains, but I'm talking about the mechanical energy (potential and kinetic) of the payload. So orbit of anything is likely just a waste of time since one cannot exceed some low speed while going in circles, at least not if acceleration is confined to 1G. Suppose the sun could be condensed into a neutron star or black hole. Presuming Joanne doesn't mind a little (a lot) of tidal stress, one could orbit such a thing at super high speed near c but it would take a lot of time to drop into this orbit. And then it gains you nothing getting to AC since all that kinetic energy must be wasted over a long time just to get back to where Earth is orbiting, all the energy being expended not accelerating to AC, but rather just climbing back out of that gravitational well you were in. When back at Earth, no net speed remains. The trip is no shorter.

6 years 1 way Earth time, 3.58 years on the ship. The figures reflect more the actual distance and not just 4 light years exactly. If it was 4 light years exactly, it would be 3.46 years on the ship.

I won't accept such cookies normally, but I pasted the link in an incognito window and it let me in without challenge. It seems to be a Newtonian calculator and yields 1.18c if I put in 10k hours at 1G Really, there are very good calculators for relativistic space travel. One of the best: https://gregsspacecalculations.blogspot.com/p/blog-page.html That one presumes fixed proper acceleration, not coordinate acceleration like a Newtonian calculator would use.

That's coordinate acceleration, and 1G of that for almost a year would kill poor Joanne, and it could not continue at all for a full year. I do realize the OP did not specify explicitly, but 'comfortable' was used, so my figure is based on a comfortable 1G proper acceleration, and that takes almost 2.7 years (ship time) to get to that speed and around 6.8 years Earth time. Fixed proper acceleration can in principle be kept up indefinitely. Oh, and your link is behind a paywall, or at least a subscribe wall. I could not view it.

I'm afraid that you cannot get to that speed (relative to Earth) in only half the distance to your destination (at least not at 1G). Perhaps a target further away like Tau Ceti, which is almost exactly how far you'd get if you got to .99c and immediately started slowing. You'd be back home in 27 years 2.3 months. To Alpha Centauri and back you'd be gone only 12 Earth years, but would only reach 0.95c Tell Joanne to skip most of the luggage and have laundry facilities onboard.

Thank you for ignoring everything that matters and finding something to nitpick about. Says the guy who doesn't back his claims. I did not claim it was a quote. Throughout the small paper, each lightning strike is treated as a single event in two-dimensional spacetime. Were that not the case, the entire conclusion would be invalid. You also only showed the first page, just before he describes the relevant bits. So I had a long car ride today and used 20 minutes of mental free time to consider the case of a 100m (proper len) ship passing really close to the observer. How big would it appear when going by at 0.8c? Turns out it appears larger. The observer would measure 165 meters in the photo he takes when the center of the ship passes him, and if he condescends to compute anything more, he can deduce 100m proper length from that. The middle of the ship (moving left to right) is directly in front of the camera as specified. The front half of it appears to extend 15m to the right and the rear half 150m to the left. Mind you, this is what the photo shows, not where the ship actually is when the photo is taken. The photo cannot show where the ship actually is since the camera cannot be everywhere present along its length simultaneously. The right kind of camera can, but we're using a wide angle camera which constitutes a single point of view (as does any telescope). We presume it takes a 360 degree image of everything with sufficient resolution to see what we need to. The further away the camera is from the middle, the longer the front appears and the shorter the rear appears, approaching 30m each if the camera is infinitely distant. That's why I put my camera a light year away in my earlier example, but during the car ride I wanted to work out the limit when the ship passed arbitrarily close. So your camera was quite close, and you claim it shows a 60m ship. You need to justify that claim with a description of how that is done without positing a magic device that just somehow knows. Use mathematics. I can do that with mine, although I have not yet done so since I did it all in my head. I used an ship with length 8 (natural units) and then scaled that up to our example of 100m. From a light year away, the photo shows the name of the ship printed on the rear of the rectangular prism shaped thing. From a photo taken nearby, this name is not visible. I think the camera needs to be at least 23 meters away for the name to be in the photo, if I computed that correctly. None of this paragraph is relevant to determining the ship's proper length except to illustrate that the photo will show a distorted image, making the task not entirely trivial.

