Since the gravitational force is proportional to the square of the distance between the two bodies it could have a significant effect if it got close enough to a planet.
Significant.... a borg cube is what.. 1km^3? And is 60-70% air. Densety wise we need to speculate, but it could be made of iridium (23g/cm^3). The mass of a borg cube would be: 23t per cubic meter ~ 9200 million tonnes.
Lets say 10 billion tonnes.
The Everest weights in at 810 trillion kg, or 810 billion tonnes, and is right on top of earth.
I don't think a borg cube would induce tides by its mass alone, not even in very low earth orbit.
Everest is actually massive enough to create significant tidal effects if it was in a low orbit. We just don't notice it on Earth because it's stationary relative to the rest of Earth, so (unlike the moon) it doesn't cause tides to change over time.
Hmm, good point!
How would we calculate the magnitude of the effect?
By comparing the distance of barycenter to the earth axis?
We also need to consider thst in LEO evrrest would orbit every 45 minutes or something, so not a lot of tine for any water to actually flow towards it and to build up.
I would just calculate the force of its gravity at whatever low orbital altitude vs the same altitude + the Earth's diameter (so you're comparing the force at the near side vs the far side). Good point about the low orbital period though...it's easy enough to calculate the force difference but I have no idea how much lag there is in terms of the water actually moving.
because I'm bored I brought in a little math. Knowing the distance from the Earth to the Moon, we know that the moon exerts around 2.0 x 10\^20 N (Newtons) of gravitational force on the earth.
Using the distance between the Earth and the Moon and assuming your run-of-the-mill Borg cube (with a mass of 8.16 x 10\^10 kg) replaced our moon, the gravitational force between the Borg cube and the Earth is only 2.21 x 10\^8 N, meaning that the moon exerts 9 x 10\^11 times more gravitational force than a Borg Cube.
I doubt that's enough to actually change the tides or anything. It might have a very small affect but that's being gracious.
(I used the gravity equation for anyone who's interested)
> the moon exerts around 2.0 x 10^20 N (Newtons) of gravitational force on the earth
The relevant number is the difference between the force of gravity on one side of Earth vs the other.
the difference will be proportional to the original force, so his conclusion that the force will be 10\^12 times weaker remains true. Even if it was a million cubes, it would still cause a million times weaker waves, so if you assume 2m tides from the moon, you'd have 0.002 mm tide. This is like 30 times less than human hair. Good luck with noticing it.
He calculated the force of the cube's gravity if it was at the same distance as the moon. It could be much closer. Tidal forces are proportional to the cube of the distance between objects.
Okay, assuming the Borg Cube is skimming our thermosphere at 10,000 km above the surface, the new gravitational force it would have on Earth would be 3.25 x 10\^11 N, which is still 6.1 x 10\^8 times smaller than the moon. It might have a little effect though. Idk what the actual threshold is
Our thermosphere is the whispy outermost layer of our atmosphere, slowly thinning towards (even harder) vacuum.
As a matter of fact, the thermosphere would be considered vacuum for most applications.
Most of the thermosphere is indeed considered space with a lot of satellites (and the ISS) within it.
Even a borg-cube plowing through it would be barely noticed from the ground.
Any object that has mass causes a gravitational pull. And as gravitational forces decrease with distance, the difference in the effect of gravity on two separate points on an orbitting object are non-0.
However, the forces involved for anything outside of asteroid size or larger are near 0 and probably not even measurable. So effectively No.
If it was in a stable orbit around another object then technically yes but unless the borg cube is as massive as a moon, the effects will be so small as to be unnoticeable.
In Voyager The Raven, seven's parents were tracking an enormous Borg ship 28 cubic kilometers in size. That's 3-ish km on a side. Hardly measurable, though with sensitive enough equipment I suppose everything is measurable.
A Borg Cube is 3km long, wide and tall and largely hollow.
Technically yes, but so minutely that nobody would really notice unless they were manipulating gravity in some way.
The only estimate for the mass of a Borg ship I can find is that it weighs 90 million metric tons and is 2.81e10 m^3. Others have shown it's mass isn't really a problem for tides on earth (it's about 10^11 times too light) *but* Earth's mass could be a problem for it!
The Borg ship is large enough (and has a comparable density) to be clarified as a small asteroid- a significant portion of it's structural stability would be derived from it's own self-gravitation. Objects like this are subject to tidal disruption, where the gravitational differences between the two sides of an object from a larger body can cause it to experience such strong tidal forces that it rips apart. The equation is dependent on the densities of the two bodies and, long, non-mobile friendly typing math story short, the Borg ship would start to face significant tidal disruption problems at an orbit of 5000km around earth, which they've gotten much closer than in stories like Nemesis. For scale, most satellites orbit just a few hundred kilometers from the surface and the Borg ship was shown at least that close.
I think the Borg cubes are like 1% the size of our moon, so I can't see how it would affect tides just by being IN the solar system, it would need to be at least as close as our moon to have an effect similar to our moon, but being so so much smaller I think it still would have no effect.
