What are the safety margins for destroying a planet from space?

Question

My empire's flagship, the Persecutor class UJBHE¹ Abhorrent, rides into the home system of my greatest enemy, Baron Obsequious, levels its Death Ray² at his beloved home, and pulls the trigger.³ Baron Obsequious' prized emerald explodes in a shower of light that would make George Lucas proud.

1. Given that the Abhorrent's defense screens can only absorb 100 TJ of electromagnetic or kinetic energy before critically compromising the ship (i.e., the ship must withstand less than 100 TJ throughout the experience or the deal's off)...

2. Assume the cross-section of the ship is one square kilometer (thanks Mark!)

3. Assume the Abhorrent is using thrust to hold a static position away from the planet. It's not in orbit. (Thanks again, Mark!)

   • How far away must the Abhorrent be to survive the blast?

   • At that distance, will the "sudden"⁴ dispersal in mass result in a change of gravity that will cause the ship to lurch backward (away from what was once Baron Obsequious)?

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¹ Representing the Universal JBH Empire dontchaknow.

² Because no name for an energy weapon can possibly be better than anything used during the Flash Gordon era.

³ Only wusses use buttons.

⁴ And here I need to make a choice. "Sudden" on a plantary scale may render the question moot. Let's assume, as impossible as it may be, that the mass disperses to basically 0% of its original density within 5 minutes of triggering the ray. That's 0.333%/second or a loss of 0.184g/cm³ per second.

Answer 1

Well, Mark Olson and Samuel provide some very respectable math that winds up with the world destroying shot being taken from about 1 AU or the distance from Earth to sun. Yes, yes; it seems very reasonable.

But what fun is that? A little flash of light, a great disturbance in the force; bah. You want a ringside seat.

I propose that your world destroying shot be taken from the distance of the moon. Then use the moon as shelter. Assume here the moon of this world is the same distance as our moon is from Earth: 1.3 light seconds. After taking the shot you have a luxurious 1.3 seconds to coast into the shadow of the moon. Then marvel at how the moon is struck into steaming silhouette by the energetic flash of its planet dissolving. Shelter there as chunks of molten core pelt the far side of the moon with those that miss streaming past you on all sides. Will the impacts from the dissolving world heat its moon to glowing? How romantic!

The path of this moon might get a little wonky as the debris cloud expands. Assure your pilot you will record it all for her to watch later - she is to keep your ship firmly in the shadow of the moon until the fireworks cease.

Answer 2

Well, at best you're depositing in the planet (or the powdered remnants thereof) energy equal to its gravitational binding energy.

According to Wikipedia, the one true source of truth, the gravitational binding energy of Earth is about 10³² J which is about as much energy as the Sun emits (in all directions) in a week. This is converted into kinetic energy of motion of the powdered planet and bulk of the planet's mass comes off at near its (pre-you) escape velocity.

Think of it as five hundred miles of rock coming at you at 5 miles/second... (At that speed whether it's solid or powder doesn't actually make much of a difference.)

Now it gets harder. The kinetic energy of 1 kg of mass moving at 5 miles/second (Mixed units. So sue me!) is 3x10⁷ J. I have no idea what size your ship is, but I'll arbitrarily assume it has a cross-section of a square kilometer. A cubic km of rock has a mass of about 3x10¹² kg.

So at near orbit the kinetic energy of the mass of rock would be 500 km times the mass of a cubic km of rock times the kinetic energy of a kg of rock. 5x10²² J. That overloads your screens by a factor of 10¹⁰.

So you need to back off so that the slower speed of impact and the lower density of the rock column give you back your safety margin. The density drops as the square of the radius, while the velocity drops more slowly. if we ignore the lower velocity, we get a factor of 10⁵ in distance or 400 million miles. So that's a very safe distance.

By 250,000 miles the velocity of the matter would have dropped by a factor of about 3 (from 7 mps at the Earth's surface to about 1.5 at 250,000 miles) so the energy of the impact would decrease by a factor of about ten, bringing the sure-to-be-safe distance to 100,000,000 miles. This shows that the main point is to get far enough from the planet that stuff launched at escape velocity has pretty much slowed to a stop. Call it a million miles or so.

