1
$\begingroup$

I wanted to check the validity of an expression for the energy required to destroy an asteroid.

Assuming that the asteroid is spherical, the gravitational binding energy can be given as

$U_{GBE} = 3/5 * GM^2/ R$.

However, this is only the energy to remove aggregates away from each other. There is a required energy to destroy the asteroid in the first place where I've assumed that by breaking a certain number of bonds, the destruction energy can be given as

$U_D = E (kJ/mol) * N * n (mol)$

where $E$ is the enthalpy of the bond (or energy required to break an atom-atom bond), $N$ is the number of atom-atom bonds and $n$ is the number of moles. Would the addition of these two energies therefore be a good way to approximate the energy needed to destory an asteroid?

$\endgroup$
2
  • 1
    $\begingroup$ (1) Destroying an asteroid isn't as useful as deflection; lots of little rocks hitting us is still bad. (2) Your addition-of-energies idea sounds reasonable, but the formula for $U_S$ isn't just multiplication because there's more than one kind of bond (ionic, covalent, metallic, hydrogen, van der Waals, each of which can be subdivided); the details vary by composition. $\endgroup$
    – J.G.
    Commented Apr 12, 2023 at 13:39
  • $\begingroup$ A rough heuristic is that the chemical binding energy must be substantial relative to the GBE, otherwise the body would be spherical. $\endgroup$
    – PM 2Ring
    Commented Apr 18, 2023 at 13:40

1 Answer 1

1
$\begingroup$

These are good thoughts, but of course everything is more complicated in practice.

The gravitational binding energy is right. After you break it up into separate atoms, that is the energy needed to move each atom an infinite distance from all the others.

Suppose you break it up into small rocks. The energy needed to separate each rock an infinite distance from the others isn't much different. You can see that by calculating the binding energy of a rock and multiplying by the number of rocks.

The energy needed to break bonds is a reasonable thought. But some asteroids are snowballs. Some are loose conglomerations of rubble. Some are solid nickel-iron where the composition is like the core of a planet. Some are solid rocks like the crust of a planet.

If you think of energy per bond and how many bonds, you are implicitly thinking about a material like a perfect crystal, where all the bonds are the same. Even in a crystal, imperfections determine the strength. A perfect crystal of iron is much, much stronger than iron we use every day.

Another problem is how much of the energy you bring to the asteroid will go into breaking bonds and separating pieces?

If you attach a rocket to each piece, you have to consider the efficiency of rockets. If you make a big explosion inside the asteroid, you might model that as suddenly converting some explosive into hot gas. How much energy goes into heat? How efficient is gas at pushing pieces?

For breaking bonds, suppose you swing a sledge hammer at a rock. How much of the kinetic energy goes into breaking bonds? If you hit soft clay, bonds are broken and rearranged, but you still have one piece. If you hit glass, the result is more what you want, but it still isn't straightforward to figure out efficiency.

Some of these things can be handled by measurement. Hit a rock and measure how much energy it takes to break it. Set off an explosive and measure the kinetic energy of flying pieces. It might be a challenge to figure out how to make these measurements. But once you do, you begin to have realistic data.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.