I mean if every variable aligns with any possible edge case that can bleed off velocity including three body interactions with the moon. Is there ever a situation where some substantial (car++) or enormous (skyscraper+++) size rock lands on the surface without explosive energy? Align stars, consult math mediums, play some ZZ Top, piss off Bary the narcissist, or conjure a primordial black hole, just land me a big rock in my yard Science Santa. I want an m-type for Maymass, but any type will do if you can land it.
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KSP player here. So, you know, ignore me.
But let’s consider how you’d rendezvous two objects. You’d want your asteroid to have an orbit around the Sun that is very nearly the same orbit as Earth’s. A perigee that just kisses the Earth’s orbital ellipse and an apogee that’s slightly further from the sun. You’d want the asteroid to approach its perigee at the same time as Earth approaches that same point in space. Then they’d have very close to 0 relative velocity, with the asteroid moving slightly faster around the Sun than the Earth. So you just bleed off some of the asteroid’s velocity through whatever magical explanation you want… such that your asteroid has 0 relative velocity with Earth, giving it the exact same orbit as Earth. I.e. from Earth’s perspective it’s just floating there motionless in space.
Problem is that this only works for a rendezvous between two very light objects with very small gravitational effects between them. The Earth is massive enough that the effects from Earth’s gravitation would overtake the Sun’s as the asteroid approaches Earth. Then, yeah, the asteroid becomes a falling rock with a lot of energy so I don’t think any of this works.
https://thatjoescott.com/2024/07/01/how-the-fifth-element-really-ended-according-to-science Joe Scott did an interesting video on the ending of “The Fifth Element”, and what would happen if you magically parked a large object in low earth orbit. You don’t get slower than stopped, and the results are still not good.
No, it’s not possible to soft-land a large falling object, even with extraordinary luck.
Our intuition serves us well in everyday situations. I think you’ll agree that, if you have a big rock 100 meters in the air, it’s going to hit the ground hard no matter how you drop or throw it. The bigger the rock, the more we can ignore things like air friction.
The same situation applies to objects in space. If it hits the Earth, it’s been falling the whole way from space. The minimum impact velocity is equal to the “height” of Earth’s gravity well, about 11.2 km/s.
The trajectories of celestial bodies are time-reversible (neglecting things like atmospheric drag and tidal forces). So any encounter would be physically valid if you ran it in reverse—which means anything approaching or leaving the earth would need to be moving at escape velocity (25,000 mph) or faster.
No neglecting allowed, but cool trick. Break a contact binary at an ideal moment/composition/spin.
I’m not a physicist by any means, but I would hypothesize that the slowest way would be to somehow get the object into a slowly decaying, nearly-geostationary orbit around Earth, closer than the Lagrange point between Earth and the Moon. Eventually its orbit will decay enough that it will just fall into our atmosphere somewhat “straight down”, making it a matter of calculating the object’s terminal velocity.
I’ll probably be wrong about many things here, but it’ll be interesting to learn when someone corrects me.
For anything that’s captured in an Earth-centric orbit, it’s never going to end up falling “straight down”. Closed Earth orbits barely decay at all, except from atmospheric drag. For anything that started from a closed orbit, the last few years are almost always a circle or ellipse that barely brushes against the upper edges of the atmosphere.
For such orbits, drag is mostly applied near the lowest point on the orbit (perigee), and drag forces applied there will mostly reduce the height at the other end of the ellipse (apogee). Even if you start from a highly elliptical orbit, this means the decaying orbit becomes more and more circular. Eventually, all parts of the orbit are inside the atmosphere and the loss in altitude starts getting faster and faster, but the velocity is still mostly horizontal.
The good news is that the concept of specific orbital energy applies to any initial orbit. For objects coming from outside Earth’s sphere of influence, terminal velocity remains a good lower bound.
Aren’t you assuming a solar system object though? Forget probabilities and statistics; what about an interstellar object with ideal keyholes to cancel out velocities coming from any vector.
I can’t get the picture out of my head of a plane stalling a few feet off the ground and coming to rest, or really the countless similar odd scenarios with a coin rolling around a large vortex toy/simulator that is used for grade school children to help illustrate physics and gravity. There are times when the coin loses all of its velocity and stops suddenly. Or better yet, if you’ve ever played golf and had a ball round the cup and get tossed back out right on the edge of the cup.
You can have any trope you’d like. Indestructible, natural ablatives, astroid made of Avi Loeb mystery meat stellar sail, ideal shapes, anything, heck detach pieces from a spinning planet killer moments before primary impact destroys everything.
I think the problem that you’re going to imagine a good analogy for this is that orbital dynamics works in sort of (but not really) an unintuitive way.
An object in an elliptical orbit around earth is moving slowest at its furthest point from the earth. Like a thrown ball that slows when it reaches the top of its trajectory. That object is moving fastest at the point that it’s closest to earth.
So you have this dynamic where if you decelerate it changes your orbit such that you’re increasing the speed you’ll be moving on opposite point of your orbit. E.g. if you decelerate at your slowest (furthest) point, it brings your closest approach point closer to earth and you’ll be moving even faster when you get there.
