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Earth is under attack, the empire has gathered thousands of battleships each has a displacement of 500,000 metric tonnes and 20 motherships with a displacement of over 15 million metric tonnes now heading towards Earth as we speak. They are groups in a tight formation around a shield generator which put up a powerful force field the size of our moon orbit and can keep out asteroids and incoming missiles. I suppose traveling at subluminal speed across large distance of space, these fleets would be attracted to each other gravitationally and hence such a tight formation is disastrous. Is there any way to overcome this problem and still maintain a tight formation around the shield generator? The emperor has restricted the budget so there can only be 1 shield generator for the entire fleet, propulsion are antimatter-matter engine and ion drive (no FTL). The plan is to overwhelm Earth defences at one go, second wave commencing at later time is the finisher but no shield generator.

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    $\begingroup$ Displacement makes sense for water ships. Since space is in a vacuum, what exactly is being displaced? $\endgroup$ – L.Dutch - Reinstate Monica Feb 27 at 9:27
  • $\begingroup$ @L.Dutch: actually nothing the empire still used that term but it is referring to the mass of the vessel. $\endgroup$ – user6760 Feb 27 at 9:32
  • $\begingroup$ I don't think it's gravity that will cause the majority of the drift. The shield will be hitting mass at speed, and that could (should) slow the shield generator down, this means that the surrounding ships will need to slow down and speed up with shield generator carrying vehicle to stay within the shield radius and not smack into the back of each other. Both propulsion methods mentioned have some form of exhaust material to control the speed of the ships, and that exhaust material will have an effect on any ships in the direction it is discharged. Handwave, clever computers, job done. $\endgroup$ – K Mo Feb 27 at 16:56
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    $\begingroup$ @L.Dutch Space isn't a perfect vacuum, so about 40 hydrogen atoms per cubic meter. As they would seem to be "submerged" in space, as opposed to floating on the surface, that would imply that the "small" 500,000 metric ton battleships would have a volume of 7.143*10^33 m^3. For perspective, about 642 of these "small" ships would have about the same volume as Betelgeuse, a red supergiant, which, if placed where our sun is, would engulf the asteroid belt, and possibly even Jupiter. This fleet might need a bigger shield. $\endgroup$ – 8bittree Feb 27 at 18:26
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    $\begingroup$ @L.Dutch The 40 hydrogen atoms per cubic meter was the direct answer to your question about what was being displaced. After that it was just exploring what that would mean about the size of these ships. Basically, in order to displace 500,000 metric tons of space, you'd need to have a volume of 7.143*10^33 cubic meters. I used Wolfram Alpha to do the math, which also gave the comparison to Betelgeuse's volume, which I thought was interesting enough to include. $\endgroup$ – 8bittree Feb 27 at 18:54
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I don't see why a tight formation would necessarily be disasterous. Gravity is an incredibly weak force, so the attraction even massive vessels feel towards each other will be minimal- small, occasional corrections should be more than enough to correct for this drift.

I imagine that compared to gravity, matching the speed and direction of each ship with respect to each other accurately will be much more significant. Even then, for a high-tech fleet this shouldn't be a problem. The ships would probably use some sort of PID control linked to the engines of each to manage the distances between the ship in the fleet, and its nearest neighbours- much like how cruise control allows a car to match the speed of a car in front of a motorway, however in 3 dimensions instead of 1.

Alternatively, if there's some central command that knows the position of every ship in the fleet, the position of each ship could be monitored and micromanaged to ensure the correct formation is kept- however from a signals intelligence point of view, this requires all ships broadcasting their position which means the signals could potentially be intercepted.

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  • $\begingroup$ Hi I was reading on how a heavy spacecraft can pull a planet killer(asteriod) gravitationally enough to derail it in it's collision course with Earth. $\endgroup$ – user6760 Feb 27 at 9:23
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    $\begingroup$ If you mean derail the asteroid's collision with earth, I imagine that would be because if the asteroid is far away enough, tiny alterations to its trajectory would cause it to miss. The asteriod has no means of correcting its orbit which is why this could potentially work. $\endgroup$ – Jack Feb 27 at 9:26
  • $\begingroup$ The ships will probably use a tight-beam transmission (e.g. laser) anyway, as they'll need to communicate without making too much radio transmissions. So any additional comms for positioning wouldn't be disastrous. But they don't need to transmit position – they'll know their position from aiming their transmissions. $\endgroup$ – Dan W Feb 27 at 16:22
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If you put those masses in the equation for gravitational force, you will see that the resulting force is negligible.

