I'm curious what factors a terrestrial planet impacting the Earth would require for the impact to cause negligible damage to the larger planet, while destroying the Earth.
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2$\begingroup$ Do you mean the tidal force of the super Earth? ;D $\endgroup$– user6760Mar 11, 2020 at 6:49
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12$\begingroup$ I think this depends on how you define "negligible". The only real possible definition would be "localized", since something the size of Earth impacting something else is going to cause damage on a "planetary" scale no matter what. To put this in perspective, if you smashed Earth into Ringworld (and assuming you don't break the ring), that's still likely to be quite noticeable... and this is a structure with a radius of (roughly) 1 AU. Per subrunner's answer, planets simply don't exist on such a scale. $\endgroup$– MatthewMar 11, 2020 at 14:55
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6$\begingroup$ For any reasonable definition of terrestrial planet, both bodies will be converted into molten blobs, with a lot of miscellaneous debris ("spatter") flying around. I doubt whether anyone would regard this as "negligible damage". See e.g. en.wikipedia.org/wiki/Giant-impact_hypothesis $\endgroup$– jamesqfMar 11, 2020 at 17:46
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2$\begingroup$ @Sean: If it's really, really big (or even just really big), it's not a terrestrial planet. I suspect that, as Matthew suggests, a large enough body of roughly terrestrial composition can't exist, because it would collapse, either into a ball of degenerate matter (similar to a neutron star), or into a black hole. Haven't done the math, though :-) $\endgroup$– jamesqfMar 12, 2020 at 3:11
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4$\begingroup$ @jamesqf I'm pretty sure a super-earth that's big enough will just wind up becoming a gas giant instead, once the escape velocity is high enough for its atmosphere to hold onto hydrogen. $\endgroup$– nick012000Mar 12, 2020 at 5:13
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6 Answers
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That really depends on what you consider 'negligible'. Is it 'Sterilization of all life, but planet is still there in one piece in the same orbit'? Or is it 'Everyone in the direct impact zone gets squashed and the rest of the planet suffers earthquakes, but after a decades-long volcanic winter we only have a mass extinction event'?
The second case might be a bit hard to achieve. Even if you use some kind of super-tractor-beam-technology to gently lower Earth onto the surface of the other planet (escape velocity or less), the two planet masses will meld together. This 'melding together' means Earth breaks apart, penetrates the comparatively thin crust of the other planet and then merges with the magma of the other planet to create a combined 'super-planet'.
Super-planet Calculations
I am going to ignore following energies:
- all the deformation energy that gets released when earth and the other planet get smushed into one piece, heating up the merging planets
- The impact energy (assume the tractor beam lowers earth onto the planet surface with next to 0 relative velocity)
Even ignoring those two things, it is not really survivable. Some Math:
Volume
Assume that $V_t$ is the total volume of the original planet volume $V_p$ plus the earth volume $V_E$:
$$ V_t = V_p + V_E $$
Radius
We assume that the combined planet will be a sphere, just like the original two planets were. With that, we can use the formula for calculating the volume of a sphere $V = \frac{4}{3}\pi r^3$:
$$ \begin{align} \frac{4}{3}\pi r_t^3 ={}& \frac{4}{3}\pi r_p^3 + \frac{4}{3}\pi r_E^3\\ r_t^3 ={}& r_p^3 + r_E^3\\ r_t ={}& (r_p^3 + r_E^3)^\frac{1}{3} \end{align} $$
Circumference
The circumference of a sphere at the largest point is the circumference of a circle with the radius of the sphere $C=2 \pi r$. The total circumference will thus be
$$ \begin{align} C_t = {}& 2 \pi r_t \\ = {}& 2 \pi (r_p^3 + r_E^3)^\frac{1}{3} \end{align} $$
Ten Times Radius - Unsurvivable!
Assume that the other planet has a radius 10 times as big as earth ($r_p=10r_E$). That means it's about the size of Jupiter and its volume is 1000x bigger than earth. For the circumference after combining that means:
$$ \begin{align} C_t = {}& 2\pi ((10r_E)^3 + r_E^3)^\frac{1}{3}\\ = {}& 2\pi (1001 r_E^3)^\frac{1}{3}\\ = {}& 2\pi r_E * 10.00333222\\ = {}& 2\pi (r_p + 0.00333222r_E)\\ = {}& 2\pi r_p + 2\pi * 0.00333222r_E\\ = {}& C_p + 0.021r_E\\ = {}& C_p + 0.021 * 6370km\\ = {}& C_p + 133km\\ \end{align} $$
What does that mean? That the combined planet will have a circumference that is 133km larger than the original planet - meaning that at the very least the tectonic plates will be ripped apart to somehow accomodate 133km more space.