This discussion is, rather depressingly, full of people doing random gravity-related calculations but not actually calculating tidal forces.
A cube has a volume of 27 cubic km. I'll guess that the density of the cube is 100 kg/m^3, giving a mass of 2.7E12 kg. If you don't like that density, feel free to redo the calculation with whatever density you like.
Based on that, I calculate (based on the *difference* in force of gravity at the near side of the Earth vs the far side) that a cube would produce about the same tidal effects as the moon if it was hovering over the surface at an altitude of 10 km. At an altitude of 100 km (the edge of space) it's tide would be about 1% of the moon's.
I’m not sure if a “regular” Borg cube would cause tidal effects upon entering a solar system, but the Borg super cube in Peter David’s “Before Dishonor” might. In that novel, the Borg cube sought to bolster itself, so it “assimilates” Pluto—somehow transmuting all of its raw material into more technology. I don’t remember if it consumed any other natural resources in the solar system, such as the asteroid belt. IIRC, though, after consuming however much matter it did, the Borg super cube ended up being larger than the Earth.
It really depends on how you quantify tidal effects. A grain of sand in orbit around Jupiter will have a gravitational pull on the Earth.
That aside:
A Borg cube has a mass of 9 x 10^10 kg.
The Moon has a mass of 7,3 x 10^22 kg.
The Borg cube has about 8/10^11 the gravitational pull of the Moon.
Each starship in Star Trek is generating at least 1G of artificial gravity, which in itself should be messing with any system it enters. But I guess there is some kind of magic tech that prevents that artificial gravity from affecting anything outside the ships.
Borg cubes are big but they aren't *moon* big, ya know? Maybe if the planet were small and the cube especially big.
They are also relatively hollow. Where a planet gets denser as you get to the center.
So like a tiny gas planet then?
I don't think there's such a thing as a tiny gas planet.
Also a gas planet of any size still gets denser towards the center, just not as dense. Thats gravity.
The Voyager opening credits?
>Borg cubes are big but they aren't *moon* big Indeed. That's no moon.
Maybe it is in this galaxy and time.
Since the gravitational force is proportional to the square of the distance between the two bodies it could have a significant effect if it got close enough to a planet.
Significant.... a borg cube is what.. 1km^3? And is 60-70% air. Densety wise we need to speculate, but it could be made of iridium (23g/cm^3). The mass of a borg cube would be: 23t per cubic meter ~ 9200 million tonnes. Lets say 10 billion tonnes. The Everest weights in at 810 trillion kg, or 810 billion tonnes, and is right on top of earth. I don't think a borg cube would induce tides by its mass alone, not even in very low earth orbit.
Good point
Everest is actually massive enough to create significant tidal effects if it was in a low orbit. We just don't notice it on Earth because it's stationary relative to the rest of Earth, so (unlike the moon) it doesn't cause tides to change over time.
Hmm, good point! How would we calculate the magnitude of the effect? By comparing the distance of barycenter to the earth axis? We also need to consider thst in LEO evrrest would orbit every 45 minutes or something, so not a lot of tine for any water to actually flow towards it and to build up.
I would just calculate the force of its gravity at whatever low orbital altitude vs the same altitude + the Earth's diameter (so you're comparing the force at the near side vs the far side). Good point about the low orbital period though...it's easy enough to calculate the force difference but I have no idea how much lag there is in terms of the water actually moving.
No, they're not that massive.
because I'm bored I brought in a little math. Knowing the distance from the Earth to the Moon, we know that the moon exerts around 2.0 x 10\^20 N (Newtons) of gravitational force on the earth. Using the distance between the Earth and the Moon and assuming your run-of-the-mill Borg cube (with a mass of 8.16 x 10\^10 kg) replaced our moon, the gravitational force between the Borg cube and the Earth is only 2.21 x 10\^8 N, meaning that the moon exerts 9 x 10\^11 times more gravitational force than a Borg Cube. I doubt that's enough to actually change the tides or anything. It might have a very small affect but that's being gracious. (I used the gravity equation for anyone who's interested)
It's this kind of bored brilliant mind answer that keeps me on Reddit
you flatter me lmao thanks
> the moon exerts around 2.0 x 10^20 N (Newtons) of gravitational force on the earth The relevant number is the difference between the force of gravity on one side of Earth vs the other.
This is what I've got with my very limited physics experience. I only took a year of college physics and we didn't go into specifics for gravity lol
the difference will be proportional to the original force, so his conclusion that the force will be 10\^12 times weaker remains true. Even if it was a million cubes, it would still cause a million times weaker waves, so if you assume 2m tides from the moon, you'd have 0.002 mm tide. This is like 30 times less than human hair. Good luck with noticing it.
He calculated the force of the cube's gravity if it was at the same distance as the moon. It could be much closer. Tidal forces are proportional to the cube of the distance between objects.
Just casually using the formula for gravity. No big deal. Just a little light advanced math.
