I am trying to find the following:
How many pieces of toast would you need to make a black hole?
From what I've learnt so far I need to find an equation for the compression force the massive amount of the intergalactic pile of pieces of toast would exert on itself to be able to compress it to within a Schwarzschild radius. As you can tell, this is not an easy or normal question.
Any help is appreciated.
Are you asking what is the pressure inside a gravitating body like a star? If so, that's a hard question because you need to know the equation of state for the object. A quick Google for something like "pressure inside star" will find you lots of helpful articles. For example I found this article, which looks pretty good.
Re your question about the black hole, the simple answer is that an object becomes a black hole when it shrinks to a size less than it's Schwarzschild radius. However I'd guess your question is a bit more general than that e.g. at what mass will a star shrink to the Schwarzschild radius under it's own gravity? Again, this is a hard question. Not only do you need to know the equation of state, but the shrinking star undergoes several phase transitions as it collapses, and the equation of state changes at each phase change. You might want to have a look at the Wikipedia article on the Tolman–Oppenheimer–Volkoff limit for more info, though be warned that you're setting out on a hard path!
Well, I can promise you that if you bring two solar masses worth of toast together, you'll probably eventually get a black hole!
I would prefer to set the density equal to the density of a black hole defined by the event horizon. There's much you can read online about the Schwarzschild radius, but in a naive sense it's the radius of the event horizon.
$$ r = \frac{2 G M }{c^2} $$
Then the density is also defined very trivially.
$$ \rho = M/V$$
$$ V = \frac{4 \pi r^3}{3}$$
Combine those equations, and solve for $\rho$. This number should be set equal (in my quick-and-easy approach) to the density of toast. I will assume this is equal to the density of normal enriched bread.
$$ \rho = 144.77 \frac{g}{\text{metric cup}} = 579.08 \frac{kg}{m^3} = 0.58 \frac{g}{cm^3}$$
That is about the specific gravity I predicted. Indeed, toast floats. So we need to solve for mass in the above equations with $\rho$ known. I will spare you the details. I found this answer:
$$ M = 3.5 \times 10^{38} kg $$
Yay, we have a giant meaningless number. Let's call upon the fact that a piece of bread is about $28 \text{grams}$. That allows us to calculate the number pieces of toast needed to make a black hole.
$$ N = 2.1 \times 10^{16} moles$$
I went ahead and divided by $6.022 \times 10^{23}$. Why? Because today is national mole day!! In order to make a black hole, you need a LOT of moles of toast. It's not quite a mole of moles, but it's not too far from it. It's roughly $(mole)^{2/3} \times (mole)$. That means, if you had a mole of moles of toast arranged regularly in a cube, the moles of toast on one surface would be sufficient to make a black hole.
The discussed $mole^{2/3}$ of moles of toast, however, could not form a Molar Eclipse (which is the theme of this year's mole day), because the radius of the black hole of toast is $3.5 AU$. If the toast was centered about the current location of the sun, it would include Earth's entire orbit, Mars, the asteroid belt, but not Jupiter.
Okay, well you could technically make a black hole with less toast, since once it was larger than a few $100 km$ in diameter it would certainly start crunching the toast, eliminating the empty space. With enough toast, it should turn into a dense plasma, but that's not how I interpreted the question, since toast turned into plasma is no longer toast. I'm thinking that the entire inner solar system is instantly filled 100% with toast. According to an external observer, both the toast and all of Earth are headed toward the (eventual) central singularity faster than the speed of light.
