High School First Interstellar Asteroid Found

[DaveC426913]

According to this other article, its length to width ratio is more like 10:1.
https://www.sciencedaily.com/releases/2017/11/171120120935.htm
So that "artist's impression" if a spearoid is not nearly as "artist's impression" as it would seem.

 

[litup]

An object with a trajectory never seen before and a shape never seen before? That's like winning the lottery 2 weeks in a row. Folks, this is an alien craft.

Well it ain't stopping here, so don't worry about an alien attack:)

 

[rootone]

Well it ain't stopping here, so don't worry about an alien attack:)

Yeah but the hyperbolic trajectory past our Sun might be just to fool us.

 

[DaveC426913]

It is already on its way out of the system.

 

[stefan r]

Long, hard, no belt, and flying free.

Seems to me, it's essentially two (or more) long, thin asteroids balancing on their tips against gravity.

Stalactites are long and thin. So are hoodoos.

This article suggests asteroids in the solar system started out with odd shapes with many elongated. They get chipped and bumped into spherical shapes over time.

 

[newjerseyrunner]

An object with a trajectory never seen before and a shape never seen before? That's like winning the lottery 2 weeks in a row. Folks, this is an alien craft.

There are several known objects with this “cigar” shape. They’re simply rare.

 

[Ernest S Walton]

There are several known objects with this “cigar” shape. They’re simply rare.

According to Sky and Telescope, this is the only asteroid - out of 750,000 - to have a light curve range of 2.5 magnitudes :

http://www.skyandtelescope.com/astronomy-news/meet-oumuamua-the-interstellar-cigar/

 

[stefan r]

According to Sky and Telescope, this is the only asteroid - out of 750,000 - to have a light curve range of 2.5 magnitudes :

http://www.skyandtelescope.com/astronomy-news/meet-oumuamua-the-interstellar-cigar/

How do we differentiate the light curve of a "cigar shape" from the light curve of a "poker chip shape"?

 

[Ernest S Walton]

How do we differentiate the light curve of a "cigar shape" from the light curve of a "poker chip shape"?

A picture tells a thousand words, remembering that the rotation takes just over 7 hours :

 

[stefan r]

A picture tells a thousand words, remembering that the rotation takes just over 7 hours :

I looked at that picture. A cigar 10 cm by 1 cm by 1cm would be at the peak when the long axis is perpendicular to the sun. 10x luminosity because 10x surface. A cookie with a face surface area 10x the edge surface area (3.2 x 3.2x1, or 5 x 2 x1) would have the same peak and trough.

 

[DaveC426913]

I looked at that picture. A cigar 10 cm by 1 cm by 1cm would be at the peak when the long axis is perpendicular to the sun. 10x luminosity because 10x surface. A cookie with a face surface area 10x the edge surface area (3.2 x 3.2x1, or 5 x 2 x1) would have the same peak and trough.

I'm trying to visualize the axial rotation.

I think a cookie shape would require more "lucky" fine-tuning than a spindle. The spindle is symmetrical about its long axis, so it only needs one axis to align every 8 hours in order to reflect sunlight toward Earth. A cookie-shape would need to be aligned on two axes to reflect properly. While it doesn't rule it out, it essentially means the cookie's chance is the square root of the spindle's.

I think.

 

[mfb]

The cookie doesn't need any tuning, while the spindle needs tuning to get a large luminosity ratio.

No matter how a cookie is oriented and how it rotates, we see its edge on twice per rotation, and its edge points towards the Sun twice as well. With a spindle you need a lucky coincidence to have any of these events.

Following the light curves while the object is moving through the solar system should give sufficient separation power between these two options.

 

[stefan r]

I'm trying to visualize the axial rotation.

I think a cookie shape would require more "lucky" fine-tuning than a spindle. The spindle is symmetrical about its long axis, so it only needs one axis to align every 8 hours in order to reflect sunlight toward Earth. A cookie-shape would need to be aligned on two axes to reflect properly. While it doesn't rule it out, it essentially means the cookie's chance is the square root of the spindle's.

I think.

We are basically back toward the sun so the aspect ratio of 10 may be all we know. We could try some improbable shapes too. A minorra could pull it off. King Arthurs sword excaliber. An eggshell, a hockey stick, a hollow ring.

Is it possible that NASA reasons like "all things are spheres until there is convincing evidence that it is not a sphere". We have "convincing evidence that there is a 10:1 aspect ratio". So one axis must be long. "All things that are long must be ellipsoid and have circular circumference until proven otherwise." ...

 

[rootone]

"all things are spheres until there is convincing evidence that it is not a sphere".

