21-Kiloton Paperclip

Physics geeks are quite familiar with the implications of Einstein’s famous equation, E=mc². Firstly, energy and matter are different guises of the same thing. Secondly, that the conversion rate between them is astronomical.

To kick off the week, let’s explore that well known equation.

The following is a slightly revised version of a Substack post I published in August of 2025. (I’m slowly reposting some of my Substack posts here since I’m no longer active there.)

Units

In science (and often in life), things come in “units”. For instance, one might have a half pint or half liter of beer (or ice cream, if you prefer). Note that, despite having the same number (½ aka 0.5), they have different units of measurement — pints and liters, respectively. So, they are different amounts of beer (or whatever).

[Half a liter is 1.05669 pints. Half a pint is 0.236588 liters. So, a pint is roughly a half liter.]

Units can vary considerably. We can measure distance in centimeters, inches, yards, miles, or light years. What matters is using the same units (and appropriate units) when comparing or discussing distances.

[Remember that, despite “year” in the name, light-years is a measure of distance — how far light travels in one year: 5,879,000,000,000 miles or 9,461,000,000,000 kilometers.]

Units can be compounded. We can measure speed by miles per hour and vinyl record spin rates by revolutions per minute. The keyword “per” tells us the units are the rate of something (the rate of speed or revolutions in this case).

Getting units right can be important. A space mission to Mars was lost because of a misaligned use of units.

Some Foreshadowing

A Joule is a standard unit for measuring energy. One Joule isn’t much energy. It’s about what it takes to lift an apple one meter.

Expressed in terms of the International System of Units (aka SI), Joules are:

\displaystyle\left[\text{Joule}\right]=\text{kg}\cdot\text{m}^{2}\cdot\text{s}^{-2}=\frac{\text{kg}\cdot\text{m}^{2}}{\text{s}^{2}}

They are expressed in terms of kilograms times square meters per square second. More generally, mass times distance over time squared. Which is a weird unit to wrap one’s head around. Suffice to say it puts energy in terms of mass, distance, and time.

As an aside, 1 Joule = 1 Watt-second.

Einstein’s E=mc²

Have you ever wondered what E really is in that famous equation? “Energy”, certainly, but what kind of energy? Horsepower? Watts? Kilotons?

We can figure it out by considering the units on the right side of the equation. The m stands for (rest) mass, and the c for the speed of light. This is science, so we’re using SI units that express mass in kilograms and speed in meters per second.

We’re squaring c, so:

\displaystyle\left[{mc}^{2}\right]=\text{kg}\cdot\left(\!\frac{\text{m}}{\text{s}}\!\right)^{2}=\text{kg}\cdot\frac{\text{m}^{2}}{\text{s}^{2}}=\text{kg}\cdot\text{m}^{2}\cdot\text{s}^{-2}

Which is what Joules are measured in, so E in that equation is the energy of the rest mass of the object in Joules.

(Note that square brackets are one way of notating “has units of”.)

The conversion rate is ferocious because c² is very big (c is big to begin with):

\displaystyle{c}^{2}=\left({299,792,458}\;\frac{\text{m}}{\text{s}}\right)^{2}={89,875,517,873,681,764}\;\frac{\text{m}^{2}}{\text{s}^{2}}

So, one unit of mass — whatever that is; nominally a kilogram — converts to almost 90 quadrillion units of energy. Mass contains a lot of energy.

How much energy is in a paperclip?

Science Geek that I am, I sometimes carry a paperclip so I can tell people I have a small atom bomb in my pocket.

The only trick would be converting every atom of it to energy. The only way we know of that will do that is combining it with an antimatter paperclip (which would double the yield, as both are converted to energy).

But let’s say we could convert one. A paperclip weighs about a gram, which is 0.001 aka 1×10⁻³ kilograms. So:

\displaystyle{1.0}\!\times\!{10}^{-3}\;\text{kg}\cdot{89.875}\!\times\!{10}^{15}={89.875}\!\times\!{10}^{12}\;\text{Joule}

That’s a pretty big number, but it’s a hard one to put in context. Since one Joule is one Watt-second, we know our paperclip can generate almost 90 tera-watt-seconds of electrical power.

Still a bit abstract, so about this: One ton of TNT is 4.184 giga-Joules, so:

\displaystyle\frac{{89.876}\times{10}^{12}\;\text{Joule}}{{{4.184}\times{10}^{9}}\;\text{Joule}/\text{ton}}={21.480}\times\!{10}^{3}\;\text{tons}

Our paperclip is a 21-kiloton device. As a comparison, Little Boy was about 15 kilotons, and Fat Man was, in fact, 21 kilotons.

So, our paperclip has the energy equivalent of that second bomb. The one we dropped on Nagasaki. In that explosion, about a paperclip’s worth of matter (plutonium) was converted to energy.

[The first one was such a simple device — essentially two just sub-critical hunks of uranium that we slammed into each other — that we didn’t even need to test it. It was the second one, the plutonium implosion bomb, that required careful engineering and testing. The technology for that become one of our key military secrets.]

What about a full kilogram? If it could be fully converted, what does it yield?

Simple, it’s just one-thousand times as much:

\displaystyle{1.0}\;\text{kg}\cdot{89.875}\!\times\!{10}^{15}={89.875}\!\times\!{10}^{15}\;\text{Joule}

And therefore:

\displaystyle\frac{{89.876}\times{10}^{15}\;\text{Joule}}{{{4.184}\times{10}^{9}}\;\text{Joule}/\text{ton}}={21.480}\times\!{10}^{6}\;\text{tons}

So, a kilogram of anything (even feathers) has 21 megatons of energy. Or about 90,000 tera-watt-seconds (aka 25 tera-watt-hours).

Either way, a lot.

One More Thing

The equation everyone knows, E=mc², is actually only part of the full equation Einstein derived for Special Relativity. The full thing looks like this:

\displaystyle{E}=\sqrt{(mc^{2})^{2}+(pc)^{2}}

Recall that m is the rest mass of the object. In the above, p is the object’s momentum, which is its velocity (in meters per second) times its (rest) mass (in kilograms). But if an object isn’t moving (relative to some observer), then its momentum is zero, which makes pc² zero, so the equation reduces to:

\displaystyle{E}=\sqrt{(mc^{2})^{2}+{0}}=\sqrt{(mc^{2})^{2}}={mc}^{2}

And hence the equation we all know and love. It applies only to objects at rest relative to us. When they’re moving, they have more energy (hence the utility of bullets).

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As a reader exercise, use your weight in kilograms to see how much energy you contain. Here’s a possibly helpful hint: 100 pounds is 45.36 kilograms, which is over 4×10¹⁸ Joule — almost one giga-ton.

Stay energetic, my friends! Go forth and spread beauty and light.

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About Wyrd Smythe

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The canonical fool on the hill watching the sunset and the rotation of the planet and thinking what he imagines are large thoughts. View all posts by Wyrd Smythe

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