[X] My Assistant

[gndonald]

Doc - a good place to go to get an idea of how big a bang an impactor can cause is H Jay Melosh and Ross A. Beyer's Crater size calculator at LPL.When you select appropriate numbers for this object (50m diameter composed of loose rock, asteroid impact at mars hitting a porous rock surface) you get somewhere between a 700m and 1.8 km diameter crater. The impact energy released is on the order of 1.2 MegaTons.

I tried putting the numbers you gave through H Jay Melosh & Gareth Collins Earth Impact Effects Program, turns out if this object hit the Earth it would airburst at 42400 ft... which is not much good for figuring out just what it would do at Mars.

Though if you do want to have some fun with this one, just put in the following:

Projectile Dia: 10km
Density: Porous Rock.
Impact: 17kms

and have it impact into sedimentary rock at 45 degrees. You'll get a rough idea of what happened to the dinosaurs... just make sure you are well away from the impact yourself (3000km).

[Doc]

I tried putting the numbers you gave through H Jay Melosh & Gareth Collins Earth Impact Effects Program, turns out if this object hit the Earth it would airburst at 42400 ft... which is not much good for figuring out just what it would do at Mars.

Pity...... oh well, at least the odds have become more favourable. Least to say it will (hopefully) be a once in a life time opportunity. Lets wait and see

[Shaka]

not much good for figuring out just what it would do at Mars.

gn,
The crater size calculator offers options for Mars impacts. You just didn't choose them.

[JRehling]

I'm rooting for an impact near a MER (however unlikely that may be). I think of the old US Civil War adage that to kill a man you needed a mass of bullets about equal to the man. The vicinity of the MER would have to be absolutely machine-gunned for a "kill" to be likely. A huge impactor might machine gun an entire region of Mars (dropping at least one secondary onto every single square meter), but this won't be a huge impactor.

If a MER can resolve an area equal to about 100 times the area of the MER itself, then we'd have far better odds of one secondary dropping within range (so it could be examined) vs. the odds of the MER itself taking a kill shot. Bring it on!

[gndonald]

gn,
The crater size calculator offers options for Mars impacts. You just didn't choose them.

Actually I was using a completely different program which was optimized to determine the impact effects of an asteroid on Earth, the Crater program is a completely different setup which only deals with the size of the crater after impact, that's not to say both produce interesting results.

One of the more sobering for me was when I ran the impact effects calculator using the Chicxulub impacter stats but having the thing hit the ground where the Wolf Creek crater was. It turns out that I probably would not survive down in Perth. It's not the impact but rather the blast wave that gets you....

Edit: I've found a simple online program that gives some idea of the ground shaking this asteroid might cause. Assuming it hits with a speed of 20km/s then it produces a 6.2 on the Richter scale.

[helvick]

Actually I was using a completely different program which was optimized to determine the impact effects of an asteroid on Earth,

I think the estimate from this, if accurate, indicates that a 50m diameter asteroid will probably make it to the Martian surface and the energy of the impact will remain substantially the same as those produced by the Crater calculator.

The Impact Effects results you link to put the initial breakup in the Earth case at around 65km with the air burst at 7km dissipating 95% of the original kinetic energy. In the Martian case the entire atmospheric column is less massive than the part on Earth that is above 65km so it seems reasonable to infer that it should reach the surface (mostly) intact and with the majority of its kinetic energy remaining.

That is if it does actually hit which remains quite a long shot.

[Moon Saloon]

.

[MarsIsImportant]

Can MRO be used to improve the orbit knowledge?

No, because the asteroid is coming from the direction of the Sun relative to Mars. There is too much glare. Only ground based observations can give us better constraints upon the orbit of 2007 WD5 until it passes Mars--or Hubble.

[SFJCody]

How great would this be if it did hit? It would be like having a free Deep Impact/LCROSS, but for Mars!

[Holder of the Tw...]

A few new observations of the asteroid have finally been reported, taken within the last 40 hours by Magalena Ridge Observatory. The reported results are a little confusing.

Apparently there is still some small chance it will impact. The nominal miss distance from Mars center is given as about 24000 miles (or roughly 38000km). The minimum possible is 1600 miles, which is below the surface.

However, the error ellipse as viewed from Earth at that time shows a possible tracking error of only 5.87 arc seconds, according to the generated ephemeris.

This would be a clear miss, as Mars itself is over 16 arc seconds big at that point. Perhaps the ephemeris is operating at a lower (90 per cent? 95 per cent?) probability.

Link

[MarsIsImportant]

The tracking error and the orbital trajectory are two separate issues. The tracking error of 5.87 secs is the error in actually plotting the asteroid in 3d space. This one measurement alone says nothing about the movement of the object. We determine the trajectory by plotting all of the observations together...to track the movement over time. Only a few observations greatly compounds the possible errors. If you have more observations, then the errors tend to gradually cancel eachother out. But you need a significant number of observations for the 'Law of Large Numbers' to have a statistical effect on the results. 30 or so observations are simply not nearly enough. I'd give it another week or maybe two for additional observations before we have any significant confidence in the numbers (that's probably multiple observations per day).

[Greg Hullender]

Law of Large Numbers says the uncertainity should decline with the square root of the number of observations, all other things being equal, so to cut that tracking error in half, it'll take about another 100 observations.

Of course if the new observations use a different technique, they could do much better than that. Also, the LLN only applies if the obervations are independent and identically distributed, and I suspect these are neither. :-)

--Greg

[Holder of the Tw...]

First a correction. I stated in my previous post that Mars would appear larger than 16 arc secs at the end of the month. It is actually going to be 12.3 arc secs as viewed from Earth. Not sure where I messed up on that one, but sorry.

The tracking error I referred to earlier is the uncertainty ellipse of where you would find the asteroid in the sky as viewed from some point on Earth. This is what is generated on the ephemeris on the NeoDys web site, and is located on the far right of the generated table of predicted positions for a particular asteroid over a given time interval.

My apologies if I wasn't clear about what I was referencing earlier. This tracking error is a two dimensional figure on the sky, not a three dimensional location in space. It says that on January 30th, the predicted position of 2007 WD5 on the sky could be in error by about six arc sec. In this case, that is mostly in two opposite directions along a very narrow ellipse (almost a line, which intersects the planet if you extend it a little).

Now, I've looked into it a little more. The sky error is only a one sigma approximation. NeoDys themselves recommend that you multiply that uncertainty by at least three to be more certain of actually locating something with a high degree of confidence.

As for getting multiple observations per day, that is a forlorn hope. The resources don't exist for an object this faint. Well, actually they do, but they're all booked up for other things.

But we only need a few more looks anyway. As the date fast approaches, and we can spot where this rock actually is in relation to Mars, we will see the error box collapse rapidly.

[Burmese]

To what extent, if any, are the calculations taking into account Mars' gravity well when the asteroid is closer? Mars may just barely fall inside the error ellipse but are these projections so far just about calculating the objects path from known observations or are they factoring in any attraction caused by Mars' gravity. How close would it have to get to be 'sucked in'? How close to be greatly deflected along a new path (but no collision)?

[Holder of the Tw...]

Yes, martian gravity is being taken into account. Small errors in the location of this asteroid when nearest Mars translate into very significant differences in the deflection of the orbit. If you carry the ephemeris forward a few days into February, you will see that the sky error dramatically increases after the close approach, assuming there is an "after". But at this point, that's a fairly safe assumption.