Not an unreasonable proposition, but ...
https://en.wikipedia.org/wiki/67P/Churyumov–Gerasimenko

 

[DaveC426913]

With a spindle you need a lucky coincidence to have any of these events.

It doesn't need to achieve edge-on or broadside orientations. Just as long as the difference between the most extreme reaches 10:1. That means it might be somewhat longer and/or narrower than the assumed 10:1.

 

[Vanadium 50]

Wouldn't a cookie shape be likely to tumble instead of spin?

 

[Filip Larsen]

Wouldn't a cookie shape be likely to tumble instead of spin?

If you by tumble mean spin like in a coin flip, then no. If you by tumble take the normal meaning of non-pure rotation, then also no.

Torque free semi-rigid bodies ends up in pure spin around the inertial major axis due to internal friction when tumbling, so for an oblate cylindrical body (cookie shape) that would be spin around the body symmetry axis and for an oblong cylindrical body (pen shape) it would be around an axis orthogonal to the body symmetry axis. The term tumbling is usually taken to mean the opposite of pure rotation, that is, a rotation around an axis that is neither the major or minor inertial body axis whereby the body elements experience acceleration variations that then via friction gives rise to dissipation of rotational energy and a drift of rotational body axis to the major inertial axis.

 

[rootone]

Spheroidal and 'cigar' shaped asterioids can be understood from Newtonian physics.
Disk like objects no so much, though disk structures are common on the scale of galaxies.

 

[stefan r]

Spheroidal and 'cigar' shaped asterioids can be understood from Newtonian physics.
Disk like objects no so much, though disk structures are common on the scale of galaxies.

I was hiking in Rocky Mountain National park a few years ago. We were in a forest fire area. Boulders were blackened except for the surface on the hottest side. There the rock was fresh and clean. On the ground was a pile of thin broken plates. The plates were black on one side and grains on the other side were an exact match to the exposed surface of the boulders. Slightly concave and slightly convex on opposite faces of the plates with the curve matching the radius of the boulder. Spall.

I am just giving an example of a natural rock that is very thin. Not aware of any evidence that spall effects have any direct relevance to Oumuamua.

 

[OmCheeto]

Has anyone done the maths on the forces on the ends?
Gravity vs centripetal force?

Using a rotation rate of 2.4e-4 rad/sec (7.3 hours/tumble), I came up with a minimum density of 1130 kg/m³ to neutralize the forces.
I'm guessing the density is probably higher, but anything lower than that, and someone standing on the end would have to hold on.

Just curious if my maths is boogered. (as usual )

I'm also curious how ductile a spinning solid asteroid would be over millions of years.
If it were a billion years old, and had spun at its current rate, then that's around 1.2 trillion revolutions.
On a rock weighing between 600 million and 4 billion kg, that seems like a lot of stress.
Like intergalactic taffy, maybe?

 

[DaveC426913]

I'm guessing the density is probably higher, but anything lower than that, and someone standing on the end would have to hold on.
On a rock weighing between 600 million and 4 billion kg, that seems like a lot of stress.
Like intergalactic taffy, maybe?

Interesting. But it would be a fine balance.Enough force to deform it - even over a long timespan - would be very nearly enough to fling anything not nailed down off into space.
They said they detected zero coma, which is what ruled it out as a comet. But it would also put an upper limit on the amount of dust and grains that would be surrounding it, if such dust and grains were to be flung off by spin.

 

[OmCheeto]

Interesting. But it would be a fine balance.Enough force to deform it - even over a long timespan - would be very nearly enough to fling anything not nailed down off into space.
They said they detected zero coma, which is what ruled it out as a comet. But it would also put an upper limit on the amount of dust and grains that would be surrounding it, if such dust and grains were to be flung off by spin.

From the site I checked for "meteor density", their range was 1790 thru 8000 kg/m³. The most common being 3400 kg/m³. So even their lightest meteor density would yield a tip gravity that exceeds centripetal force by 60%. Which kind of implies that nothing should ever get flung off by its spin, at least recently.

Perhaps tomorrow I'll spin the creature back down to a spheroid, a billion years younger, and see how the maths works out.

I also wonder if anyone would know how to model a full metal asteroid of this size. Say, create a 40 meter diameter, 400 meter long nickel-iron cylinder, support it on one end, and see how far it droops over time. Metallurgy is one of my least studied subjects.
Plasticity, looks like it might be a good candidate, though even the wiki description makes my head hurt. I put the odds at 1000:1 against me being able to get anything useful out of that.