I didn't say anything of the sort. I know, but it can't be very close. Far away removes a lot of distortion. You want to keep the light travel time from all parts to you as similar as possible, and it's not if you're far closer to the middle than you are to the ends. So no need for your pilot to get it close, especially since it is inertial and not under pilot control. We have a good telescope, so distance is not an issue. OK, but Einstein relied on timing of very explicitly defined light pulse events and not on pictures taken by observers in different frames. All the measurements of lengths (proper or otherwise) were computed, not directly measured. I still don't know how one can make a direct measurement of anything that doesn't involve a computation at some point, but you seem to disqualify a measurement of the proper length of a moving object based on the fact that the measurement involves a computation. Sure, but that doesn't say how its length gets measured, coordinate or proper. The train is receding when your picture is taken. Let's assume there is a red stripe mid-ship and the camesra snaps a pic just as that red stripe is centered in the frame. If you're close enough, the train end will still be getting closer when that shot is taken, but the middle and front will both already be receding at that time. The observer can use a simple clock to determine how long it takes for the train's length to pass. Sure, that involves a computation, but it's a measurement nonetheless. Is that direct? Don't know your criteria for that. Two flashes, each leaving a mark both on the train and the platform, each pair of marks presumed to be spatially separated by negligible distance. OK, so you do accept computation as a valid form of measurement, meaning we do actually measure the proper distance of the ship in the frame where it's moving. all presuming we've measured its coordinate length, which has yet to be described. The photo we took (or a series of them if that helps) didn't convey it, it only conveys one subtend angle, and that only works if the ship is quite a distance away and we know that distance, something the telescope doesn't yield. Not saying it can't be done, just saying that you need more than a telescope which only measures angles, not distances. What, a moving ruler? Only works if we know it is at the ship, which is why I had suggested the glue. A ruler halfway between moving ship at ship speed will make the ship appear around 50m long, and will appear to be moving relative to the ship. I am assuming zero time shutter speeds. But for a nearby ship, the flight time of the light can be 10x longer from the ends as it is from the middle, leading to that massive distortion in the picture. The front will appear far shorter than the rear half of the ship. So we let it pass at a great distance where the light travel times are almost identical. That's why I suggesting it passing a light year away.

That was my suggestion. I thought about what was trying to be conveyed, and I think the ship is photographed when it isn't approaching, but rather moving perpendicular to the line of sight, hence exposing its contracted length. My apologies for getting thrown off by the description of it 'approaching' which made me think the motion was of an object getting closer at the time the photo gets taken when it is actually moving away when the photo gets taken. Yes, the ruler along side (or glued to it as I had suggested) would be a nice way to almost directly measure its proper length. The ship will need to be significantly far away (hence the need for the telescope) else the middle will be significantly closer than the two ends, effecting light travel times and making the ship not appear centered. I presume a rectangular ship. The thing will appear deformed, and the rear of the ship will be visible in the photograph taken when the middle of the ship is centered in the telescope view aimed at the point of closest approach. The front of the ship will not be visible, so from that asymmetry, one can likely deduce from a still shot the direction it is moving. Engineering problem irrelevant to this topic. Equivalently, you can just ask the ship occupants how long their ship is, even if the occupant is somebody you put there with a tape measure. Somehow that feels like a measurement in the ship frame though, a violation of what we're trying to do. Actually no. Say the ship is a light year away at closest approach, and I aim my telescope at that closest approach point in space (my frame). The time to take the shot is a year after the ship was there because that's when the ship will appear in the picture for a microsecond or so. You want to take the shot when it appears closest, not when it actually is closest. What if it's the Earth that's moving, and it's you with the telescope that needs to return back to the inertial ship that was never any proper length other than 100m? I say this because the comment above smacks of a suggestion of a preferred frame. Bell's Spaceships The Bell spaceship scenario nicely illustrates proper distance vs proper length. You have to small identical ships stationary in frame X and separated by a light day in X. They're pointed east, and the east ship has a light-day string attached behind it reaching to the other ship but not attached. At time 0 in X, both ships accelerate at a proper 1g for 1 year ship time (about 1.18 years in X) at which point they run out of fuel. In X, the two ships are at all times separated by a coordinate distance of one light day but their proper separation (not really defined until fuel runs out) becomes larger. The string is moving in X and its contracted length no longer reaches the west ship. Were it attached there, it would break, which is the point of Bell's scenario. In the inertial frame in which both ships eventually come to a halt, the string is stationary and is of coordinate length 1 light day, but the ships are now separated by 1.58 light days, and since the ships become mutually stationary when both their fuels run out, that is a proper separation. Point is, proper separation of ships went up (not contracted), but the coordinate length of the string went down in X (length contraction). The string is treated as a rigid thing and its proper length never changes.

