Halc

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.

Yes, it's a neat coincidence, and memorized since it's a useful fact for such discussions. I imagine it means: Reachable before you die. In an inertial frame, any distance is reachable in finite proper time given enough speed, so with a speedy ship, I can get to the other side of the galaxy (where the dinosaurs were) before I die. I say dinosaurs because some of the most popular ones were around when our solar system was half a galactic orbit prior. In a universe with dark energy, the event horizon is the limit and there is no reaching that or anything beyond, regardless of the power of your ship or your life expectancy.

It's outside Jupiter's Roche limit, so we will remain a large moon and not break up into rings. As Janus points out, the tidal forces would be incredible, and the heat generated from it would more than compensate for lack of sunlight. The oceans I'm guessing would boil away as long as Earth's spin was there to generate the heat. The earthquakes and volcanoes would probably be bad enough that you wouldn't much notice what the weather was trying to add to it. No sun, but still plenty of heat to drive atmospheric convection, so yea, lots of wind but no rain. Nowhere safe to hide. We'd be pretty close to the orbit of IO. Enough for them to interfere? IO is long since tide locked and doesn't get the geothermal tide heating that a spinning Earth would.

It is a current proper distance measured along a line of constant cosmological time. Light emitted from here will never ever reach a comoving object 17 GLY away. There's no hypothetical about that. That's what an event horizon is. Communication is not possible across one. SR does not apply since it only applies to inertial (Minkowskian) frames, and the universe is not Minkowskian. Under SR and the flat spacetime under which it is relevant, yes, light will eventually get to any coordinate in any given inertial frame given enough time. But even in SR and flat spacetime, light will never catch up to a sufficiently distant object that undergoes constant proper acceleration (as does said object 17 GLY away), so in a way, SR rules are sufficient to illustrate the effect. If an object is accelerating away from us at say 1G, it only needs to be a light year away from us and any signal we send will never reach it. A star 16 GLY away (as measured in our inertial frame) need accelerate away at far less than 1G to put us beyond its event horizon. Sorry for slow reply. I don't log on here every day.

Not anything beyond what is now about 16 GLY away, which is where the event horizon currently is, even though we can see it (and they can see our past). I assume the OP was speaking hypothetically, assuming an arbitrarily powerful ship and a totally clear path. The point wasn't an engineering one. Well actually you experience nothing special, per the first postulate of special relativity. In the frame that you're experiencing, you're stationary. But yes, all the galaxies and stuff are moving fast past your and, being things in motion, are length contracted. So you're correct that in an inertial frame, there is no limit to the distance (in a given frame) that one can travel in a human lifetime. Unfortunately the universe is not described by any inertial frame, hence the event horizon limit referenced above. An infinitely powerful ship cannot reach a star 17 GLY away (proper distance along a line of constant cosmic time), even though one could see it (presuming it was a super long lived star that was visible both now and for as long as the attempt takes). But if the distant star system launched a similar ship in our direction today, the two ships could at least meet.

But the numbers were unrealistic, and you never used them anyway, except to attempt an invalid force relation. Using real numbers gives some plausibility to the scenario. Using exact numbers is probably not useful. To illustrate: Theia was (supposedly) 10 times the lunar mass, not 1.5 times. The extra mass either escaped (high speed object, mild glancing hit), in which case where did it go? or it hit more directly and was completely absorbed except for the ejecta that managed to coalesce in orbit. Anything not already in orbit has to be moving at a minimum of around 40000 km/hr relative to Earth, so that's a lower bound to any impact from an external object. Mass and speed do not determine force, and force is not directly relevant to the scenario. Force is different everywhere, and is not one value. Fluid dynamics must be invoked. Yes, it was an off-center hit. A straight on shot would have resulted in ejecta with minimum net angular momentum, and thus no moon that requires it. The offset hit would have significantly altered the spin of both bodies, resulting in an Earth will say a 10 hour (or less) day. How long did it take for the moon to get tide locked? Does the simulation answer that sort of thing? Certainly not until any second moon merged into just the one. You don't seem to contribute after this. All your recent posts seem focused on personal attacks against those trying to help. I suppose I will also be the eventual target of that.

