Subject:
|
Re: Hypothetical design question
|
Newsgroups:
|
lugnet.space
|
Date:
|
Tue, 24 Jun 2003 02:56:34 GMT
|
Viewed:
|
603 times
|
| |
|
|
In lugnet.space, David Laswell wrote:
| |
In lugnet.space, Jonathan Mizner wrote:
| |
If I understand physics correctly, it doesn’t make a difference whether it
is the ship traveling at .9 c or the hydrogen atom. The energy released is
the same.
|
That’s essentially correct. Two cars hitting each other head-on at 30MPH is
effectively the same as one car hitting a stationary vehicle at 60MPH.
Obviously the two accidents would not be perfect mirror images of each other,
but the level of damage will be similar between the two.
| |
Thus, that atom is effectively dealing far, far more energy than 1.5E-10
watts.
|
How do you figure? It’s the smaller mass that determines the total energy
generated by impact, not the larger mass. A single hydrogen molecule
traveling at .9c does not cause more damage to a Star Destroyer than it would
to an X-Wing just because the ISD is bigger.
|
I’m not sure on the physics, but we can assume either to be interchangeable; the
ship impacting a motionless particle at .9c, or a particle at .9c impacting the
ship. Kinetic energy is derived from mass and velocity. So it is, in effect,
a molecule traveling at .9c, and with the kinetic energy of such (which is far
more than 1.5E-10 watts).
| |
| |
Not to mention it will also have significant “drag” effects (not that it
matters if it vaporizes your ship first).
|
One individual hydrogen molecule will not cause much drag on anything large
enough to fit a human inside, but hitting a whole mess of them (like
travelling through a nebula) will result in accumulation of drag. It’s kinda
like hitting a brick wall one brick at a time. Without any means of
increasing your speed, it will eventually bring you to a standstill.
|
Exactly. But except for the issue Shawn mentioned of deflection of particles
(which wouldn’t deflect the kinetic effect very much, but anyways), it doesn’t
matter if aerodynamic or not.
| |
| |
Remember, E=mc^2 is only accurate for matter at rest.
|
I thought the whole idea of the Theory of Relativity was to predict the
behavior of matter as it approaches the speed of light, not as it sits on the
living room couch scratching itself and eating potato chips.
|
E=mc^2 is only accurate standing alone when applied to matter that is
non-stationary. At velocity, the equation is E=((mc^2)/(1-V^2/C^2)).
In essence, energy is equal to kinetic energy + mc^2. For more detail, check
out http://www.btinternet.com/~j.doyle/SR/Emc2/Derive.htm#Kinetic%20Energy.
|
|
Message is in Reply To:
 | | Re: Hypothetical design question
|
| (...) That's essentially correct. Two cars hitting each other head-on at 30MPH is effectively the same as one car hitting a stationary vehicle at 60MPH. Obviously the two accidents would not be perfect mirror images of each other, but the level of (...) (21 years ago, 23-Jun-03, to lugnet.space, FTX)
|
57 Messages in This Thread:
  
 
- Entire Thread on One Page:
- Nested:
All | Brief | Compact | Dots
Linear:
All | Brief | Compact
|
|
|
|
