Influence of Schwarzschild radius and event horizon of ordinary objects As far as I know, every body (with mass) has a Schwarzschild radius and therefore an event horizon. Thus, take an orange with radius $R$ and mass $M$ for example, then the Schwarzschild radius is given by:
$$R_S = \frac{2 G M}{c^2}$$
but what happens when I cross this radius let's say by cutting it in half with a knife?
And in the orange, are there particles that do not experience moving time?
And what happens inside this pretty small but existing inner ball (inside the event horizon)?
Thinking about these questions, it seems counterintuitive to be so, but using the physics the way I understand them, it does make sense.
I hope somebody can point out my logical error.
P.S.: I am new to physics and just a little bit confused.
 A: The Schwarzschild radius as an event horizon only applies if all the mass is interior to that radial coordinate. Thus your orange does not have an event horizon at its centre.
In other circumstances, the Schwarzschild radius is simply a parameter that appears in the Schwarzschild metric, which applies outside any static, spherically symmetric mass distribution.
It is useful, because GR effects will become strong as the radial coordinate of a body approaches the Schwarzschild radius. i.e. When we cannot assume that $r \gg r_s$. Of course for objects other than black holes the surface of the object is encountered before something can get very close to the Schwarzschild radius.
A: When you calculate the Schwarzschild radius $R_S$ of your orange, then you have the radius of the event horizon if there was a black hole with the mass of your orange. It has nothing to do with your orange itself.
Your orange is not a black hole and its radius R is considerably larger than the Schwarzschild radius of a black hole of the same mass, $R >> R_S$.
There is no event horizon inside your orange and you also cannot cut it in half, i.e. nothing spectacular will happen, unfortunately.
When people calculate "the Schwarzschild radius of an everyday object", then this is done merely to get an intuitive understanding to what incredibly small size you have to compress something if you wanted to turn it into a black hole: if you compressed your orange into a ball with radius $R_S = \frac{2G M_{orange}}{c^2} = 3\times 10^{-28} m$ (if your orange weighs $M = 200g$), it would be a black hole.
Compare this to the size of an atomic nucleus, whose radius is of the order of $\approx 10^{-15}m$, to see how small your orange would need to be in order to turn into a black hole.
