Gravitation force vs. Gravitational Constant
I recognize that I am splitting hairs (or perhaps in this case atoms) describing the way that objects fall, but I feel that I need to point out, again, that gravitational force (FsubG) and the Gravitational Constant (G) are not the same. FsubG is a function of G.
Implying that the small difference in weight between two objects relative to that of the earth makes differences if FsubG negligible is an incorrect assumption. We are not comparing the weight of the object compared to the Earth (which I agree is incredibly small), we are comparing the weights of the objects compared to each other.
To calculate FsubG:
FsubG = (G*M1*M2)/R^2
Assume G, M1 (mass of earth), and R (distance) to be constant
Comparing F1subG to F2subG would reduce to
F1subG / F2subG = M(hammer)/M(feather)
This would be a function of weights only, which obviously would show that the hammer would fall VERY MUCH faster than the feather.
Therefore, if we assume that the hammer weights at least 10 times more than the feather we should subsequently see the hammer hit the ground much earlier than the feather. The video I linked of the astronaut on the moon testing the Galileo's hypothesis shows this is not the case. Even divding by 6 to account for the moons lesser gravity a noticable delay in the drop of the feather should be observed.
(again the link:
http://www.youtube.com/watch?v=5C5_dOEyAfk)
Please read the Physics website link below if two PhDs write that they fall at the same rate then that's what I'm sticking with. If someone else can show the math and prove it then go for it. I will definately entertain any challenges.
Physlink.com: Do falling objects drop at the same rate?
http://www.physlink.com/education/askexperts/ae6.cfm
To continue to beat a dead horse:
Why is the rifle projectile more effective on a human target? I understand rifle rounds travel faster, but what does that mean to the target?
Generally, because they travel faster and have more kinetic energy behind them. Actual terminal effects are determined by bullet shape, weight, velocity, composition, etc. It is a lot more complicated than pure kinetic energy, but generally speaking that's what does it.
So I understand that a ("standard") rifle is considered more effective than a ("standard") handgun. Say we compare a 9mm to a 5.56: What about a .308, why would that be more effective?
Again, generally speaking case size (i.e. powder charge), bullet weight and diameter, and velocity (which is the major contributor to KE) affect terminal outcome. KE is a good "rule of thumb", but it should be taken in measured doses when bullet weight, range, target, etc. are considered.
Why are some rifle rounds considered adequate for varmints, some for large game, and some for dangerous game? What makes the (more powerful?) rounds more effective?
Honestly, the question has to deal more with the type of bullet generally loaded for a specific cartridge rather than the cartridge itself. Broadly speaking use a fragmenting bullet for varmits, an jacket expanding bullet for large game, and a very heavy solid bullet for dangerous game. This has to do with terminal ballistics more so than external ballistcs. Generally, bore size increases on a gradient between game types, and likewise case size. Therefore, KE usually increases with larger sized cartridges, which follows the rule of thumb above.
Regarding relativity:
Keep in mind, you also age more slowly while you're driving, because relative time slows down as your velocity increases. It's just not by any appreciable amount while moving at terrestrial speeds (even when on commercial flights).
I could go so many directions with this, but I'm not going to. Wait, yes I am. To expand on the above: "When two observers are in relative uniform motion and uninfluenced by any gravitational mass, the point of view of each will be that the other's (moving) clock is ticking at a slower rate than the local clock." This of course doesn't apply to anything that we're discussing. Don't forget to blame all your misses on the DeBroglie wavelenth of the bullet.
