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 kinetic energy and its kinetic energy depends on its mass and how fast this will be lost is 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.