I don't know if I'm a proponent of any particular explanation. A large glancing collision might result in a big molten planet which spins too fast for stability, giving time for all the material to mix and become somewhat one composition. I don't know the dynamics of spinning too fast, nor the possible dynamics of a slow collision, especially if two non-mutually orbiting moons are to collide. It's a little easier to envision if they're orbiting each other, but Roche limits would tear apart both bodies before they coalesced I'd think. Not unreasonable to say that happened. 8 BY is less than the time before the moon turns around and starts approaching Earth again, and peanuts to the amount of time (trillions of years) before it hits Earth. It seems the sun will swallow the pair before then, unless the sun loses sufficient mass to let Earth's orbit get sufficiently far away, and I don't think it will lose enough. Damage deeper than that, yes, but the crater is 3 km, filled with maria (lava). The cooled lava field (the dark stuff we see) is measured at 3 km at worst. Yay for them knowing this. What, do they set up seismic sensors all over the place? I don't know what all Apollo brought with them. I don't think the moon 'hit' us. If there was a collision, the moon was formed by the ejecta of that collision, but the collision was with something not-moon. If this is the case, yes, where did it come from and where did it go? Was the matter entirely absorbed by Earth such that no significant chunk retained escape velocity? What was Earth's orbit before that hit, and why is its orbit so 'undisturbed' now?? That question seems to be one ignored by proponents of the Theia idea. Yes, of course Earth had as many craters as any other object, but like Jupiter, Earth doesn't retain them long, the smaller ones leveled by weather more than heat. You can see a lot of the really large ones (Hudson bay being one), but none as big as the one creating the moon. No crater from that, and it erased all prior craters. Plenty of the current moon craters (the ones on top) are from hits from material that is not 'returning', but rather new unrelated objects. 4 billion years is plenty time to clean up the original bits that flew off that collision, but slow enough to 'come back'. Every single object that retains its craters is pretty much covered with them everywhere. It is just a show of lack of erosion, as studiot points out. If that were true, the debris wouldn't have left 'the moon' in the first place. Earth is a bigger target and larger gravity source, but so is the sun, and I imagine not much of the ejecta found its way there, so I admit that it being bigger isn't a valid argument. But returning material would move too slow to make much of a crater. Most of the big damage seems to be done by things not in Earth orbit. Moon forming within hours. It would almost have to, no? I mean, if it isn't ejected in hours, it's not going to eject. Coalescing in hours? Maybe... Depends on what percentage coalescs into one or a limited number of small objects before somebody decides "yes, that's 'the moon' now". The simulation needs to produce an Earth-moon pair of current mass and spin. They don't just pick random values (except as guesses). So to counter their proposal, you need to run a different simulation with different masses, speeds, and offset, one that produces an outcome more similar to what we see today. So it's not all unknown, since most guesses result in wrong results. If Theia was not totally absorbed (something continued on, not in Earth orbit), then yes, it should be out there. Scars on it? Heck no. Those would be gone for the reason you yourself gave. Melted away. But we don't see any object that fits the bill. I was involved in another topic about this sort of solid anomaly we have at our core. The center of Earth has asymmetrical lumps of different material. Related?