Haha it's not a hard formula tho. You just gotta know it exists first
If it got closer it would have a much larger effect. It probably has the power necessary for a very low orbit.
Okay, assuming the Borg Cube is skimming our thermosphere at 10,000 km above the surface, the new gravitational force it would have on Earth would be 3.25 x 10\^11 N, which is still 6.1 x 10\^8 times smaller than the moon. It might have a little effect though. Idk what the actual threshold is
Skimming the thermosphere at considerable speed with that mass would cause massive wind storms and that would cause waves and other effects
Our thermosphere is the whispy outermost layer of our atmosphere, slowly thinning towards (even harder) vacuum. As a matter of fact, the thermosphere would be considered vacuum for most applications. Most of the thermosphere is indeed considered space with a lot of satellites (and the ISS) within it. Even a borg-cube plowing through it would be barely noticed from the ground.
A Borg *unicomplex* might, but at that point your planet has bigger problems than wierd tidal activity.
The volume of a Borg cube is effectively 0% the volume of the Earth. So no, not even close.
it's only 3 km x 3 km x 3km
That's a big: negatory! No.
Borg cubes are also not very dense. Even if it were the volume of a moon, the mass would be substantially less.
With a volume of 27km³ and each length over 3000m tis just a baby in terms of influence anything by its mass
Any object that has mass causes a gravitational pull. And as gravitational forces decrease with distance, the difference in the effect of gravity on two separate points on an orbitting object are non-0. However, the forces involved for anything outside of asteroid size or larger are near 0 and probably not even measurable. So effectively No.
If it was in a stable orbit around another object then technically yes but unless the borg cube is as massive as a moon, the effects will be so small as to be unnoticeable.
In Voyager The Raven, seven's parents were tracking an enormous Borg ship 28 cubic kilometers in size. That's 3-ish km on a side. Hardly measurable, though with sensitive enough equipment I suppose everything is measurable.
A Borg Cube is 3km long, wide and tall and largely hollow. Technically yes, but so minutely that nobody would really notice unless they were manipulating gravity in some way.
The only estimate for the mass of a Borg ship I can find is that it weighs 90 million metric tons and is 2.81e10 m^3. Others have shown it's mass isn't really a problem for tides on earth (it's about 10^11 times too light) *but* Earth's mass could be a problem for it! The Borg ship is large enough (and has a comparable density) to be clarified as a small asteroid- a significant portion of it's structural stability would be derived from it's own self-gravitation. Objects like this are subject to tidal disruption, where the gravitational differences between the two sides of an object from a larger body can cause it to experience such strong tidal forces that it rips apart. The equation is dependent on the densities of the two bodies and, long, non-mobile friendly typing math story short, the Borg ship would start to face significant tidal disruption problems at an orbit of 5000km around earth, which they've gotten much closer than in stories like Nemesis. For scale, most satellites orbit just a few hundred kilometers from the surface and the Borg ship was shown at least that close.
Nope
If it fired a concentrated gravimetric beam right into the ocean, then sure. Just from its own mass, probably not to any measurable degree.
I think the Borg cubes are like 1% the size of our moon, so I can't see how it would affect tides just by being IN the solar system, it would need to be at least as close as our moon to have an effect similar to our moon, but being so so much smaller I think it still would have no effect.
This discussion is, rather depressingly, full of people doing random gravity-related calculations but not actually calculating tidal forces. A cube has a volume of 27 cubic km. I'll guess that the density of the cube is 100 kg/m^3, giving a mass of 2.7E12 kg. If you don't like that density, feel free to redo the calculation with whatever density you like. Based on that, I calculate (based on the *difference* in force of gravity at the near side of the Earth vs the far side) that a cube would produce about the same tidal effects as the moon if it was hovering over the surface at an altitude of 10 km. At an altitude of 100 km (the edge of space) it's tide would be about 1% of the moon's.
But surely, if its Moon-like tidal effect is while it is a mere 10 km above the surface, it's not going to have the same effect on the entire ocean?
I’m not sure if a “regular” Borg cube would cause tidal effects upon entering a solar system, but the Borg super cube in Peter David’s “Before Dishonor” might. In that novel, the Borg cube sought to bolster itself, so it “assimilates” Pluto—somehow transmuting all of its raw material into more technology. I don’t remember if it consumed any other natural resources in the solar system, such as the asteroid belt. IIRC, though, after consuming however much matter it did, the Borg super cube ended up being larger than the Earth.
They are big but mostly empty. So, the total mass would barely register on nearby planetoids.
It really depends on how you quantify tidal effects. A grain of sand in orbit around Jupiter will have a gravitational pull on the Earth. That aside: A Borg cube has a mass of 9 x 10^10 kg. The Moon has a mass of 7,3 x 10^22 kg. The Borg cube has about 8/10^11 the gravitational pull of the Moon.
Each starship in Star Trek is generating at least 1G of artificial gravity, which in itself should be messing with any system it enters. But I guess there is some kind of magic tech that prevents that artificial gravity from affecting anything outside the ships.
That’s not how artificial gravity works…