A 500,000 metric tonnes ship and a 15 million metric tonnes ship will attract each other with a force of 500 N at a distance of 1 km.

That gives $1 \cdot 10^{-6} m/s^2$ acceleration to the lightest ship.

Not zero, but surely manageable.

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  • $\begingroup$ (+1) Much less than "manageable", probably negligible. Normal activity inside even the smaller ship (say, some small forklifts for cargo management) probably would generate much more (random) variations on the ship's acceleration. $\endgroup$ – Lorenzo Donati supports Monica Feb 27 at 15:40
  • $\begingroup$ @LorenzoDonati: The center of gravity of the ship will not move when things are happening inside it. Also, the trajectory of the center of gravity will not move (no speed changes, no acceleration). If you would move the content of the frontal magazine to the rear magazine (let's say 5000 rockets of 10 tons each over a length of 200 meters), the outside of the ship will move some less than 22 meters - but the speed of the ship will not change, so over let's say a year of travel won't get any closer than those 22 meters from its companionships, as compared to distances without rocket transfer $\endgroup$ – Calin Ceteras Feb 27 at 16:21
  • $\begingroup$ @CalinCeteras I know the center of mass wouldn't move, but yep, I neglected the fact that the ships are meant to maintain formation for long time spans (years?). So you are right, that acceleration can actually cause a significant drift of the relative positions of the center of mass of the ships over several months. $\endgroup$ – Lorenzo Donati supports Monica Feb 27 at 16:36
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Assuming 3000 battleships, your entire fleet weighs 300 million + 1500 million tonnes, or 1800 million tonnes.

The moon orbits at about 400,000 km. If we assume the masses are uniformly scattered over a 200,000 km radius sphere, the "surface gravity" will be GM/r^2, where G is $6.67408(31)×10^{−11} m^3⋅kg^{−1}⋅s^{−2}$.

Plugging those values in, we get a surface gravity of 3 * 10^-15 m/s^2.

If the fleet coasts 100 light years at 1% of light speed (so a 10,000 year journey), items left "loose" at the edge of the fleet of ships will be moving at 0.000946707779 m / s. However, over the 10,000 years, they will drift 149,375.937 kilometers towards the center (assuming constant gravitation).

Ie, the 200,000 km radius would "fall" into a 50,000 km radius sphere.

In that 200,000 km radius sphere, the 3000-odd ships each have 10^13 km^3 of space, or a spacing of about 20,000 km. After they "fall" towards each other, they'll have a mere 5,000 km spacing.

While large, a 500,000 metric tonne mothership probably has long axis under a km.

What more, the amount of effort to "course correct" and spread out again is going to be trivial.

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You have, I posit, a problem greater than gravitational concerns.

It would take light about two and a half seconds to go from one end of the sphere to the other. So whatever a ship saw or sensed, would not be where it was seen or sensed. That is, ships would have to maneuver BEFORE they saw the other ship maneuver, in order to maneuver at the same time. But that would mean that the fleet would not APPEAR to be maneuvering in sync. But if the ships were so close together, NOT maneuvering in sync means the possibility of massive collisions, give the high relativistic velocities and delta-v. Not to mention time dilation.

And, of course any central coordinating computer would be getting signals perhaps a second after they were executed. Of course, they could always use GPS. Oh, wait, they can't.

SO how would any ship know where any other ship is, if they can never see or sense or communicate the change until a second after? 'That ship is here - wait, no they changed direction a second ago, so they are ... wait, they just sent us a message saying h=they were changing course ... wait, that is old news, they changed a second ago ... they are now here, but where is here? There are no coordinates in space unless you plot with reference to a stationary object - wait, there ARE no stationary objects in space except ME ....

See the problem? How do you sync movement in space at such distances?

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