Astronomers discover 3I/ATLAS – Third interstellar object to visit Solar System

This one is coming in fast, it has an eccentricity of over 6 with the current fits. For point of reference, 1I and 2I have eccentricities of 1.2 and 3.3.

Right now it is mostly just a point on the sky, it is difficult to tell if it is active (like a comet) yet. If it is not active, IE: asteroid like, then the current observations put it somewhere between 8-22km in diameter (this depends on the albedo of the surface). From what we know, we would expect it to likely be made up of darker material meaning given that range of diameters it is more likely to be on the larger end. However if it is active, then the dust coming off can make it appear much larger than it is. As it comes in closer to the sun and starts to warm up it may become active (or more active if its already doing stuff).

It will not pass particularly close to any planet. It will be closest to the sun just before Halloween this year at 1.35 au, moving at 68 km/s (earth orbits at 29-30 km/s). It is also retrograde (IE, it is moving in the opposite direction of planetary motion), for an interstellar object this is basically random chance that this is the case.

Link to an orbit viewer: https://ssd.jpl.nasa.gov/tools/sbdb_lookup.html#/?sstr=3I&vi...

The next couple of weeks will be interesting for a bunch of people I know.

Source: Working on my PhD in orbital dynamics and formerly wrote the asteroid simulation code used on several NASA missions: https://github.com/dahlend/kete

From the simulation you linked looks like it is passing closeish to the Mars... but I do know that space is big. However, I am curious of what would happen if an object of this magnitude hit mars at 90km/s.

Assuming it’s at the upper range of the size estimate above, and of average rocky density, the kinetic energy of the impact would be something like a 10 billion megaton nuke.

If we could steer it to hit one of Mars’s poles, it might do a bit of terraforming for us!

Where did my math go wrong? I got about 50,000 megatons. Assuming the high-end of 22km and a rocky/metallic density of 5000 kg/cubic meter (and assuming it's a cube):

  kinetic energy = 1/2 m v**2 = 1/2 * size * density * v**2
  = 1/2 *(22000 m)**3 * (5000 kg/m**3) * (90 m/s)**2 / (4.184E15 J/megaton)
  = 52,000 megaton

If it's an icy comet then the density is more like 500 kg/cubic meter, or 1/10th that number.

1040 x more energy that the Tsar Bomba.

Or 5-ish Tsar Bomba per country on Earth.

Or 3466 Hiroshima nukes.

Or 17 Hiroshima nukes per country.

In light of the error in the parent comments math, I retract my previous comment and substitute the following bit of awkward silence:

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