[QUOTE=science_man_88;502666]strength of gravity ? Also 6.25% over an area of 4*Pi*22500 square light years isn't a lot (~22 ppm on average per square light year)[/QUOTE]a) Irrelevant because gravity affects Fe-60 and Al-26 equally.
b) You don't need much to be detectable.

[QUOTE=science_man_88;502678][B]NOTE[/B]: once you approximate the earth as a perfect sphere of radius 6300 km, and and assume it deposits the ratio of that area to a square light year you can decrease your estimate roughly 16 orders of magnitude further. So to see 1 ppm of earth's
current mass be deposited in the isotope you would need on the order of 2^18 earth masses to begin with.[/QUOTE]
What? Why the hell do you need anywhere near that much. Your estimate is so crazy it's not even wrong.

First, a SN will deposit its ejecta as a thin layer on the surface of the earth, so your 1ppm of the earth as a whole would leave the surface coated in a thick layer of highly radioactive material.

Let me be generous and assume you meant 1ppm of a thin layer, perhaps a centimetre thick, and not 1ppm of the earth as a whole. Even so, you seem to have no clue as to how sensitive radioactivity detectors have become. I'm not going to do the calculations for you, you seem to like that activity, but I'll inform you that one radioactive decay per day per kilogram of material is readily achievable. Given the 0.7Ma half-life already stated, work out the concentration of Al-26 needed to give a usable signal.

[QUOTE=xilman;502719]What? Why the hell do you need anywhere near that much. Your estimate is so crazy it's not even wrong.

First, a SN will deposit its ejecta as a thin layer on the surface of the earth, so your 1ppm of the earth as a whole would leave the surface coated in a thick layer of highly radioactive material.

Let me be generous and assume you meant 1ppm of a thin layer, perhaps a centimetre thick, and not 1ppm of the earth as a whole. Even so, you seem to have no clue as to how sensitive radioactivity detectors have become. I'm not going to do the calculations for you, you seem to like that activity, but I'll inform you that one radioactive decay per day per kilogram of material is readily achievable. Given the 0.7Ma half-life already stated, work out the concentration of Al-26 needed to give a usable signal.[/QUOTE]

okay, even at 1 part per quintillion you would need about 1 billion Earth's assuming nothing collects it before it hits earth's distance from it.

[QUOTE=science_man_88;502678]... you would need on the order of 2^18 earth masses to begin with.[/QUOTE][QUOTE=science_man_88;502728]okay, even at 1 part per quintillion you would need about 1 billion Earth's assuming nothing collects it before it hits earth's distance from it.[/QUOTE]Erm, you do realise that 2^18 is less than 1 billion, right? So your estimate has risen?

[QUOTE=retina;502733]Erm, you do realise that 2^18 is less than 1 billion, right? So your estimate has risen?[/QUOTE]

opps 10^18 not 2^18 doh. makes sense why my numbers seem off now( been around Mersennes too much)

[QUOTE=xilman;502659] <snip>
Anyway, how did you expect the Fe-60 to get here and not the Al-26 in the same time period?
<snip>
There are large signals from the Al-26 decay easily detectable towards the galactic centre and out Cygnus way. Why none (or very little) around here?[/QUOTE]
OK, let's see here:

1) [sup]60[/sup]Fe has been found here on Earth, and is very likely a supernova remnant.

2) [sup]26[/sup]Al has been detected in interstellar space, and theory predicts it is also produced in supernova explosions (If I'm reading [url=http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.872.7023&rep=rep1&type=pdf]this paper[/url] right, it's generally produced in greater amounts, too.)

3) The paper first mentioned deals with a possible mass extinction 2.6 million years ago, thought to have been due to a huge flux of muons. If I'm reading this correctly, the muons are created by cosmic rays hitting the atmosphere. This is almost necessitated by their having a half-life of only a couple of microseconds. They are often created high up in the atmosphere, and reach the bottom of the atmosphere before decaying only because of relativistic time dilation due to their great speed relative to Earth.

4) The more recently-cited paper features Geminga, which is thought to be the remnant of a supernova explosion about 300,000 years ago. That's out of the time frame for the mass extinction that got Megalodon, but it still happened. The [sup]10[/sup]Be in the Mediterranean is thought to come from it. However, in this case too, no [sup]26[/sup]Al seems to have shown up.

