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Author Topic: This might be a little different: If E=Mc^2, then does c^2 = E/M ??  (Read 5286 times)

Offline the_big_m_in_ok

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Take the famous Einstein equation and divide both sides by 'M'.   You get c^2 = E/M.
       The story problem and question...
Dividing E = Mc^2 by M implies more and more propulsive force might ger a rockship to the square of lightspeed?   As an alternative:  c = (E/M)^1/2.   The square root of energy addition the preceeding equation only gets one the the speed of light.
 
Einstein did say, theoretically, it was impossible?   What do you think?   Me?   This is a purely conjectural question.   I really don't know if it's possible, but are there any suggested methods to achieve this feat?
 
--Lee
 

Free Energy | searching for free energy and discussing free energy


Offline s3370389

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Re: This might be a little different: If E=Mc^2, then does c^2 = E/M ??
« Reply #1 on: September 07, 2014, 03:09:00 PM »
To put this into perspective for you the_big_m_in_ok, you would require approximately 9 exa Joules of energy to accelerate a 100 {kg} mass to the speed of light. That is 9x10^18 {J}. According to wikipedia the total world energy consumption in 2006 was 8.3 exa Joules.

I think I will defer to Einstein's standpoint and assume that reaching the speed of light is not possible. If it is possible, it will be a long long time before we are able to generate and store that amount of energy to be used in a space craft. Probably never.

Offline the_big_m_in_ok

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Re: This might be a little different: If E=Mc^2, then does c^2 = E/M ??
« Reply #2 on: September 09, 2014, 10:24:24 PM »
To put this into perspective for you the_big_m_in_ok, you would require approximately 9 exa Joules of energy to accelerate a 100 {kg} mass to the speed of light. That is 9x10^18 {J}. According to wikipedia the total world energy consumption in 2006 was 8.3 exa Joules.
       I think I will defer to Einstein's standpoint and assume that reaching the speed of light is not possible. If it is possible, it will be a long long time before we are able to generate and store that amount of energy to be used in a space craft. Probably never.
       Oh, yes, it does take a great deal of power to get the mass to lightspeed.
Okay, think of it this way:  Using differential equations, relating, say, acceleration that's been generated by electromagnetic field strength,  and/or/to  the power required to produce said fields... and the kinetic energy contained in the mass of the object being accelerated---is there more or less efficiency in one method or another to achieve the speed required?
       There's something else?   Reactionless, inertialess drive units to accelerate mass highly efficiently?   (I admit the equations above may not apply for these types of propulsion units?)

That's what I can come up with in a few hours' time of theoretical mental concentration.   What is your opinion, anyone?

--Lee

Offline the_big_m_in_ok

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Re: This might be a little different: If E=Mc^2, then does c^2 = E/M ??
« Reply #3 on: September 12, 2014, 12:29:00 AM »
That's what I can come up with in a few hours' time of theoretical mental concentration.   What is your opinion, anyone?

--Lee
       There's something else, finally:
I looked at the equation in the thread title and realized I think I originally wrote it backwards...


Offline the_big_m_in_ok

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Re: This might be a little different: If E=Mc^2, then does c^2 = E/M ??
« Reply #4 on: September 12, 2014, 12:34:11 AM »
       There's something else, finally:
I looked at the equation in the thread title and realized I think I originally wrote it backwards...

Rather than C^2 = E/M , I should have wrote:   E/M  =  c^2
       The theory as a story problem...Adding more and more energy as thrust to mass in space will result in an eventual astounding rate of speed, namely, the speed of light squared.   That is, regardless of the possible lightspeed barrier at the speed of light.

--Lee
 

Free Energy | searching for free energy and discussing free energy

Re: This might be a little different: If E=Mc^2, then does c^2 = E/M ??
« Reply #4 on: September 12, 2014, 12:34:11 AM »
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