🚀 go-pugleaf

In article <3364189B.38B8@nbnet.nb.ca>, lonewolf@nbnet.nb.ca wrote:
>Jack Jiang wrote:
>
>> You guys are confused by the �reference FRAME s� problems.
>> 
>> A similar question is: If a meteoroid having mass m and velocity v, hits on
>> Earth having mass M. What is the amount of kinetic energy caused by this
>> collision?

Thank you, I should ask: What is the amount of energy caused by this 
collision?

>> 
>> According to meteoroid�s frame, the answer is 1/2 Mv ^ 2. But, this is a crazy
>> amount and obvious false!
>> The true answer should be about 1/2 mv ^ 2, based on Earth FRAME centered at
>> Earth�s center. Every body agrees?

I said *ABOUT* 1/2 mv ^2 . The difference is negligable, OK?

>
>Nope.
>
>First of all kinetic energy is not caused by the collision.  Kinetic energy
>already exists prior to the collision and is changed into something
>else.

Kinetic energy already exists? Well, that depends...
What if the meteoroid did not hit anything, and you sit on another galaxy 
receding from Earth at same speed as that meteoroid, so you see that meteoroid 
is stationary. In your observation, that meteoroid has zero Kinetic energy 
because speed is zero. Right?

>
>Furthermore the *total* kinetic energy in the collision is 
>
>             .5mv(1)^2 + .5Mv(2)^2
>
>M is mass of the Earth, 
>v(1) is the velocity of the meteorite
>V(2) is the velocity of the Earth
>
>Obviously v(1) and v(2) are measured in the same reference plane in
>order to account for a head on collision as opposed to a "catch up"
>collision.
>
>There is no "frame of reference" in a situation like this.  The energy in
>the system is cumulative, although of course the effect on the Earth is
>negligable.
>

You are right! So, I said *ABOUT* 1/2 mv ^2 because 1/2 Mv(2)^2 is negligable.


Jack z.c. Jiang
Email me at jiang400@themall.net
Free article: "New Model of Universe" is in 
http://www.themall.net/~neocosmo/