What you wrote was not understood and seemed contradictory. I was attempting a guess as to what was meant. I apparently guessed wrong. The word 'approach'. Please be more clear as to the actual trajectory. I am now imagining a ship angling in so it is both getting closer (approaching), but heading somewhere other than toward the observer such that its path crosses the fixed orientation of the telescope. I probably have that wrong as well, but a repeat of a quote isn't much of a clarification. A photograph takes an image of a ship It is not a measurement of length. I cannot determine the size of a bird from a photo of it against a blank background. So kindly clarify how this 60m was 'observed' instead of just repeating the assertion. OK, but it can't be done when it's stationary either. Almost all measurements are indirect, if only by the fact that involve a computation, and you seem to suggest that any computation invalidates the value computed. If you think that is possible please describe how it might be done. I don't know what you consider to be a direct measurement. I would have glued a tape measure onto the ship, but that involves subtracting the number on the tape where the front of the ship is from the value at the back, and subtraction is a calculation which you disqualify as a direct measurement. OK, you've barely given enough clue to actually get me to put in search criteria that makes sense. I've never heard of stadia before. From https://en.wikipedia.org/wiki/Stadiametric_rangefinding and my bold "The stadia method is based upon the principle of similar triangles. This means that, for a triangle with a given angle, the ratio of opposite side length to adjacent side length (tangent) is constant. By using a reticle with marks of a known angular spacing, the principle of similar triangles can be used to find either the distance to objects of known size or the size of objects at a known distance. In either case, the known parameter is used, in conjunction with the angular measurement, to derive the length of the other side. " Point is, if you already know the size, you can determine the distance. If you know the distance, the size can be determined. Problem is with the ship we're looking at, we know neither. Hence my utter confusion as to how any kind of telescope can be used to determine the size of something at an unknown distance. Maybe I'm stupid and there's a way to determine these when both are unknown, but the article there didn't seem to spell that out. I also still don't know the path the ship is moving, or what the orientation of the ship is relative to either that path or relative to the line of sight of the telescope. Is the nose of the 100m approaching rocket in front like in cartoons and Hollywood movies, or is the engine-end in front like it typically is in reality, or maybe sideways? I suppose it depends if it's planning to accelerate, and which way.

Proper length of a rigid inertial object can be directly measured. If it's accelerating, it's arguably a computation no matter how it's done. Proper distance between a pair of events seems to be a computation in any frame. Notice that I didn't need to specify a frame to say that. A statement of coordinate length would need such a specification to be unambiguous. You say it's defined only in the ship's inertial frame (if accelerating, ship frame isn't even an inertial one), implying that the statement, lacking the reference, is ambiguous. But that's your assertion. The wiki site defines the proper length in other frames (said computation above, which is defined in any frame). Don't know what stadia means here, but I'm find with infinite resolution telescopes that can count spider legs from 1000 LY away. The ship is approaching, so it is always centered on the telescope, not just passing in some instant. Perhaps you mean something else like it is moving from left to right or something, which would not be 'approaching'. If you see it end-on, you cannot even tell that it has length. Telescopes don't measure length. They measure subtend angle, so an approaching ship appears to be getting bigger, but without knowing how far away it is, no size dimension can be inferred from a photograph. Maybe 'stadia' is some kind of magic that lets you do this. Anyway, if it is approaching, one can only see the front, not the depth of the ship, regardless of its orientation. No size is obvious even from subtend angle. OK, the ship has a coordinate length of 60m so it is moving in the direction of that length (and not sideways). The telescope cannot measure that, especially since it can only see one end of the thing. So you do not observe any length at all, nor a distance to the thing. I agree with this statement of coordinate length, but not that it was something observed. Not sure how the telescope can figure the velocity either unless one already knows say what color the lights are and what redshift is observed. I'll allow that since decent spectography can do that. So I'll grant that one can work out (not directly measure) its approach velocity. Not sure of the whole point of the example. We already stated the proper length, and I agree that there are ways to figure it out (with a tape measure, not a telescope) in any frame. You assert that since it might involve a computation, that it doesn't count as it having a proper length. That's your choice, but unless it has a different proper length in some other frame, the proper length is still frame invariant, which is what I'm asserting, and which the bottom line of your recent wiki quote says. Totally agree there. c is not a velocity at all, let alone a constant one. It is a scalar constant, not a vector constant. c is even a constant in alternative theories that do not posit light moving at that speed as does SR. In some alternative theories, light speed is not frame invariant.