That's interesting that no coma was seen. Thinking about it though, it kind of makes sense. Something traveling between stars or galaxies wouldn't have the swarm of dust and stuff you find spinning lazily around a solar system. I'm guessing most everything it met was going 94,000 kph! (60,000 mph)

More things for me to think about in the morning: Interstellar and intergalactic hydrogen densities.

Lot's of interesting stuff to think about!

 

[stefan r]

...
Using a rotation rate ... to neutralize the forces... that seems like a lot of stress...

No, would be 0 stress. An elongated non-rotating asteroid would have more compressive stress at the center. Force from rotation would have to exceed gravity in order to cause tensile stress.

...
I'm also curious how ductile a spinning solid asteroid would be over millions of years.
If it were a billion years old, and had spun at its current rate, then that's around 1.2 trillion revolutions.
On a rock weighing between 600 million and 4 billion kg, that seems like a lot of stress...

You might be thinking of cyclical stress. If you bend a bar back and forth it can harden and become more brittle. The material in the asteroid should not be bending at all. Especially in interstellar space were there is nothing (very little?) to give it tidal forces.

It would have billions of years (or insert age) of ionizing radiation. Asteroids in our belt would have same/similar cosmic radiation and a lot more solar wind. "More brittle" does not mean the same thing as "weaker" and sometimes the opposite. We use ion implantation to toughen steel drill bits.

 

[OmCheeto]

No, would be 0 stress. An elongated non-rotating asteroid would have more compressive stress at the center. Force from rotation would have to exceed gravity in order to cause tensile stress.

Ouch! I do believe you are correct. I'll have to spin this thing backwards in time to see what happens.

You might be thinking of cyclical stress. If you bend a bar back and forth it can harden and become more brittle. The material in the asteroid should not be bending at all. Especially in interstellar space were there is nothing (very little?) to give it tidal forces.

It would have billions of years (or insert age) of ionizing radiation. Asteroids in our belt would have same/similar cosmic radiation and a lot more solar wind. "More brittle" does not mean the same thing as "weaker" and sometimes the opposite. We use ion implantation to toughen steel drill bits.

Like I said, metallurgy and rocks are two of my weakest points. Where are @davenn & @billiards when you need them.
What the heck could shape a rock like that in outer space? Or did it start that way, like some giant splinter?

 

[rootone]

Or did it start that way, like some giant splinter?

https://www.physicsforums.com/members/stefan-r.615251/
said something related to that.
https://www.physicsforums.com/threads/first-interstellar-asteroid-found.930191/page-4#post-5889097

 

[OmCheeto]

https://www.physicsforums.com/members/stefan-r.615251/
said something related to that.
https://www.physicsforums.com/threads/first-interstellar-asteroid-found.930191/page-4#post-5889097

Yes, I saw that. It's kind of big to be a "splinter", so that would be my last choice as to why it's shaped like it is.

But I did finish my retro spin analysis. All I can say is that the universe gave us a great homework problem.

When centripetal acceleration is greater than gravity(at the ends), then there will be a tensile stress.

Bumping the density up to that of dwarf planet Haumea, bumps the "g" curve up, yielding a less elongated equilibrium length to diameter ratio:

Looks like 6.5:1 to me.

One can only scratch one's head about the evolution of such graphs.
Did Oumuamua partially evaporate on its journey?

Further reading: Jacobi ellipsoid

 

[sophiecentaur]

(Does not imply the visitor was stationary itself.)

We're all stationary, relative to ourselves but what reference frame did you assume for the visitor?

 

[DaveC426913]

We're all stationary, relative to ourselves but what reference frame did you assume for the visitor?

I believe NR is saying,
- never mind whether or not the object is moving or stationary wrt any other FoR.
- Is it possible the object is not gravitationally bound to either our system or another nearby system.

I'd have to track it down, but yes, there was a hypothesis that the object could have formed independently of a star system, such as perhaps part of a rogue planet.

 

[sophiecentaur]

- never mind whether or not the object is moving or stationary wrt any other FoR.

If it is stationary relative to our Solar Sysyem, it won't be approaching us so we would never observe it going through the SS. It would not surprise me to find objects like that, in deep space, that have been 'tracking us' for millions of years but that's not what the thread has been discussing.
The train and the fly are heading towards each other and the same damage still be done at the collision, whatever FoR we use. I don't think that 'stationary visitor' is a meaningful description.

 

[tony873004]

I made a new simulation where you can pan around. It's similar to the one I already posted, except it includes a path for Oumuamua. The previous one didn't because the code only knew how to superimpose Keplerian ellipses, not hyperbolas. So I had to hand-code Oumuamua's path.
http://orbitsimulator.com/gravitySim.../oumuamua.html