The moon was not tide locked shortly after the Theia impact, nor was it even 'the moon', something it only became after most of the pieces coalesced. There was a likely merger of two moons, but what was that like? Two moons in a similar orbit will set up a horseshoe orbit and never touch. There are several objects that are in the same solar orbit as Earth in such a horseshoe arrangement. So what else? Two moons in mutual tide locked orbit might be small/solid enough to do a soft touch, but only if they're in a tight orbit, so lots of angular momentum. There can be no sustained bulge from tides, but you still get the density asymmetry that we see today. Why is the dense side on the Earth side? Tide will tend to align a linear mass vertically, but with either end up, not necessarily the low density end. All questions, not so much answers. Yes, interesting discussion. NASA says "The Moon’s core is proportionally smaller than other terrestrial bodies' cores. The solid, iron-rich inner core is 149 miles (240 kilometers) in radius. It is surrounded by a liquid iron shell 56 miles (90 kilometers) thick. A partially molten layer with a thickness of 93 miles (150 kilometers) surrounds the iron core." https://science.nasa.gov/moon/facts/ Unsure how they measured this. Yes I am. Having something the size of Mars hit us will do that. There is an alternate (far less consensus) theory that there was no impact, and Earth just turned so fast that it spun off a big gob of material. Or even one week of simple fluid dynamics. I cannot leave a scar on the ocean after the waves have dissipated and currents have mixed any dyes and other evidence of disturbance. That they do. Many of the pieces either escaped or fell to Earth. Many of the small pieces in orbit were cleaned up by the largest lump, which eventually formed one moon, but perhaps two of them for quite a while. Assuming we wait long enough (far far more than the current age of the universe), and assuming unreasonably that the sun will not consume the Earth/moon system, the moon will eventually fall to low enough orbit to break up into rings like Saturn, and then fall as small bits to Earth. Yes, it's currently receding, but only as long as kinetic energy exists to fuel that.

Yes, the biggest evidence is that the surface has similar composition to Earth, which is true of no other planet's moon. The larger impacts impart enough energy to melt vast volumes of rock, and this liquid spreads out and covers significant area. The largest impacts might have created a crater perhaps 3 km deep, but the lingering energy from the event also continues to trigger volcanic activity for some time, and this adds to the maria, and also originates from much deeper than just a few km. I've not seen a consistent explanation for this, but the best explanation seems to be the thickness of the crust. Where the crust is thin and dense, large impacts can melt rocks instead of just eject lighter thick material sort of like a child jumping into a ball pit. Other way. Thicker (and less dense) on the far side, sort of like it's the continent over there, and the near side having thin, more dense crust like our ocean floors. The near side is the dense side, which is why it faces Earth. They're the stuff on the far side actually. The small merged bits are the low density stuff. Yes, the theory is that there were probably two primary moons for a while, both formed from the same event, which eventually did a slow motion collision/merge, with the smaller less dense moon forming the thick crust on what is now the far side. It is unclear if the moon becoming tide locked occurred before or after (or because of) this slow merge. Yes, there would have been shrapnel everywhere, which eventually collected into a limited number of moons whose gravity cleaned up all the little pieces. Much of it of course achieved escape velocity and didn't ever come home. Theia itself? One would think that it would still be around if it survived, so I think the material merged with the pre-Earth to become one planet-moons system, the material of which was composed of matter from both objects. The pre-impact planet was not Earth particularly. It had a different mass, spin, and orbited somewhere other than where Earth is now. The post-imact Earth orbited similar to where it is now, but more eccentric, with a spin of perhaps 10 hours, similar to Jupiter. I don't think it is meaningful to refer to the 'side which struck Earth', or for that matter, refer to the place on Earth where the Theia impact took place. Both bodies were reduced to molten balls, a bunch of smaller molten balls, and a lot of ejecta. A liquid ball does not have a meaningful location of where something occurred on it. There's not going to be a crater or other scar, or a place where cooling takes place faster. Not if you get it hot enough.

