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Are these math equations really the same?

F=ma & E=mc²

If so © is a set rate of acceleration!

Then why is © squired since Einstein him self said

that nothing can accelerate faster than ©?

Wouldn’t that also go for all particles in a nuclear explosion?

If you drop the (c²) and just have © then you can get very close the real

speed / force of the particles moving out from a nuclear explosion.

At (E=mc) with a mass of (4³in. of u-234) you will get a total

change reaction /or evaporation area at ground zero

of roughly a 5 mile radius.

So is (E=mc) the same as (F=ma) just at the quantum level.

Alpha-137:-)

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No they are not the same.

Newton actually said $F = \frac{d(mv)}{dt}$ but this can be simplified to F=ma... but that's kinda irrelevant.

F=ma is about the force required to accelerate an object.

E=mc² is about the conversion of mass and energy and is an equation allowing us to calculate the conversion of one to the other.

c is the speed of light. It is a constant, everywhere, in any situation. It has nothing to do with acceleration.

You cannot change the c because then the equation is incorrect for the units it was intended for.

Also in E=mc² we are dealing specifically with rest mass. If you start accelerating things then we need to use the full equation which takes into account momentum: E² = (mc²)² + (pc)²

Also I don't really follow your post or your reasoning. You seemed to say "assuming it does..." well I'm not going to assume it does. Why don't you show it does rather than making the assumption and moving on.

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No;

I did not assume anything!

I just put the question out after reading the article on

that I gave in my posting.

I was sure to what answers I would get in reply, and I was right.

I know that E=mc² is right.

I just wanted to see what others would say.

Alpha-137

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Yeah:

Light is pure energy and energies nature is squared relative to mass.

The full equation is E=mc2 + pc2, p being kinetic energy.

E=mc2 only works for matter at rest, so really the force it takes to turn inertial matter into motion really has little to do with how much energy a mass contains.

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the link gives a 404. in F=ma m and a are both variables. in E=mc^2 m is the only variable. c is a constant. hmmm... these don't look the same. in eqn.1 we have 2 variables and eqn.2 we have 1 variable. not the same.

Thats true.

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The full equation is E=mc2 + pc2, p being kinetic energy.

Actually, it´s $E^2 = m^2c^4 + p^2c^2$.

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Actually, it´s $E^2 = m^2c^4 + p^2c^2 [/math']. If you wanna go that way it is actually [math] E^2 = m^2c^4 + p^2c^4$.

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Nope. Just take a look at the units to see that E = pc, not pc². Looking at the units is an easy, powerful and usually underestimated quick check if an equation/statement makes any sense at all. In fact, it would also be a good idea for Alpha-137 to take a look at the units and reconsider if the speed of light really is an acceleration.

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you mean it isn't?

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mimefan599 & insane_alien

Have you ever sat in weapons class and had to work out the radius

of ground zero of each of the ten warheads of a Polaris Missile.

Well' date=' I will assure you that none of them will equal the total energy that E=mc² says that there is in mass.

If we did ever make a bomb with that matched the total power of E=mc² it would vaporize roughly 19% of our Solar-system.

Aphla-137[/quote']

no i don't do weapons, but isn't it inly the change in mass thats used since it isn't a total conversion weapon. and ehh isn't it roughly a hiroshima sized bomb for every gram converted. thats not enough to vapourize 19% of the solar system.

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I did not assume anything!
You assumed that the two equations could be linked just because they both had an m in them. This is an incorrect assumption.

The full equation is E=mc2 + pc2, p being kinetic energy.
That's not technically correct, as pointed out already. But I did say the proper equation in my original post so I beat you all! Also p is momentum and not KE.

Well, I will assure you that none of them will equal the total energy that E=mc² says that there is in mass.
That's because not all of the mass is converted into energy. This isn't a flaw in the equation, it's that the value for m you need to use is the total mass which is going to be converted into energy, this is not the whole of the missile. In fact it is a very small amount. In Horishima the Mk I "Little Boy" nuclear warhead had 0.7kg of U-235 which underwent nuclear fission.
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5614 & insane_alien

The Mk I or Little Boy only did lest than 1/10 of it’s’ ability at that time, but today’s

can do about 87% of their ability. But even so today’s 50megiton bombs only uses lest

than a very small fraction of 1% of E=mc² total energy.

It was first pointed out to me in a mini lecture by Nobel winner Dr. Leon Lederman

and also in his book “The God Particle” about midway in the book.

Alpha-137

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mimefan599 & insane_alien

Have you ever sat in weapons class and had to work out the radius

of ground zero of each of the ten warheads of a Polaris Missile.

Well' date=' I will assure you that none of them will equal the total energy that E=mc² says that there is in mass.

If we did ever make a bomb with that matched the total power of E=mc² it would vaporize roughly 19% of our Solar-system.

