Your question leaves out two rather important variables to give accurate answers. You've given us a size, but not a density or shape. Because orbital decay is related to the loss of an orbiting objects potential energy and its potential energy depends on its mass and how fast the energy is lost will be related to its shape.
For a more step by step explanation you can read through the math of the Wikipedia explanation here (https://en.wikipedia.org/wiki/Orbital_decay), its actually pretty good explanation which avoids getting too deep into the weeds.
This is pretty abstract stuff talking about dimensionless values and such, but it gets a lot more obvious what's going on when you focus just what aerodynamic drag is caused by. Atmospheric drag on an object, regardless of velocity, and even for the tenuous wisps of molecular gasses above 400km altitude, is determined by the ballistic co-efficient (https://en.wikipedia.org/wiki/Ballistic_coefficient) which can be understood as (( how dense is the object * how long is the object ) / an special adjustment factor for the shape of the object ).
So lets get to how this affects your question. You've asked about a 1cm piece of debris, the easy way to show how this is affected is probably to show a little table for comparison. Using average density from wikipedia and the basic drag coefficient values https://en.wikipedia.org/wiki/Drag_coefficient lets get some rough ballistic coefficients for various hypothetical bits of debris of about 1cm in size.
Material |
Shape |
Approximate Ballistic Co-efficient |
Frozen Water |
Sphere |
18.3 |
Glass |
Cubic |
23.8 |
Mylar Insulation |
Square (Flat) |
0.0029 |
Steel |
Square (Flat) |
7870 |
Steel |
Sphere |
175 |
Steel |
Cube |
78 |
Steel |
Short cylinder (Imagine the sheered off end of a broken bolt 1cm long) |
71.5 |
Now these are extremely rough numbers. But you can see from how large the range is depending on the density of the materials when you look at the flat square and how massive the difference is between light mylar insulation and steel, and then how simply changing the shape of the steel debris in the last few lines from our square to sphere, the cube, and then to our hypothetical chunk of a broken bolt. The changing shape has a large difference in the final ballistic coefficient and this is what drives how quickly our object will shed its kinetic energy and thus how fast its orbit will decay. When you look at how this plugs into the math on orbital decay you can see this parameter means that for a given size of debris, 1cm in the case of your question, without knowing what its made of and having a better idea of the shape, you can't analytically predict ahead of time how long it will take to decay. When you add the additional factor of how different the mass will be between a thin flat piece of broken of sheet, compared something like a cube or sphere or cylinder this further magnifies how large of a range you are looking at.
When people talk about approximate times for things to decay they are often just looking at the statistical averages of real world decay times. One of the best examples is cubesats which because they have a known shape and all fit in a fairly common total mass can be approximated rather well, most 1U 2U or 3U Cubesats orbiting under 500km should re-enter in under 25 years based on modelling such as https://iopscience.iop.org/article/10.1088/1742-6596/641/1/012026/pdf But random space junk is not uniformly shaped and its material is usually not well known for the smallest pieces, consequently people just fall back to taking the tracking information we do have and building statistical approximations which can be used as a best guess, leading to approximations like the one put out by UNOOSA, referenced by one of the other answers here https://space.stackexchange.com/a/55995/19695