# Energy to destroy our solar system?

I was told to ask more impractical and unrealistic questions in the world building SE community, so here's my question.

Say we have a hypothetical solar system that we want to destroy by separating all of the mass of the major bodies into infinity.

The scenario is that a well placed, but powerful enough explosive is placed in the center of the sun that can not only destroy the sun, but also have enough energy left after propagating to Neptune's orbit to be able to destroy it and all the planets before it by breaking there gravitational binding energy.

Of course ignoring variables like planets intersecting each other and absorbing some of the force, and other unpredictable variables.

So how much energy in TNT Megaton force would this explosive need to yield to be able to achieve such a feat, assuming this is isn't unquantifiable?

• Are what you thinking specific for our Solar System or do you want to apply it to generic planetary systems? Jun 25, 2016 at 0:24
• TNT Megaton force - is no go, even Sun la-type explosion will be not enough to evaporate Jupiter as example, but it may be barely enough to move it from system. But 10-50 of La type or one Hypernova explosion will be probably enough. 10 for Jupiter and 50 for Jupiter and Saturn, and one Hypernova for them all Jun 25, 2016 at 3:23
• Tri-Lithium: memory-alpha.wikia.com/wiki/Trilithium Jun 25, 2016 at 5:05
• Assuming perfect efficiency then the energy is the gravitational binding energy of the star. (+ a negligible amount for the planets) Jun 25, 2016 at 21:15
• Is it actually necessary to destroy the planets? Unbinding the star will sterilise their surfaces and cause them to fly off into interstellar space, there no longer being a star there for them to orbit. Jun 25, 2016 at 21:41

TL;DR: TNT-megatons aren't the unit. You can use a variation of joules, or comparison units like hypernovas of a certain star size.

Energy needs a medium. Explosives work by producing a lot of gases, which shoot out everywhere. If you wanted to use conventional, gas-creating explosives placed in the center of the Sun to destroy the planets in our solar system... Let's just say the amount that would be needed to decimate even Venus is uncountable. Space is huge, and empty. You would need enough gases to make gigalightyear-scale volumes of pressurised gas. The explosives would probably be an object larger than the Sun. And that's just for Venus.

Alternatively, some kind of radiation would have to be used to heat up all the objects in the Solar System so much that they would explode. Atomics can do that, and is more feasible. You can get calculating how much does Sun heat up the planets, and count how many years of that would need to be released in one moment, then go find estimations of how much energy does the Sun output every year.

That's true, I didn't account for the medium, and the forms of energy. how ever this past hour I've been doing some reading on energy and how it propagates(etc.). I haven't really fiddled with it how ever this is what I've managed to get.

The inverse square law of radiation is $I= s/4πr²$, where s is the source intensity and 4πr² is the surface area of a sphere. How ever what I don't know is the source intensity, what I do know is what I'm trying to find, which is Neptune's gravitational binding energy, and it's approximate distance from the sun, or In this case the origin of the blast.

If Neptune's gravitational binding energy is $\approx 1.7\times 10^{34}$ j, and its distance from the source is $\approx 4.49\times 10^{12}$ m

 1.7×10^34≈ s/4π(4.49×10^12)²

1.7×10^34≈ s/2.53×10^26

S≈ 4.3×10^60j


And my final result comes out to this, which I can rework the problem and the 2 figures match up, which brings me to a stand still.

$1.7 \times 10^{34} \approx \frac{4.3 \times 10^{60}} {4\pi(4.49 \times 10^{12})^{2}}$

I ended up doing more looking into this to make sure this is right, because $4.3 \times 10^{60}$j is a stupidly high number,being more than the estimated mass energy of the milky way galaxy by almost one order of magnitude! . This how ever here has conflicting results from the figure I found .his yield being roughly $2.12 \times 10^{47}$ j, ($2.12 \times 10^{54}$ ergs), which is almost 13 orders of magnitudes off from mine

• Note (cc @MolbOrg) that we have $\LaTeX$ on Worldbuilding. Here's how to use it. Jun 25, 2016 at 19:13
• Yes, but probably, I pointed not what I had to, value which you have calculated, it energy of source, which it have to be, for that tough $1m^2$, will be flow 1.7e34J , because you took $s=1m^2$. @HDE226868 thanks for reminding, second link useful, searched for small text option today, I wrote that way intended, for easy paste to bc Jun 25, 2016 at 20:07