5) The question has been raised: If [sup]60[/sup]Fe from recent supernovas made it to Earth, why hasn't the [sup]26[/sup]Al been detected? I don't know. The Geminga paper says,

[quote]In this regard, it is interesting that Cini Castagnoli et al. (1992) indicate that they performed two [sup]26[/sup]Al measurements on their Mediterranean core at the peak regions. They do not give a quantitative result, but cite this measurement as evidence against a contribution from cosmic dust. The implication is that there was not a large [sup]26[/sup]Al signal. While this certainly does not strengthen the Geminga hypothesis, it cannot rule it out. For example, [sup]26[/sup]Al might not be a supernova product (though interstellar γ -ray line observations argue against this), [b]or the direct ejecta may have been excluded from the inner solar system[/b][/quote]

This last suggestion seems to be ruled out by the likely detection of supernova remnant [sup]60[/sup]Fe, unless some mechanism is proposed that preferentially keeps [sup]26[/sup]Al out of the inner solar system.

So -- the question remains. Where's the [sup]26[/sup]Al? I don't know. I swear to God I didn't take it. Absent a mechanism that preferentially keeps it away from Earth, I can only assume it actually made it to Earth. I can suggest two possibilities. One, it was where it was being looked for, but wasn't detected. Two, it wasn't there, because it wound up somewhere else; that is, once it got here, it was separated from the iron. How? I don't know. In addition to not being an astrophysicist, I'm also not a geochemist.

[QUOTE=science_man_88;502728]okay, even at 1 part per quintillion you would need about 1 billion Earth's assuming nothing collects it before it hits earth's distance from it.[/QUOTE]Please show your working.

Assume that the incoming Al-26 is collected on the earth's surface uniformly and in the subsequent 2.6Ma becomes mixed into a layer 1cm thick. (Note that the much older KT boundary layer is at most 1cm thick, 66Ma after it was deposited, so 1cm will be a ball-park figure you can use here.) That is calculate the mass of a 1cm thick layer of rock, density 3 tonnes per cubic meter, which covers the entire surface of the earth. Express your result in units of kg and earth-masses

Next, calculate the mass of Al-26 which will produce one radioactive decay per day. Express your result in units of kg and of earth-masses.

From the two above computations, calculate how many earth-masses of Al-26 the surface of the earth needed to collect for the Al-26 still to be detectable now. (Remember that over three half-lives have elapsed so the primordial amount will be a factor ten or so greater than what is still around).

Assuming the SN emitted 3 earth-masses of Al-26 isotropically, compute its maximum distance for the signal to be detectable here and now. Express your result in metres, earth-radii and light-years.

[QUOTE=xilman;502760]Please show your working.

Assume that the incoming Al-26 is collected on the earth's surface uniformly and in the subsequent 2.6Ma becomes mixed into a layer 1cm thick. (Note that the much older KT boundary layer is at most 1cm thick, 66Ma after it was deposited, so 1cm will be a ball-park figure you can use here.) That is calculate the mass of a 1cm thick layer of rock, density 3 tonnes per cubic meter, which covers the entire surface of the earth. Express your result in units of kg and earth-masses

Next, calculate the mass of Al-26 which will produce one radioactive decay per day. Express your result in units of kg and of earth-masses.

From the two above computations, calculate how many earth-masses of Al-26 the surface of the earth needed to collect for the Al-26 still to be detectable now. (Remember that over three half-lives have elapsed so the primordial amount will be a factor ten or so greater than what is still around).

Assuming the SN emitted 3 earth-masses of Al-26 isotropically, compute its maximum distance for the signal to be detectable here and now. Express your result in metres, earth-radii and light-years.[/QUOTE]

[CODE]
(630000001/630000000)^3*4/3*Pi
%1 = 4.1887902247330110390704975067
%-1
%2 = 3.1887902247330110390704975067
? %*6300000^3*4*Pi
%3 = 10019763305289618992911.776874[/CODE]

times the last result by 1000 to get metric tons.

But I get the point. I should stop doing any calculation at all.

[QUOTE=science_man_88;502764]But I get the point. I should stop doing any calculation at all.[/QUOTE]No! You do not get the point at all. You most emphatically should do the computations. Why do you think I spent quite some time and effort in my previous post on guiding you how to do them? It would have been so much easier for me to announce that you were a mindless jerk who will be first up against the wall when the revolution comes.

What you must do, IMAO, is set down your assumptions clearly and make plausible order-of-magnitude estimates for any numerical quantity which you can't look up to a greater precision.

Then, and only then, perform the calculations.

As stated above, please show your working.

[QUOTE=xilman;502802]No! You do not get the point at all. You most emphatically should do the computations. Why do you think I spent quite some time and effort in my previous post on guiding you how to do them? It would have been so much easier for me to announce that you were a mindless jerk who will be first up against the wall when the revolution comes.

What you must do, IMAO, is set down your assumptions clearly and make plausible order-of-magnitude estimates for any numerical quantity which you can't look up to a greater precision.

Then, and only then, perform the calculations.

As stated above, please show your working.[/QUOTE]
because like most on here, you are on the expert in everything scale.

[QUOTE=science_man_88;502821]because like most on here, you are on the expert in everything scale.[/QUOTE]

Do you think people are simply born with expertise?