Surface-area/volume ratio and physics (fancy speak for science)

Ok so if a cube has 6 1cmx1cm sides, the surface-area will be 6cm^2 and the volume will be 1cm^3 (the ratio will be 6:1)
Now, I'll do the exact same thing, but in mm... 6 10mmx10mm sides means a surface-area of 600cm^2 but the volume will be 1000mm^3 (giving a ratio of 3:5).
I have a feeling that this has to do with powers of 10, but somebody explain it to me... Please? ;P
I had a similar problem when I was reading something about an object (fictional) that would ^3 the force of the strike. I thought "that doesn't make sense if you ever use unit conversions, you would have to have a standard of measurement and keep it." As an example if you put 1 newton of force into the object, you would get one newton out. If you convert that to nanonewtons, you are putting in 1000000000 of force, but getting out 1E27 nanonewtons, or 1E18 newtons of force... tf? For the power input output problem, I might be able to explain it by the formula/conversion (when you convert N to nN you increase by 1E9, you increase that number by ^3. After the ^3, you can't immediately change back to N because of the ^3 (pls keep in mind I'm just typing words, I don't know whether or not they're right), you have to reduce it by ^-3 again (my math teacher explained it a lot better with conversions using volumes and surface-areas for US-customary (it went along the lines of 1^2 foot isn't 12^2 inches, it's actually 144^2 inches))).
Somebody tell me everything I did wrong in this YouTube comment, b/c I looked up the formulas and probably messed something up:
MATH TIME!!! (Before we start, I'll start with four givens, the ball is ~*1/8 inches in diameter*, pi is 3.14, not 22/7, the density of the stainless steel used is 7.85 g/cm^3 (or 4.54 oz/in^3), and the ball does NOT change the axis of rotation
First let's find the circumference using 2 pi r
2(3.14)(0.0625)
6.28(0.0625)
0.3925 inches (this is how far the ball will travel for each rotation)
Now that we have the circumference, we have to find the RPM
9,800(60)
588,000 RPM (how many times the ball will travel the circumference in 1 minute)
To find the ROTATIONAL SPEED of the ball, the formula is rate (RPM/how many cycles are completed per minute/588,000) multiplied by distance (circumference/the distance traveled each cycle/0.3925 inches)
588,000(0.3925)
230,790 inches per minute
19,232.5 feet per minute
WOW that must take a lot of energy right?
No.
Volume is (4/3)(3.14)(r^3) You can do the math (probably incorrectly ;p), the only variable you have to find out is r^3, which is 2.44140625 E-4
The volume ends up being 1.0226538585904274 E-3 inches^3 (I'm simplifying it to 1.022 E-3 inches^3)
The mass is density by volume
1.022 E-3(4.54)
0.00463988 oz is the mass
To find the Kinetic energy of rotation (KEr = I * ω2 / 2), we have to know the moment of inertia (the mass and distance from axis) and angular velocity
Moment of inertia ends up 3.3149728397292E-10 kg/m^2
Distance from axis and mass have been converted to metric
And the kinetic energy of rotation is:
0.62843717247491 Joules which, to put in perspective, is about the same amount of energy as 0.23900574 Calories (big).
P.S. I didn't factor in friction or anything other than accelerating the ball in a vacuum.

Enough math, I have a physics (fancy science) question, if the speed of light (and the EMS in general) decreases when travelling through matter, what happens when it travels through anti-matter?
(for the pic, the biggest to smallest is galaxy, solar system, planet, moon, satellite, with arrows showing direction) Let's say there are 2 galaxies heading toward each other. In the galaxies, there are 2 solar systems, both orbiting their galaxy so that they are heading straight toward each other for a while. In those 2 solar systems, there are planets orbiting and heading straight for each other. The planets have moons etc. when you get to the smallest level, probably either a moon or satellite for the moon, wouldn't those two be going faster than light (relative to each other)?