Better worded, thanks. The wiki site explains the pressure, due to electrons not being classical particles: "The pressure exerted by the electrons is related to their kinetic energy. The degeneracy pressure is most prominent at low temperatures: If electrons were classical particles, the movement of the electrons would cease at absolute zero and the pressure of the electron gas would vanish. However, since electrons are quantum mechanical particles that obey the Pauli exclusion principle, no two electrons can occupy the same state, and it is not possible for all the electrons to have zero kinetic energy. Instead, the confinement makes the allowed energy levels quantized, and the electrons fill them from the bottom upwards. If many electrons are confined to a small volume, on average the electrons have a large kinetic energy, and a large pressure is exerted." Yes, the degeneracy pressure is what supports the stars, and the effect is related to the Pauli exclusion principle. My point still stands: Such large dense things squashed together by gravity are arguably a single nucleus of some unfathomable element. It certainly isn't a collection of small separate atoms that are just really close to each other. For one, the one large atom has far higher neutron-to-proton ratio than any known element. It's a low-energy solution to the otherwise problem of where to go with all the extra electrons forming the degeneracy pressure.

Arguably, any neutron star is a giant nucleus, continuously changing the number of protons contained, but on a percentage basis, not by much. It is the Pauli exclusion principle which supports their stability, not so much any particular 'balance of forces'. Somehow I think this reply is not what the OP was looking for.

From what I've read on such noted reluctance to mergers of large masses, it seems that the energy/angular-momentum decay needed to merge is mostly dissipated by gravitational drag, the flinging of nearby matter away at high speeds, as is typical of 3 body dynamics when one of the bodies masses much less than the other two. Eventually, the local area is cleaned out and there's nothing left to fling, leaving a fairly stable 2 body orbit, decaying at least by gravitational waves, but that is trivial at that distance. Dark matter is just more [small] mass, just like regular matter. If it gets close it gets flung away and is cleared out like anything else. I don't see how dark matter is therefore any sort of solution to the effect. I don't know if 1 parsec is typical, or can be derived from any math. It would seem to be a function of the masses of the respective black holes and of the density of other matter near them. Clearly black holes do merge since our own galaxy is a collection of dozens of smaller galaxies, many of which contributed their own SMBH. We have but the one, so they've merged at some point. Some of then recent swallowed galaxies may have central masses that are en-route to the center, but none orbits particularly nearby at this time. Sgr-A has plenty of stars orbiting within that 1 parsec radius, so there is available mass to be ejected during any upcoming merger. Notably, in perhaps 10-20 billion years, Sgr-A will be merging with something 10 times its own mass. I wonder how far away you need to be from that for it not to be dangerous.

Using a dish would focus on signals only from a very specific direction, and would not amplify signals from elsewhere. Using a straight solar array would receive signals from all directions, great if you're just after electricity, but awful if you're trying to filter the noise away from the signal.

That seems completely inappropriate. I don't get the bike wheel thing, whose perimeter was never at the axle. There is no coordinate system that ties events in our universe to whatever constitutes a point in another. There's not even units for that. In our universe, the big bang, in an idealized naïve model, might be the event (0). Just one coordinate. No space coordinate because as MigL says, it's everywhere. But beyond that it doesn't work. For instance, there is not coordinate system that gives the absolute location of you. To do that (10 meters west of my mailbox), one must first identify an origin for your coordinate system (the mailbox). This is fine and relative, but it isn't an absolute coordinate because the location of the mailbox hasn't been specified. You can't use the big bang reference as your spatial reference because it lacks a location in space. Not true of any other event, but true of that one. It isn't like Utah where all you need to do to get a letter somewhere is give two coordinates. Three in Hong Kong.

Sorry, this makes no sense. Likewise there are no straight trajectories - every time it will be a moving object aiming to meet a moving object where their gravity-curved trajectories cross. That comment had nothing to do with straight trajectories. It has to do with preventative action against an object whose orbit crosses the orbit of Earth. Of course orbits are curved. Such an object will intersect Earth given enough time. It is officially dangerous. Point is, if you deflect the object while the object is at Earth's orbit, it will return to that location regardless of the new trajectory you send it (unless you impart escape velocity from the sun to it). Orbits are regular visitations to the same locations over and over. So the idea is to deflect the dangerous object when it is well away from Earth order, deflecting into a new orbit that does not intersect Earth's orbit. My comment above stated that doing it at the wrong time accomplishes nothing.