# Kamehameha!

How much energy does it take to destroy a planet?

A planet is held together by gravitational attraction; every part of a planet attracts every other part, with a force that varies inversely with the square of their separation. In order to dissemble a planet, we have to overcome this binding force, and this requires the input of a considerable amount of energy on our part.

We’re going to imagine breaking thin shells off the planet, gradually decreasing its radius until nothing remains. To get a clean answer, we’re going to assume that

• the planet is perfectly spherical throughout its deconstruction
• the planet is of uniform mass density $\rho$

Let’s first evaluate the work we need to do to peel off a shell of radius $R$ and differential mass $dm$.

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To lift the shell away from the planet, we need to counter the force of gravity acting on every part of it. It can be shown using Gauss’ law that the gravitational field a distance $r$ from the planet’s centre is

$\displaystyle \bold{g}=-\frac{GM}{r^2}\hat{\bold{r}}$

where $M$ is a planet’s mass and $\hat{\bold{r}}$ is a unit vector which points directly away from the planet’s centre. This means the force acting on some point object of mass $\delta m$ a distance $r$ from the planet’s centre is

$\displaystyle \bold{f}=\delta m\,\bold{g}=-\frac{GM\delta m}{r^2}\hat{\bold{r}}$

To overcome this gravitational attraction, we have to apply this force to the object in the opposite direction. That is, our applied force must be

$\displaystyle \bold{F}=-\bold{f}=\frac{GM\delta m}{r^2}\hat{\bold{r}}$

Hence the total work we do in moving this object from the planet’s surface to somewhere an infinite distance away is

$\displaystyle \delta w=\int \bold{F}\cdot d\bold{r}$

$\displaystyle \delta w=GM\delta m\int_R^\infty \frac{1}{r^2}dr$

$\displaystyle \delta w=\frac{GM}{R}\delta m$

This means the work done in removing a shell of differential mass $dm$ is

$\displaystyle dW=\frac{GM}{R}\,dm$

It might seem like a trivial statement, going from one line to the next, since I’ve just replaced the mass $\delta m$ of the point object with that of the shell, $dm$. The reason I didn’t do so immediately is the shell is not a point object but an extended body, so it would not make sense to write

$\displaystyle dW=\int \bold{F}_{\text{shell}}\cdot d\bold{r}_{\text{shell}}$

But I hope you will agree that, since the shell can be treated as comprising many point particles, on each of which we do the same amount of work, we can simply sum up the work done on each point to give the total work:

$\displaystyle \sum \delta w = dw$

$\displaystyle \frac{GM}{R}\sum dm=dw$

$\displaystyle \frac{GM}{R}dm=dw$

That was step one; now we have to add up the work we have to do to remove each shell to give us the total work done.

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We have the equation

$\displaystyle dW=\frac{GM}{R}dm$

It’s important to remember that, as we dissemble the planet, both the mass of the planet $M$ and the mass of the next shell to be removed $dm$ will change. That is,

$M=M(R)$

$dm=dm(R)$

What are these explicitly? Well, assuming the planet (and indeed the shell) has uniform mass density $\rho$,

$\displaystyle M=\frac{4}{3}\pi\rho R^3$

$dm=4\pi\rho R^2\,dR$

In each case, we simply multiply the density of the body in question by its volume. For the planet, we use the volume of a sphere of radius $R$, and for the shell we take the volume to be its area $4\pi R^2$ multiplied by its thickness $dR$.

Sticking everything together, we get

$\displaystyle dW=\frac{4\pi G}{3}4\pi\rho^2 R^4\,dR$

We add up these contributions from a shell of radius $R_0$, the planet’s initial radius, to one of radius 0, when the planet is finally dissembled:

$\displaystyle W=\frac{4\pi G}{3}4\pi\rho^2\int_0^{R_0} R^4\,dR$

$\displaystyle W=\frac{4\pi G}{3}4\pi\rho^2\frac{1}{5}R_0^5$

This is not wholly enlightening, so let’s rewrite the planet’s density $\rho$ in terms of its initial mass $M_o$ and radius $R_0$:

$\displaystyle \rho=\frac{M_0}{\frac{4}{3}\pi R_0^3}$

$\displaystyle W=\frac{4\pi G}{3}4\pi\left(\frac{M_0}{\frac{4}{3}\pi R_0^3}\right)^2\frac{1}{5}R_0^5$

which after simplification gives

$\displaystyle W=\frac{3}{5}\frac{GM_0^2}{R_0}$

That’s our answer! It’s sensible to check: does it have the dependence on $M_0$ and $R_0$ we would expect?