There's a trivial formula given on the wiki site to compute the proper length of a moving object. It produces the same answer regardless of choice of frame, which is what is meant by the proper length being frame invariant. Do you dispute this? Does a ship with a 100meter proper length in a frame where it is stationary have a different proper length in a frame where it is moving forward at say 0.8c? If so, what is that different proper length? Clue: In the 2nd frame, it has a coordinate length of 60m. Yes, I agree. For instance, Earth and Alpha Centauri are separated by a frame invariant proper distance (only if you presume incorrectly that they have the same velocity, but they're reasonably close), whereas a rigid ship moving (inertially or accelerating) between the two would have a frame invariant proper length. Two spacelike separated events also have a proper distance separation, and with that specification there is no need for the mutually stationary bit since events don't have a defined coordinate velocity, being physical points in spacetime. If two events are timelike separated, then their proper separation is a proper time, not a proper distance.

The wiki article is accurate and says nothing of the kind, saying instead that "proper distance, provides an invariant measure whose value is the same for all observers." meaning it is frame independent. The rest length of an object doesn't change in a frame where it is moving. The length does, but if it's moving, that's not its rest length. The article says that the length and the proper length are the same not only in the rest frame, but in any frame where the motion is perpendicular to the length dimension being measured. I did however make a mistake with my reply to KJW A frame invariant proper distance between two spacelike-separated events is the coordinate separation between those events in any frame where they are simultaneous. The two events mentioned are still spacelike separated and have a proper separation under 2 light years, not just 'somewhat less' than the proper separation between the 2nd pair of events. But I said it was meaningless, which it wasn't. Both pairs of events were ambiguously identified, but I presumed the temporal references were relative to Earth's frame.

Quite right, and I didn't suggest otherwise. This goes for proper length as well. Length of an object or distance between two objects is very much frame dependent, but proper length and proper distance is not, which is why it was unnecessary for me to provide a frame reference when stating that our two hypothetical planets were separated by a proper distance of 2LY. The former is meaningless. Distance/length is not a measure of spatial difference of a pair of events at different times, else I could meaningfully say that my car is 200 km in length because the front is at home today and the rear of it in the next town tomorrow. Sure, there's a 200 km difference in the spatial coordinates of those two events in the frame of the ground, but it is not in any way the length of the car, proper length or otherwise. It is true that the distance between our two planets is frame dependent, and that each frame would compute that distance as a spatial separation between different events, but only if the pair of events have the same time coordinate in the chosen frame, which your description did not. Length contraction is arguably a coordinate effect, but there are examples that demonstrate it to be physical similar to how differential aging demonstrates that linear time dilation is also physical and not just a coordinate effect. This seems to not explain things since distance is not a measure of an interval. The interval between two events is frame independent, but the difference in spatial coordinates of two events is not. I think you know that, but your statement seems to imply otherwise, and you meant to say that length of some object measurements in one frame compare different pairs of events than the events compared for a length measurement in another frames.

Everything is stationary in its own frame by definition. If it isn't, it's probably because it doesn't have a defined frame. I said 'proper distance', which means a distance as measured by rulers moving at the same velocity as at least one of the objects. That works for special relativity In cosmic coordinates (not relevant to this thread), proper distance is usually measured by rulers that are stationary in the cosmological frame, that is, they have zero peculiar velocity. Peculiar velocity is meaningless in special relativity. Short answer, 2LY apart in Planet frame, and since the ship speed is 0.866 relative to that, the planets are 1 LY apart in ship frame and that's not a proper separation.