You can decelerate at your closest approach point but eventually it brings the opposite end of your orbit closer to earth than you are, and then you’ll fall and of course speed up again. There’s no real way around this. You’re going to be moving fast when you approach earth unless you’re doing a lot of very active deceleration.
I totally get that part of orbitals. I guess I am just bad at explaining the magnitude of abstraction I am asking here.
Like I am not at all asking about simple, average, or conventional objects and orbits.
So let’s say the Sol system is a large valley of a gravity well. That means Earth is perched somewhere up the side of that valley. If I’m an interstellar object, I’m likely to be captured into an orbit but my vector is not as constrained to orbital norms. The question is about edge cases so all possible scenarios are relevant and there are no statistical simplifications.
As I enter the Sol system, I must travel down into the valley to intercept Earth. Obviously, if I do so on the down hill side of gravity, I am going to hit with tremendous velocity regardless of my interstellar velocity vector. But I can have any configuration of factors or composition too. I do not need to park in this crazy lemming’s yard as a whole object or even land softly. My only constraint is that I must put a rock on the ground without a big explosion. So let’s call that landing the world’s lightest car as a minimum, and doing so with less energy than a ton of TNT and hope this rat has a basement bunker or he’ll be riding the sunset clouds with Science Santa.
Instead of intersecting the Earth on the downward slope of the Sol gravity well, let’s go much deeper and graze Sol itself. Perhaps through ablation, perhaps through particle interaction, or perhaps through some breaking and shedding of materials I drop enough velocity to loop back up the Sol gravity well and into Earth’s vicinity.
Now, I am faced with a new substantial but much smaller gravity-valley. I could use Luna to slow me to a crawl when I cross cislunar space, but there is still a gravity chasm below me.
I am a conglomeration of materials. Perhaps I am a spinning contact binary, perhaps I have a planetesimal fragment once a magma chamber on an alien world with unusual fractional crystalization. Perhaps I have a pancake core of well fused pumice or scoria surrounded by other materials fused or not.
So perhaps I cannot descend from cislunar directly, but I have no constraint on my origin like some object assumed to be on the orbital plane of Sol. I can enter the system from any orientation and while elliptical dynamics are a statistical probability, this post question is about exceptions not statistical simplification.
So I am going to single shot this thing. As I fall into the gravity-valley, any means of slowing is possible. So let’s say I have an ideal shape to skip across the atmosphere and some composition capable of ablation without turning me into a bolide air burst. I’m able to travel abnormally deep into the atmosphere while ablating some unusual volcanic layer formed on my former anoxic stellar alien world of our galaxy. The energy builds to impart a spin as I enter deep into Earth’s atmosphere. As I reach perigee, forces within fracture and release an ultralight core fragment, a pancake shaped disc of an ultra low density pumice at the ideal moment to cancel considerable velocity. The rest of me that continues is irrelevant. I am now this ultra low density aerodynamic pancake released into the atmosphere. All I must do is get to the ground without more than a ton of TNT in energy.
For every rule; an exception. The post is a request for abstractive absurdity such as this. Perhaps this is still flawed. At some level, isn’t the theory of panspermia ultimately an assumption of such a scenario. It was a thought experiment for fun, and a litmus test of Lemmy’s overall engagement and Machiavellian spectrum in a random time slice. My offbeat sense of humor probably doesn’t help here, like calling la grange points listening to ZZ Top, or Barycenter Bary-the-narcissist – as a hint that I am not asking for anything normal, conventional, or basic, or at least that is how I thought of these. It is all for kicks and giggles with digital neighbors in the public commons anyways. So, thanks for responding.
No, there isn’t anything like that. Big heavy objects fall. Falling objects are moving fast when they hit the ground. Further details are irrelevant.
Are you absolutely sure that you’ve seen a coin in those vortex demos just stop? Coins falling over doesn’t count, there’s nothing like that in space. Otherwise, they are moving at quite a clip when they reach the bottom of the funnel.
The one possible exception is when you detach part of the mass. If your vehicle removes some mass and launches it, you can use the reaction force to slow down. This is what a rocket engine does, for example. (Note the propellant will still impact at very high speed, as evident from the plume of any rocket landing.) The higher the relative velocity of the reaction mass, the less you’ll need to come to a complete stop. But at this point we are talking about a manmade vehicle, not a naturally occurring rock.
9.8m/s?
That’s the acceleration due to gravity, 9.8m/s per second. The speed at which it hits depends on how long it’s been accelerating at that rate.
Assuming it was just sitting there and the earth encountered it, wouldn’t that be the bare minimum speed that it could impact the planet?
9.8m/s^2 isn’t a speed. It’s a rate of acceleration; it tells you how fast your speed is changing. The minimum speed an object can impact the planet is equal to “escape velocity”—the speed you’d need to launch an object from the surface so it escapes the Earth’s gravity permanently. In the Earth’s case, that speed is about 40,000 km/hr or 25,000 mph.
Ah, gocha. I’ll be doing calculus in the fall and physics in the spring. I’m not quite there yet.
That’s an acceleration. The speed would increase the longer it had been influenced by the earth’s gravity.
Any idea what the impact speed would be? I haven’t gotten to that point in they math/physics department (yet).