Let’s first look at its dependence on the planet’s mass. We find that it is quadratic in $M_0$. That $W$ increases as $M_0$ increases is hardly surprising. There are two effects at work here: first, the greater $M_0$, the more mass we have to remove from the planet, so the harder we have to work; second, the greater $M_0$, the harder the planet pulls on that mass we’re removing, so the harder we have to work.

What about the dependence on $R_0$? It says that, for a planet of a given mass, the greater its volume, the easer it is to dissemble. This makes sense too; if the planet is initially minuscule, its constituent parts, which are very close, exert a powerful gravitational force on one another, making the planet more tightly bound and  therefore harder to break apart. If $R_0$ is greater, the spatially separated parts of the planet are weakly bound to one another, so the planet is easier to destroy.

Just to get an idea of what sort of energy scale we’re talking about, let’s take the Earth for instance.

$M_0=5.97\times 10^{24}\,\text{kg}$

$R_0=6.37\times 10^6\,\text{m}$

$\implies W=2.24\times 10^{32}\,\text{J}$

It’s difficult to convey the enormity of this number; here are some feeble attempts nonetheless:

• $W$ exceeds the total energy output of the Sun per day by a factor of 10
• $W$ exceeds the total kinetic energy of the Moon in orbit as measured in Earth’s frame by a factor of 104
• $W$ exceeds the total annual energy consumption of Earth by a factor of 1012

I encourage you to plug in some values of $M_0$ and $R_0$ of your own.

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The explanation above was thorough, but to a great extent unnecessary. We can end up at the answer much quicker using dimensional analysis. We make a list of all the physical quantities the answer could possibly depend on, and then find the only possible way they can combine to give an answer with the correct units.

What could $W$ depend on? Well, it definitely has to depend on the initial mass $M_0$ in some way, since the heavier the planet is, the more mass we have to lift, and the more tightly bound the planet is. It should also depend on $R_0$, since if, for a a given planetary mass, the planet is bigger, the strength of its gravitational binding is smaller, so the planet will be easier to break apart. We also expect the gravitational constant $G$ to feature somewhere in our answer. So we propose a solution of the form

$W=\kappa G^\alpha M_0^\beta R_0^\gamma$

where $\alpha$, $\beta$ and $\gamma$ are constants to determined and $\kappa$ is a dimensionless constant prefactor (like 7 or $\pi$). For this equation to be viable, each side of the equation must have the same units, or dimensions. To find the appropriate values of $\alpha$, $\beta$ and $\gamma$, we can just write out the units each side must have, according to our ansatz:

$\text{kg}\,\text{m}^2\,\text{s}^{-2}=\left[\text{m}^3\,\text{kg}\,\text{s}^{-2}\right]^\alpha\left[\text{kg}\right]^\beta\left[\text{m}\right]^\gamma$

$\text{kg}\,\text{m}^2\,\text{s}^{-2}=\text{kg}^{-\alpha+\beta}\,\text{m}^{3\alpha+\gamma}\,\text{s}^{-2\alpha}$

Equating the time parts gives

$-2=-2\alpha$

$\alpha=1$

Equating the length parts gives

$2=3\alpha+\gamma$

$\gamma=-1$

Equating the mass parts gives

$1=-\alpha+\beta$

$\beta=2$

Hence

$W=\kappa G^1 M_0^2 R_0^{-1}$

$\displaystyle W=\kappa\frac{GM_0^2}{R_0}$

So we’ve arrived at exactly the same dependence as predicted above. Note that there is no ambiguity here: there is no other conceivable way of combining $G$, $M_0$ and $R_0$ to give you an answer that makes sense. Granted, we couldn’t find the missing factor of 0.6, but this doesn’t matter too much since our answer will be of the right order of magnitude. Furthermore, our assumption that the planet is of uniform density means our previous answer will be unrealistic anyway, so a factor of 0.6 either way doesn’t destroy an otherwise perfect answer.

A final comment: we could imagine carrying out the destruction process in reverse. That is, we could picture a colossal cloud of gas of mass $M_0$ collapsing to form a planet of radius $R_0$. The quantity $W$ is still relevant; it quantifies the internal energy of the planet after it is formed. You can in principle work out the temperature of the planet as a function of $R_0$ – you might like to work out the maximum radius of an iron planet such that its interior is molten.