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What does the asteroid belt look like
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13-10-2016 3:38pm#1
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Another distance anecdote, this time on a fairly small scale: The Perseid meteor shower in August occurs when the Earth crosses the dust trail left behind in the orbit of comet Swift-Tuttle. The typical meteor is caused by a particle the size of a grain of sand as it burns up in Earth's atmosphere. Since meteors burn relatively low down in the atmosphere ( ~ 100 km) what you see is a relatively narrow path through the meteoroid stream in space, since meteors displaced horizontally far from your location will be below your horizon. So if we count the meteors we get a reasonable idea of their straight line separation in space. An average Perseid shower maximum (which only lasts a couple of hours) has a zenith hourly rate (ZHR) of about a hundred, or one per 36 seconds. The earth is ploughing through the stream at 30 km/sec. So there is about 1,000 km between grain-of-sand sized particles. That's a linear separation, so we might surmise that there is one grain of sand per million square kilometres (or per billion cubic kilometres). Since most things in the solar system orbit in a flat plane, we might be better sticking to square measurements. Remember, this example is for the very narrow maximum of the most active annual meteor shower, so in general space is far, far emptier.
We could do another back-of-an-envelope guesstimate of the average separation between objects in the asteroid belt, bearing a few things in mind. The asteroid belt has no well defined boundaries, and there are different populations of asteroids such as the ones at Jupiter's Lagrange points, and the Earth-crossing asteroids, as well as the main belt between the orbits of Mars and Jupiter. In order to construct a toy model, we'll assume some well defined circular region of constant density between Mars and Jupiter. Another problem is that our knowledge of the density decreases as we go down the asteroid size scale. But various counts of asteroids have been made, as well as estimates of the distribution of sizes, which we will make use of.
In general the size-frequency distribution (SFD) follows a power law. That is, increasing the asteroid size by some multiple increases the number of asteroids of that size by some power of the multiple. (You've come across power laws in things like gravity's inverse square law -- it has a power law index of -2 since increasing the distance by n changes the force by n⁻². The index for asteroid size vs. frequency will likewise be some negative number, corresponding to the increase in number of asteroids at decreasing sizes).
For a toy model let's start with Wikipedia's picture of the main asteroid belt and turn it into a homogeneous distribution with the asteroids laid out in a square lattice like this:
Our toy asteroid belt is a ring whose area is the difference between the areas of an outer circle and an inner circle. We completely ignore inhomogeneities like the Kirkwood gaps. We then divide by the number of asteroids N of a given size to get the area per asteroid, and take the square root to get the average separation d:
We've rearranged to separate out N because everything else in our model will be constant. We put the inner edge of the belt at the orbit of Mars and the outer edge half way between Mars and Jupiter, for inner and outer radii of 230m and 500m kilometres respectively. Putting these numbers into our equation gives:
Next we must consider asteroid brightness. The scientists use the concept of H magnitudes. If you know about absolute magnitudes, it's a way of normalising the apparent brightness of a star to how it would look at a standard distance of 10 parsecs. H magnitudes are a similar scheme for asteroids, normalised to a distance of 1 AU (= 150m km). NASA has supplied us with a handy table of H magnitudes, translating them into asteroid diameters depending on albedo or reflectivity. (Unfortunately we usually can't measure asteroid size directly but have to estimate it based on brightness).
Wikipedia tells us that there are 140 main belt asteroids with diameters greater than 120 km. According to the most optimistic estimate of albedo, our table translates this into H magnitudes greater than 6.5. As luck would have it, that's just the limiting magnitude that we can see with the naked eye on the darkest night. Similar rules apply out in space -- we need to make sure we're not looking toward the Sun, or dazzled by a nearby bright object like the side of our spaceship. So let's put our 140 asteroids into our formula:
So according to our toy model, the nearest asteroid greater than H magnitude 6.5 is about half an AU away, and should therefore be visible. On the other hand, there is an uncertainty of about 2 magnitudes, or a factor of more than 6 times less brightness if our asteroid is one of the less reflective ones. It turns out that more than half of asteroids are of low or intermediate albedo. So let's just say that there is at best a fighting chance you could see one asteroid in the asteroid belt with the naked eye if you were out there among them. Of course, it would look like one of the dimmest stars you'd see on a dark night from earth, and as such would be completely lost among the actual stars unless you knew exactly where to look.
How about smaller asteroids? They'd be dimmer still, but there are more of them so there's a chance one of them would be closer to us and so have a greater apparent brightness. Estimates of the number of asteroids greater than 1 km diameter range up to about two million. These have minimum H magnitudes between 16.5 and 19.0. Remember, this is the absolute magnitude M at 1 AU. To bring these up to naked eye apparent magnitudes m of less than 6.5 we can use a modified version of the distance-luminosity formula for stars, which gives the required distance in AU as:
Putting in our magnitude ranges gives a distance of 0.01 to 0.003 AU in order for these asteroids to be visible to the naked eye. Meanwhile, using our other formula for average separation of two million objects gives:
This lies between the extremes of our naked eye distance range. So once again we can't say for sure whether anything would be visible, but it's possible. Given that the nearest asteroid seems to be hovering on the edge of naked eye visibility at both the scale of the largest objects and 1 km objects, we should consider the power law size-frequency distribution in the context of our toy model:
The luminosity of an object is proportional to its surface area and thus the square of it's radius:. But according to the inverse square luminosity distance relation:
. So if the distance increases, to maintain constant luminosity we must increase the radius linearly with distance:
. But then from our separation formula we have:
and thus:
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In short, if the size-frequency distribution's power law index is less than -2 then asteroids should become more visible as we go down the size scale. Due to properties of logarithms, the power law index is equal to the slope of a graph plotted on a log-log scale. So here is a graph of the logs of asteroid size versus number:
The slope is in fact not far off -2, but it varies across the size distribution. The upshot is that in general, if we were in the asteroid belt we might expect to just about be able to see the nearest asteroid to us with the naked eye under dark conditions. It would be one dim point of light amongst the thousands of visible stars. It absolutely definitely would not look like this:
Apart from the Wikipedia entries I've linked, here are some papers I looked up, which make interesting further reading:
Tedesco, E., "The Statistical Asteroid Model. I. The Main-Belt Population For Diameters Greater Than 1 Kilometer", The Astronomical Journal, 129:2869–2886, 2005 June (link)
Gladman et al., "On the asteroid belt’s orbital and size distribution", Icarus 202 (2009) 104–118 (link)0 -
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That's crazy. Would the gravity of the other stars and the probable black hole at the centre of these galaxies cause some effects or are the gaps even that big that the two would pretty much pass through each other unaffected?
Even though individual stars are unlikely to collide, the collision will effectively destroy both galaxies and create a new one.
At the moment, both galaxies are broadly similar spiral galaxies. Andromeda looks like this:
And the Milky Way is thought to look something like this:
But after the collision is complete and things have had time to settle down, they'll probably have formed a single, miserable looking elliptical galaxy like this:
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. But according to the inverse square luminosity distance relation: 
. So if the distance increases, to maintain constant luminosity we must increase the radius linearly with distance: 
. But then from our separation formula we have: 
and thus: 
.
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