That? Can't be healthy. Or survivable. (If the life forms haven't been incinerated before simply due to the deformation energy heating everything up)
Hundred Times Radius - a habitable Planet?
$$ C_t = C_p + 1.334km $$
Adding a whole kilometer to the circumference - that doesn't sound too bad or unsurvivable.
On the other hand - your planet has 100x the radius of earth - meaning a volume of 1 million times that of earth.
Even assuming the other planet is 'only' a comparatively light gas giant like Jupiter (still has 10x the radius of Jupiter, meaning it's 1000x bigger than Jupiter!), it will still have more gravity than Jupiter's 2.5g. And considering that Jupiter is just a bit too small to become its own sun, your planet would probably have achieved fusion!
If it's not made from fusible materials (similar composition to earth), it will be a lot heavier than 1000xJupiter - meaning the gravity on the surface should be in excess of 50g (too lazy to do the gravity calculations but I'd say it's a good guesstimate). I dare you to find a life form that can survive something like that...
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15$\begingroup$ "I dare you to find a life form that can survive something like that..." Next question: "How could a lifeform survive 50g?" $\endgroup$ Mar 11, 2020 at 15:20
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6$\begingroup$ Did a google, many bacteria can survive far more than 50g, some 400.000g. $\endgroup$– MarkMar 11, 2020 at 16:03
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7$\begingroup$ Earthquakes cause massive damage from continental plates moving a few feet. Suddenly moving over a kilometer would be a massive change. $\endgroup$– SkylerMar 11, 2020 at 16:47
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6$\begingroup$ I like it tha tthis argument even applies if there is no impact to begin with, but, say, the Daleks teleport Earth and make it appear out of the blue in a just touching position at relative rest as harmless as possible (as if they would) $\endgroup$ Mar 11, 2020 at 19:41
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2$\begingroup$ If it's made of materials that are not fusible, then it may collapse into something much more compact because of such high gravity. A body that is 10x the diameter of Jupiter is the same size as the sun. If it is made of something more dense than helium, then it likely exceeds the minimum mass to create a black hole. So the 50g surface gravity is the least of your problems, as the core will have enough pressure to collapse until it either starts fusing something, or creates neutronium and eventually a black hole. $\endgroup$ Mar 11, 2020 at 23:53
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If the two planets are not on a direct collision course, but they pass very close to each other, the bigger one might destroy the smaller one with its gravitational attraction. I mean that the gravitational force is different on different parts of the planet depending on the distance from the center of mass of the bigger body, this difference is often enough to break into pieces an orbiting body. After that the bigger planet will have to withstand a shower of very big fragments, but not a full impact.
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As far as we know, there was an impact event between the Earth (12700 km diameter) and a 10-15 km size object about 66 million years ago. This caused non-negligible damage, wiped out 75% of the species (non-avian dinosaurs are the most well known).
So an impact with a meteorite with merely 0.1% of diameter (0.0000001% of volume or mass, if we assume roughly similar density) changed a lot. If you replace that meteorite with Earth and Earth with "Planet XXXL" and assume a same proportion of size causes a similar (non-negligible) damage, you get that "Planet XXXL" has a diameter of 12700000 kilometers, or about 10 times the Sun (1390000). Your planet is a star, and even that would get damaged.
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2$\begingroup$ ...or about 0.1 AU, which is in the same neighborhood as my prior comment that Ringworld would notice such an impact. (Ringworld is about 1 AU in diameter, but, well, a ring; a 1 AU ring and a 0.1 AU sphere seem roughly comparable.) I suspect this is the answer; anything short of a 1 AU Dyson Sphere is going to notice, and that assumes your Dyson Sphere isn't punctured by the impact (because it'll notice that). $\endgroup$– MatthewMar 11, 2020 at 20:37
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Assuming that the impact will happen at velocity $v$ of one planet with respect to the other, the energy of the impact will be $1/2m_p v^2$, where $m_p$ is the mass of the planet.
Depending on your definition of negligible, you can get a ballpark figure on the mass of the planet.
If a 1% is negligible, you "just" need a planet 100 times more massive than Earth to achieve it: the planet will get from the impact 1% of the energy it is giving to Earth.
