Black hole with NO radiation

Question

I've been recently thinking about black holes and...,

...let's say we have a black hole that doesn't radiate anything. Then you will drop this black hole to some solid surface (let's say the black hole has relatively small mass -> $0{,}5$ $\mathrm{kg}$, which implicates very big density and let's also say that this surface is Earth surface). We know that anything below Schwarzschild radius ($R_s=2G_EM_d/c^2$) cannot escape from the black hole. And here's my question: How can I compute how many of surrounding matter can black hole swallow in some given amount of time (for example 3 hours or 2 days)? Or how long does it take for the black hole to swallow $2$ $\mathrm{kg}$ of matter.

Thanks a lot!

Edit: In reaction to the first comment...I would abandon any geothermal and thermodynamical processes. You're currently at Earth's surface and black hole you just made, you recently dropped. Gravitional field is heading towards the geometric center of Earth and that is where the hole is heading (free fall?). Earth has almost none density compared to our black hole, so that black hole will fly threw it like nothing, yet cannot leave Earth because of gravity. That way it will be oscillating. I want just a model of how this could be computed (amount of time in which the black hole can swallow a particular amount of matter). Let's assume ideal conditions.

Answer 1

As a the comments say you are setting up a rather complicated scenario. There are a few factors that make the calculation tricky, and one of the biggest is the rigidity of the earth itself. Imagine for instance that you would create the black hole at the centre of the earth; yes, matter would fall into the black hole, but the speed at which this happened would depend on the "surface tension" of the in-falling matter. The core earth is solid, but the pressure of the center of the earth is immense, and the walls of the just created cavity (by the black hole) would melt and boil once the pressure is relieved, this vavour would then fall into the black hole also. That being said, the BH could release so much energy that it keep matter away (turning it's neighborhood into plasma in the process) and slows accretion a lot.

But now we put the black hole at the surface. Let's make things simple and just assume that everything inside of the horizon gets eaten and anything outside is unaffected. This may sound strange but is (in my opinion) a quite reasonable approximation since the black hole is small and thus has a very weak gravitational field at long distances.

As a fist approximation we can also assume that the black hole obeys Newtonian dynamics, that is $F=ma$ (I know this is now strictly true but it gives us something to work with). In this case the acceleration at earth's surface is just $a=g$. As the black hole tumbles towards the earth's centre it it will increases in mass (it literally eats it's way through the earth). However, since $a=F/m$ the acceleration will not depend on the mass of the (still small) black hole, but only on the mass of the earth at smaller radious. Assuming constant density (which is wrong) the acceleration would go as $$ a = g \frac{r}{R} $$ where $R$ is the radios of the earth. As you say the black hole will oscillate between the earth surface through the core and out again.

EDIT: Thanks to @KevinKostlan for the comet on the earth's core. Corrections have been made.

Answer 2

A tiny black hole with $M=0.5$ kg has a theoretical time to evaporation by Hawking radiation of $\sim 10^{-17}$ s. It won't accrete anything, it will disappear in a puff of hard gamma rays.