And has to make allowances for whatever frame type has been chosen. There is no one 'actual velocity' of one thing relative to another. No, the ship is moving relative to the planet, and thus the departure planet is moving away from the ship. This is using the inertial frames of the planet and the ship respectively, and the scenario is presumed to be a 2LY proper distance between planets, and the ship is ballistic (not accelerating except at beginning and end) at speed 0.866 for a dilation factor of 2. In planet inertial frame, yes. The planets are separated by a proper distance of 2LY, and 1LY is halfway in that frame. I would use 'speed', but yes. The velocity of the planets relative to the ship is equal but opposite the velocity of the ship relative to either planet. No numbers there, but yes, that's kind of a frame independent fact and nobody has to actually see it for it to be true. That is a frame dependent question, but since you said 'ship reckon', I can presume a reference to the ship's inertial frame. The ship is stationary in that frame for the whole trip, so the midpoint is the same as all the other points and is just 'here'. So at any time, the spatial distance to the midpoint of the trip is zero. This entire post has answers that are true even in Galilean relativity (except the bit about the dilation factor, which wasn't an answer to any of the questions).

The velocity near c is relative to the Earth frame. In the ship frame, the ship is stationary and Earth (and presumably some destination object) are moving fast, and yes, assuming the destination object is reasonably stationary relative to Earth, the distance between them in the ship frame is contracted. That is, on the worldlines of the two planets, the events of the respective worldlines of the two planets that are simultaneous (relative to the ship frame) with the ship's midway event, have less spatial separation in the ship frame than do those same two events in the planet frame relative to which those same two events are not simultaneous. A mouthful, but hopefully followable. The Earth observer is computing (not directly measuring) a spatial distance between two different events which happen to be simultaneous in the Earth frame. Relativity of simultaneity is probably what you're looking for. In the planet frame, the planets are always stationary, and the midway point of the ship journey is a fixed event at a fixed location in space, so it is half the proper separation of the planets, say 2 of the 4 LY separation of the planets. The ship is moving at 0.866c relative to the planets at that event, so the Lorentz factor is 2. The synced clocks on the planets read 0 at departure of the journey, as does the ship clock which makes the whole trip at that ballistic speed, so it takes 2.31 years to get to that midpoint event, and 4.62 years total. The ship clock reads 2.31 at the destination and 1.155 at that halfway event. In Earth frame, the two planet clocks read 2.31 and the ship clock reads 1.155. These three events are simultaneous in the planet frame. Relative to the ship frame, at the ships midway event, the ship is stationary and Earth has been receding for 1.155 years at 0.866c, putting Earth 1 light year behind, the event where the Earth clock reads 0.577. The approaching destination planet is 1 LY away and will reach the stationary ship in 1.155 years. It's clock reads 4.043. So the three simultaneous events in that frame are Earth@0.577, Ship@1.155, Dest@4.043. These are three different events (well, the middle one is the same) as the 3 events simultaneous in the planet frame above. Point is, different frames label different sets of events as simultaneous, which is what relativity of simultaneity is about. Those different events have different spatial separations from each other, and in the ship frame, the two simultaneous planet events are spatially separated by only 2 light years. In that same ship frame, the three events that were simultaneous in the planet frame are no longer simultaneous, and the spatial separation of the two planet events is 8 LY, more than their proper separation. But that's not a length since a length is by definition computed between simultaneous events.

The perspective of any observer is always being at rest. One is always at rest in one's own frame. So length contraction is the contraction of the length of any object moving in that inertial frame. It is an inertial frame thing, and different rules apply to different kinds of frames. To be as exact as I can, The front and back of some object each traces a worldline through spacetime. Points along these worldlines are events, and events are objective, frame independent. Different inertial frames (coordinate systems) assign different coordinate values (spatial location and time) to each event. In a frame where the object is stationary, all the events on each worldline have the same spatial coordinates but different time coordinates. The difference in the spatial coordinates of those two worldlines is the object's proper length. In a frame where the object is moving, each event is at different spatial coordinates. At any given time in that frame, the spatial coordinate of each worldline corresponding to that particular time differ by less than the proper length of the object. That's length contraction. It is a coordinate effect, but there are ways to manifest it physically. There are no external observers. That wording makes it sound like there's an objective, preferred reference frame, which relativity denies (but alternate theories do not). I don't think you meant that and I agree with the post other than that bit. About 'sees', I just has somebody ask what recession velocity is observed in a particular scenario, and it turns out that velocity isn't observed, it is computed. It is entirely frame dependent, and the same observer would 'observe' very different velocities for the same object in his own frame depending on the kind of frame chosen (cosmic frame, inertial, accelerating frame, rotating frame, etc). Each results in a different value for the same observed distant object.