E=MC2 my friend.
If the M is low, the C must be high to do that damage. So if this 30meter wide object (There isn’t really a 3 dimensional measure at that range.. so it may be a lawn dart or it may be a dime) is going fast enough, it’ll have the E to release.
Imagine diving into a pool from the top of the building. The pool itself has surface tension and as people say “it’ll be like hitting pavement” (It won’t) but imagine instead of just falling from a building you’re shot at the water so fast that the heat of your body touching the water causing it to boil and vaporize in front of you (Super cavitation).
A human body would be travelling at a few tens of thousands of miles per hour, which is about an asteroid of this Apollo type is doing right now. Only it’s about 100 feet across. When it hits the atmosphere it will super heat the 60 some miles of atmosphere we have over our heads and the cavitation of the air in front of it will make a boom easily equivalent to an atomic bomb of one type or another.
Hits water ? it could cause a tsunami like the one that hit Oceania a few years ago. Hits land ? Tunguska to the extreme. The tunguska event (If it was a meteor) never reached the ground. The air under it became so full of energy that the asteroid itself puffed into dust. (There is no crater, only a focal point of energetic explosion)
This is really, really close. I mean at .04 LD, if the accuracy is even in that margin, this is within geo-stationary equipment range.
“This is really, really close. I mean at .04 LD, if the accuracy is even in that margin, this is within geo-stationary equipment range.”
Yea .044 is real close. So close Im surprised we are not hearing more about it
Isn’t C constant?
that was actually explained really well in layman terms.
Not easy to do. Thanks :)
F=MA
That is for objects at light speed, for bullets and such, it is E=.5MV2