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Re:Voyager 1 hears the hush of interstellar plasma by willis936 on Tuesday May 11, 2021 @07:58AM (#27116994)
It is E=MC^2 though. 0.1% of the mass is converted into energy.
Re:Voyager 1 hears the hush of interstellar plasma by whatshisface on Tuesday May 11, 2021 @11:09AM (#27118995)
Windup toy cars are also E=mc^2.
Re:Voyager 1 hears the hush of interstellar plasma by prewett on Tuesday May 11, 2021 @12:14PM (#27119935)
E = mc^2 implies that mass (m) is being converted to energy (E), and a windup toy car is definitely not converting mass to energy. Windup cars get their energy from torsional springs, which are E = (1/2) k theta^2.
Re:Voyager 1 hears the hush of interstellar plasma by whatshisface on Tuesday May 11, 2021 @01:05PM (#27120554)
>and a windup toy car is definitely not converting mass to energy

Ah, but it is. E=mc^2 is always true in the reference frame where the object is stationary. (We assume that we are co-moving with the windup car.) A spring under torsion is heavier than an unwound spring.

Re:Voyager 1 hears the hush of interstellar plasma by Chris2048 on Tuesday May 11, 2021 @02:18PM (#27121516)
> A spring under torsion is heavier than an unwound spring

Does it have more (invariant) mass?

Re:Voyager 1 hears the hush of interstellar plasma by whatshisface on Tuesday May 11, 2021 @03:45PM (#27122598)
Yes.
Re:Voyager 1 hears the hush of interstellar plasma by varjag on Tuesday May 11, 2021 @08:09AM (#27117102)
No, it's E=(0.001M)C^2
Re:Voyager 1 hears the hush of interstellar plasma by willis936 on Tuesday May 11, 2021 @08:22AM (#27117190)
That is absurd. It is E=MC^2 where M is 0.1% of the total mass of the reactants.

100% of the mass difference is converted into energy.

Re:Voyager 1 hears the hush of interstellar plasma by varjag on Tuesday May 11, 2021 @08:48AM (#27117430)
The GP was talking in relation to W/Kg, and the output per Kg is definitely not e=mc^2. It helps reading things in the context.
Re:Voyager 1 hears the hush of interstellar plasma by StavrosK on Tuesday May 11, 2021 @10:47AM (#27118707)
You are correct, though it would help to clarify that you're correcting the GP on the "per kilo" part, and not the "energy" part.

I agree, if you're deploying 1 kg of fissile material, only 0.1% of that will be converted to energy. We should talk about energy per mass of fissile material, not energy per mass of "mass difference".