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1$\begingroup$ Ummm... what is v then? And why not considering the mass/KE of the impacted instead of that of the impactor? (my point: if you are gonna go all sciency with formulae, how about going it full way and explain what system of reference you use and why that system is appropriate. You know, head on collision and catch-it-from-behind may be a bit different). $\endgroup$ Mar 11, 2020 at 6:47
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3$\begingroup$ Velocity is always escape velicity - tens of kilometer per second for super-earth $\endgroup$– ksbesMar 11, 2020 at 9:18
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3$\begingroup$ @AdrianColomitchi: It’s always at least escape velocity, if not higher. And impact from behind vs impact from ahead aren’t actually that different in the end unless your impactor is already moving at well above escape velocity, reason being that if your impactor is in a tail chase it has more time to accelerate than if its head on. Human intuition about velocities gets a bit squiffy at interplanetary scales. $\endgroup$ Mar 11, 2020 at 9:56
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$\begingroup$ @ksbes velocity of which body, the impactor or the impacted? Velocity relative to what reference system (where is your origin of spatial coordinates): the impactor, the impacted, the star of the planet system, the baricenter of the impactor/impacted? $\endgroup$ Mar 11, 2020 at 10:09
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$\begingroup$ @AdrianColomitchi Escape velocity is always relative to whatever body it is you're escaping. The planets will collide at a relative velocity of at least the escape velocity of the super-earth. $\endgroup$ Mar 11, 2020 at 15:33
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The incoming planet has giant rings that serve as a shock absorber.
https://earthsky.org/space/huge-distant-planet-has-rings-200-times-bigger-than-saturns
The colossal rings and many moonlets around this planet act as a giant shock absorber. The incoming earth is battered by each ring in turn as Earth draws closer to the planet. These impacts slow the earth, robbing it of its kinetic energy. By the time Earth has traversed the entirety of the ring system, it is barely moving. It will gently kiss up against this other planet and they will form a shared atmosphere binary planet.
The planet does not care about the loss of its rings. Earth, on the other hand, has converted most of its relative momentum into heat via the impacts of all this ring material, and it has become quite warm.
hmm... sharing atmospheres with a ball of magma might warm things up for your other planet. But it was chilly there before, so all good.
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3$\begingroup$ Interesting idea, but I'd imagine the damage would still be significant even if you put both planets touching one another with zero kinetic energy and then let them merge into one. The binary planet idea is cool, but it doesn't cover the OP's question of planetary impact. $\endgroup$ Mar 11, 2020 at 15:28
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4$\begingroup$ There's no such thing as "planets gently colliding in space with no kinetic energy", because planets attract each other gravitationally. In a theoretical ideal situation where the rings somehow managed to absorb all of the movement of the incoming Earth...you're still dropping the Earth from orbit. That's going to leave a mark. $\endgroup$ Mar 11, 2020 at 20:10
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1$\begingroup$ @IndigoFenix - yes, losing momentum to the rings would move Earth to a different orbit. Having another celestial body close enough to collide probably would alter the orbit too, even if they did not collide. No hard science tag on this one. Best not to think too much. $\endgroup$– WillkMar 11, 2020 at 20:12
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1$\begingroup$ Even if you somehow manage to pull off the colliding gently trick, those rings are moving at orbital velocity, and will slam into the Earth, causing considerable damage. Consider what one little comet (Shoemaker-Levy) did to Jupiter: en.wikipedia.org/wiki/Comet_Shoemaker%E2%80%93Levy_9 $\endgroup$– jamesqfMar 12, 2020 at 3:18
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As the other answers already explained, your idea of planets colliding with negligible damage to one is pretty much impossible, so we'll need to think out of the box.
If your population lived on a large dyson sphere - or maybe an interconnected dyson swarm - surrounding a large star, the earth-like planet colliding with it could crash through the surface and fall into the star in the center. Sure, you'd lose an earth-sized chunk of the surface and probably quite a bit more, not to mention the solar flares from the earth-star-collision, but a large enough dyson sphere or dyson swarm could survive the collision with some of the population alive if it was thin and brittle enough to take up little energy from the collision. You could think of it like a needle piercing an egg shell. You lose some of the shell, but most of it stays in one piece.
It might cause the dyson object to become unstable and tumble into the sun after some time, though if it's big enough, it could take millennia during which scientists have time to find a solution.
I do not have the knowledge to calculate how big and thin it would have to be, I don't even know if it would be viable at all, but it's the only possible survivable planet-to-"planet"-collision I can at least imagine.
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$\begingroup$ I don't want to put it here as a separate aswer. Planets can't collide with "negligible damage" because of defenition of the planet. Gravitation forces of the planet are so strong (by definition!) that it can't deviate from spherical form more then 1/100 - 1/1000 (i.e. highest mountain or deepest depresion can't be more than 6-60 km on Earth). It means that colliding planets will form one planet or couple of new planets and this process could not be called "negligible" by any scale. $\endgroup$– ksbesMar 12, 2020 at 11:57