———

today’s 50megiton bombs only uses lest

than 1% of E=mc² total energy[/quote']

You are making the wrong measurement here. Nobody (with any knowledge on the matter, at least) contends that all of the mass, or anywhere near that, is converted to energy; that happens in matter+antimatter reactions, which fission/fusion are not. If you measure the mass difference, it will account for the energy released.

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You are right it could only happen in an matter + antimatter reaction, I was just trying to

show that we have not gotten even close to the total energy in mass, and we surly do not

want too.

Alpha-137:)

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You are right it could only happen in an matter + antimatter reaction' date=' I was just trying to

show that we have not gotten even close to the total energy in mass, and we surly do not

want too.

Alpha-137:)[/quote']

Who was saying anything about that except you? In any case, fission and fusion do convert a small amount of matter to energy, as does every exothermic reaction.

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mimefan599 & insane_alien

Have you ever sat in weapons class and had to work out the radius

of ground zero of each of the ten warheads of a Polaris Missile.

Well' date=' I will assure you that none of them will equal the total energy that E=mc² says that there is in mass.

If we did ever make a bomb with that matched the total power of E=mc² it would vaporize roughly 19% of our Solar-system.

Aphla-137[/quote']

You like making things up dont you. Nobody who has even an inkling of what they're talking about would say that 100% of any mass inside a nuke is converted to energy. Its not even close, anyone who knows how nukes works knows this.

And vaporize 19% of the solar system? Dont pull idiotic numbers out of your ass and pass them off as true around here, people notice.

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The mass conversion of an H-bomb is connected to going from Lithium deuteride into He. The Li stores the D as a solid. Besides storing the D as a solid at room temperature, the lithium was found to give an extra kick being converted to He along with the D. The mass change is quite small since all the original protons, neutrons and electrons are conserved. Only a tiny mass conversion is needed for the nuclear force change.

Potential energy such as the potential energy in rest mass E=MC2 equals force x distance.

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mimefan599 & insane_alien

If we did ever make a bomb with that matched the total power of E=mc² it would vaporize roughly 19% of our Solar-system.

Aphla-137

Let's check that figure. The Sun converts 5' date='000,000 tons of mass to energy every second. This is considerably more mass than we could ever put in a bomb, so vaporizing 19% of the Solar system is out.

In fact, considering that the gravitational binding energy of the Earth alone is in the order of [math']2 x 10^{32}[/math] joules, it would take the conversion of 2,444,444,444,444 tons of matter to energy to "vaporize" just the Earth alone.

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some how i don't think this guy does a weapons class either. i mean simple physics explains everything here.

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OK; People if we were able to build a bomb to equal E=mc²

To make it as simple as I can we will look at just one particle’s action at first.

This particle will travel at c² = 8.987551787^16 meters per sec.

Convert this to feet per sec. = 2.948671846^17 feet per sec.

Convert this to miles per sec. = 5.584605769^13 miles per sec.

Remember that is just one particle going in one direction.

So this 5.584605769^13 miles = the radius of all of the bomb’s original particles actions.

So this 5.584605769^13 miles x 2 =1.116921154^14 miles in diameter of the original

mass particles.

Now multiply 1.116921154^14 miles in diameter by millions of times for the

chain-reactions that will come about due to all of the nuclear mass that is here on Earth

both man made and natural along with the other planets within this area.

Basically a new star would be made.

Just a part of a mini-lecture by Admiral Rickover to the crew of SSBN-598

USS Gorge Washington in the mid-60’s

PS: Tycho?

At no time did I say that we use 100% of any mass energy. I have only been trying to show just the reverse.

Alph-137:)

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OK; People if we were able to build a bomb to equal E=mc²

To make it as simple as I can we will look at just one particle’s action at first.

This particle will travel at c² = 8.987551787^16 meters per sec.

Convert this to feet per sec. = 2.948671846^17 feet per sec.

Convert this to miles per sec. = 5.584605769^13 miles per sec.

Remember that is just one particle going in one direction.

So this 5.584605769^13 miles = the radius of all of the bomb’s original particles actions.

So this 5.584605769^13 miles x 2 =1.116921154^14 miles in diameter of the original

mass particles.

Now multiply 1.116921154^14 miles in diameter by millions of times for the

chain-reactions that will come about due to all of the nuclear mass that is here on Earth

both man made and natural along with the other planets within this area.

Basically a new star would be made.

This is nothing but mathematics run amok. There is no physical meaning to the calculations.

Just a part of a mini-lecture by Admiral Rickover to the crew of SSBN-598

USS Gorge Washington in the mid-60’s

Unless you present some evidence to back up your story, I will not accept it. I do not believe that Rickover had such a piss-poor grasp of the scientific principles involved, as it runs counter to prior experience (I was an instructor in the navy's nuclear propulsion program for ~five years).

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