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The question stems from watching a scene from ID2 and seeing the massive alien mothership landing/crashing on the North Atlantic.

What mass would a starship, or any other artificial object constructed in low earth orbit, need to be to start to create noticeable gravitational disturbances for the planet or spatial object it is orbiting?

By 'noticeable', I mean something like higher tides that can clearly be linked to the object in orbit or a slight shift in the orbit of the planet. Not something that only some precise instrument can measure and that scientists can determine through mathematical formulae. Something that would have the leaders of the planet think that it is time to get the construct away.

Is there a general rule, like 'at 10% of the mass of the planet', or does it vary depending upon what you are orbiting? Ie '10% the mass of a planet but 25% the mass of a G-type star'?

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    $\begingroup$ What do you mean by "gravitational disturbances"? $\endgroup$ – RonJohn Oct 15 '18 at 13:38
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    $\begingroup$ What's a "noticeable gravitational disturbance"? Because if you're talking about what we can measure that's tiny but if you want to feel it or to see an effect on the tides or something then you need much larger masses. $\endgroup$ – Ash Oct 15 '18 at 13:39
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    $\begingroup$ Can you define your "noticeable"? Submarines can be noticed using their gravitational disturbances, for example. Could be since 1989, so now probably much smaller items can. $\endgroup$ – Mołot Oct 15 '18 at 13:44
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    $\begingroup$ Added a precision. $\endgroup$ – Sava Oct 15 '18 at 13:47
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    $\begingroup$ @Mołot The very last line of the article you cited 'The concept of detecting submarines by means of detecting gravitational anomalies they produce, should be abandoned.' $\endgroup$ – Justin Thyme Oct 15 '18 at 14:51
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Assuming an object in lunar orbit 10% of the mass of the moon, or 7.342×1021kg, will cause a 5% variance of total tidal amplitude, a difference of some 55mm overall. That will definitely be broadly noticeable, to anyone looking at tide gauges at least, rather than only being noted by scientists who study those effects in detail. For an object in LEO, where we are more likely to be building, for ease of access, that number will be considerable smaller.

Newton's Law of Universal Gravitation states that:

F=G(m1m2)/r2

We can rearrange to solve for m2 at a given amount of gravitational influence, F. F for an object 10% the size of the moon, in lunar orbit, is 1.9x1019N so for an object only 2042km away (in Low Earth Orbit) we get a mass of 2x1017kg or a little more than the mass of Saturn's moon Prometheus.

Apparently the above is nonsense, thanks AlexP, sorry it took a while to get back to this but the formula I should have been using is Tidal Force and not general gravitation. So using that new formula we get a rather different, and more correct outcome:

Tidal effect of an object in lunar orbit of 10% the mass of the moon is actually only 6.6x1017N so an object in Low Earth Orbit would need to be 1.1x1015kg, approximately equivalent to 1.4x1011m3 of construction steel.

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  • $\begingroup$ I was in the middle of of a similar answer. You failed to consider Roche's limit. $\endgroup$ – Gary Walker Oct 15 '18 at 14:44
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    $\begingroup$ @GaryWalker Nope I just assumed that, given it's an artificial construct, I couldn't accurately apply said limit and ignored it. I understood that most, if not all, artificial satellites orbit within the normal Roche limit for objects of their mass. $\endgroup$ – Ash Oct 15 '18 at 14:45
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    $\begingroup$ @Mołot It's not a complex question, anything else I might add would, in my estimation, be waffle and thus irrelevant. $\endgroup$ – Ash Oct 15 '18 at 14:59
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    $\begingroup$ @PSquall - I guess my reason for raising the issue of Roche's limit for an artifical satellite is that is would be such a large artificial satellite. Roche's is an approximation for gravity-only binding, which is never actually true. I did not expect an exact calculation (complex and too many undefined conditions) just that it could be an important consideration at the scale of the satellite. $\endgroup$ – Gary Walker Oct 15 '18 at 16:13
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    $\begingroup$ No it isn't. The tidal force is the difference in gravitational force between your head and your legs, for example. In a first order approximation the tidal force is inversely proportional with the cube of the distance. $\endgroup$ – AlexP Oct 15 '18 at 16:31
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I suspect the answer is somewhat different from Ash's, in part because the required effects are still not well-defined.

First, what exactly is "low earth orbit"? A common definition is low enough to produce more than 11.25 orbits per day, which works out to 1269 miles altitude for a circular orbit. Let's call it 1000 miles or less just for convenience.

It's well-known that for a uniform spherical body of uniform density the surface gravity is proportional to the radius. An orbiting body (presumably spherical) at 1000 miles altitude cannot have a radius greater than 1000 miles, and if it has the same density as earth, its surface gravity will be just about 1/4 g. Then the effective gravity directly beneath it will be 0.75 g. This would noticeable, alright, but I'm not entirely sure how objectionable it would be. It would also not have a simple effect on tides. On the one hand, tidal patterns are complex and non-intuitive, as the way the water sloshes around the planet is profoundly affected by the coastal and sea-bed conformations. On the other, an LEO satellite moves fast, by definition at least 12 times faster than the tidal cycle. I'm simply not up to calculating the effect on the tides, but it wouldn't be simple.

Nor can the satellite have a simple effect on the atmosphere. The air underneath it will tend to expand upwards, but since that point is moving at about Mach 3 there won't be much in the way of a "dragging" phenomenon.

Also note that making the satellite smaller really reduces effects. Cutting the diameter by a factor of two, to 1000 miles, reduces the mass by a factor of 8, and the gravity differential as it passes overhead drops to 3%.

Of course, at these scales Roche's Limit does rear its ugly head. While it's important to realize that structural materials such as steel are far better in tension than rock is (and it's tension rather than compression that counts here), the scale of the forces involved probably makes this a minor factor. But all of that is (or can be) pretty much moot. If the satellite is built to the proper shape, and its spin adjusted to make it effectively tidal-locked from the git-go, it will already be in the minimum-energy conformation and will not be affected.

Actually, I suspect that the most important immediate effect that would trigger a response is its effect on other satellites. Even a fairly small object would grossly distort all orbital dynamics for any other LEO satellites and would make such orbits unstable. This would probably extend to destabilizing geosync satellites as well, so we'd lose both communication satellites and things like GPS.

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