### Re: How big an explosion would it take to destroy our star system?

Date: Sat Nov 3 19:01:13 2001
Posted By: John W. Weiss, Grad Student in Planetary Science
Area of science: Astronomy
ID: 1004644292.As
Message:

I really enjoyed answering this question. It isn't every day that I get to ponder blowing up a whole solar sytem!

The answer depends on whether you have a smart device which beams its energy at its targets reasonably carefully or if it is a simple explosion, which will waste a lot of energy since most of it will miss the planets altogether and coast off into space. I'll do both calculations, here, because it's fun and easy to do with a spreadsheet.

Now, what does it mean to destory a planet or star? What does it take? What it means is that you've pulled the whole body apart, working against its own force of gravity which tends to hold it together. (I'm ignoring internal forces. In reality, rocky bodies like Earth would be somewhat more work to tear apart because rock doesn't like to come apart. The gas bodies like the Sun and the Jovian planets fit my model well, though.) It turns out that you need an amount of energy equivelent to

 E = 3 5 G M2 R
(1)

where G is the gravitational constant, M is the mass of the body and R is its radius. So all we need is mass and radius for the bodies of the solar system (I'll stick with the Sun and the 9 nominal planets). The following table shows those numbers and the needed energy (under Raw Energy)

 Body Mass Radius Distance from the Sun Raw energy for explosion Total Sun-centered explosion [grams] [cm] [cm] [ergs] [ergs] Sun 2.00E+33 6.96E+10 2.30E+48 1.14937E+15 Mercury 3.30E+26 2.44E+08 5.791E+12 1.79E+37 4.03E+46 Venus 4.87E+27 6.05E+08 1.0821E+13 1.57E+39 2.00E+48 Earth 5.97E+27 6.38E+08 1.496E+13 2.24E+39 4.93E+48 Mars 6.42E+26 3.39E+08 2.2794E+13 4.86E+37 8.76E+47 Jupiter 1.90E+30 7.14E+09 7.78463E+13 2.02E+43 9.61E+51 Saturn 5.68E+29 6.03E+09 1.42675E+14 2.14E+42 4.80E+51 Uranus 8.68E+28 2.62E+09 2.87107E+14 1.15E+41 5.53E+51 Neptune 1.02E+30 2.52E+09 4.49832E+14 1.66E+43 2.12E+54 Pluto 1.27E+25 1.14E+08 5.90646E+14 5.68E+34 6.13E+48 Needed Energy [ergs] 2.30E+48 2.12E+54 Needed Energy [tons of TNT] 5.50E+31 5.09E+37 Needed Mass of Matter/Antimater [grams] 2.55E+27 2.35E+33

The total needed energy is therefore 5.50x1031 tons of TNT (to use the units of weapons development). Or, being a Star Trek far too, I looked at it as a matter/anti-matter bomb problem. That still takes 2.55x1027 grams of matter and anti-matter. That's about half of Earth's mass, or one big bomb.

As I alluded to above, this isn't the whole picture. Most bombs go off and spread their energy out in all directions equally. This means that most of the blast energy will miss the planets. Each planet has to intercept the required amount of energy from the blast over its cross-sectional area (pi R2) when the initial energy has been spread out of 4 pi a2 surface area (where a is the distance to the Sun). Working this out, you need 4(a/R)2 times more energy to blow up each planet to account for the wasted energy. This energy is computed in the right column of the table.

Luckily for our hypothetical space-faring demolition men, we don't have to add all these energies up. We really only need out explosion to be as large as the largest energy needed (plus the energy needed to destroy the Sun, which is relatively negligible). This is because energy wasted by missing Mercury can hit Venus, Earth, Mars or any of the other planets. And the same goes for the wasted energy with each other planet (except Pluto, of course). It turns out that Neptune really sets the amount of energy we need. (In reality, if we had totalled the last column, the result would be pretty much the same as the energy needed to destroy only Neptune.) As you can see, it's 5.09x1037 tons of TNT. Or, in other words, if you took a mass of anti-matter a bit larger than half the mass of our Sun and added it to the Sun, causing a terrible matter/anti-matter reaction, it would release enough energy to destroy the Sun and all of the planets.

A matter/anti-matter reaction is the most efficent explosion we know of today. This doesn't mean that we won't find a more efficient one sometime before Voyager's supposed era (we have nearly 500 years, so there's time). But as you can see, that's still one heck of a weapon by today's standards!

In fairness, I should acknoweldge that I got the planetary data from Solar System Dynamics by Murry and Dermott, one of my all-time favorite textbooks. You can see the derivation of the graviational binding energy I used in Stellar Interiors by Hansen and Kawaler. I got the conversion from ergs to tons of TNT from http://www.matter-antimatter.com/energy.htm. And I used tth, availible at http://hutchinson.belmont.ma.us/tth/ , to convert LaTeX into equation (1).

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