# FeaturedB First Interstellar Asteroid Found

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1. Nov 21, 2017

### DaveC426913

2. Nov 21, 2017

### litup

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

3. Nov 21, 2017

### rootone

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

4. Nov 21, 2017

### DaveC426913

It is already on its way out of the system.

5. Nov 21, 2017

### stefan r

Long, hard, no belt, and flying free.

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.

6. Nov 22, 2017

### newjerseyrunner

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

7. Nov 22, 2017

### Ernest S Walton

8. Nov 22, 2017

### stefan r

9. Nov 22, 2017

### Ernest S Walton

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

10. Nov 22, 2017

### stefan r

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.

11. Nov 22, 2017

### DaveC426913

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.

12. Nov 23, 2017

### Staff: Mentor

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.

13. Nov 23, 2017

### stefan r

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." ...

14. Nov 23, 2017

### rootone

15. Nov 24, 2017

### DaveC426913

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.

16. Nov 24, 2017

Staff Emeritus

17. Nov 24, 2017

### Filip Larsen

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.

18. Nov 24, 2017

### 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.

19. Nov 24, 2017

### stefan r

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.

20. Nov 24, 2017

### 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?