So, assuming you used the absolute minimum of energy to destroy the planet, your safe distance is probably on the order of a million miles.

If you give in to your (perfectly understandable) wish to do something really spectacular, you'd better get a lot further away.

There was a question about the gravitational effect of the explosion on the ship. Oddly, there would be none at first. From the outside, any spherically symmetric distribution of masses acts like all the mass is concentrated at the center of the sphere. So until the first ejecta got to the ship the gravitational field would be unchanged. After that, the amount of mass inside the sphere closer to the (former) center of the planet than the ship would decrease smoothly, so the effect of the (former) planet's gravity would decrease smoothly also, going to zero when it had all passed. (Changing gravity would be the least of your problems!)

And remember: Blowing up planets is dangerous, so be sure to wear your safety goggles. And friends don't let friends blow up planets.

Answer 3

This is going to depend on the surface area facing the radiated energy from the planet.

If it takes $2.5\cdot10^{32}$ joules of energy to completely destroy an Earth-like planet as suggested here. Then we can assume that an equivalent amount of energy is going to be radiated from planet being destroyed. So, this boils down to calculating the radius you need to be from that expanding sphere in order for your surface area to collect 100 TJ or less of that energy.

Since we're dealing with energy and not power, we can have this happen as quickly or slowly as we like. Using one second seems simple enough. We can treat the planet like an isotropic radiating antenna and you're an antenna collecting that energy.

$$100\cdot10^9\ \frac{\mathrm J}{\mathrm{m^2}}=\frac{2.5\cdot10^{32}\ \mathrm J}{(4\pi d^2)}$$

$$d = 1.41\cdot10^{10}\ \mathrm m$$

This means at a distance of $1.41\cdot10^{10}$ meters from the planet, you'll receive 100 TJ of energy per square meter of exposed shielding. Multiply that by $\sqrt{A}$ where $A$ is the surface area facing the planet and you've got your minimum safe distance.

You won't notice much gravitational effect. The planet will be at least 47 light seconds away, about a tenth the distance from the Earth to the Sun. The gravity you would be experiencing is in the nano-gees and would reduce relatively smoothly as most of the planet moves away from you and a small portion moves toward you.

Answer 4

A planet wouldn't "explode in a ball of light" unless it had been converted into energy somehow.

If it has not, then the total quantity of energy emitted can vary wildly: for example the light could be just the atmosphere igniting and reducing the world to a cinder. In that case the total energy reaching the Abhorrent would be negligible even at comparatively short distances.

But if it has, and your remark about gravitational effects makes me think so, then how much matter has become which energy? (Neutrinos would be the least dangerous, even if enough of them can still kill).

The worst case scenario has the whole mass of the planet turn into electromagnetic energy. 6 10¹² tera-kilograms of m multiplied by 9 10¹⁶, yielding 4.5 10²⁹ TJ of E. The Abhorrent can only absorb 100 of those, so I declare the energy release to be 4.5 10²⁷ abhorrences (it's also about 0.005 foes).

To be able to survive, the Abhorrent must be outside a spherical shell with a surface area of 4.5 10²⁷ km², so that all that energy gets spread thin enough. This assumes that the energy release is omnidirectional and isotropic; if (as it's plausible) the energy release is greater where the Death Ray hits, then the ship has to keep farther away.

This means that the radius of the shell must be $\sqrt{\frac{4.5 \times10^{27}}{4\pi}}$ or 1.8*10¹³ kilometers, or about 1.9 light years. At that distance, any gravitational effects would be negligible.

This is the total deposited energy, not the radiated power intensity (which would be a much less spectacular figure). The reason the number is so large is that we're calculating an absorbing surface of one square kilometer. The normal Sun light, at 1500 W/m² irradiance, would already deposit on it one and a half million kilojoules per second, i.e. 1.5 terajoules per second; which means that the Abhorrent's survivability in direct sunlight at 1 AU is little more than one minute, raising legitimate suspicions of vampirism against the Universal JBH Empire.

A supernova, around 200x times more powerful, has been estimated to be lethal to Earth from any distance below 30 light-years. There are actually some worries about IK Pegasi B, a potential "induced" supernova.

However, due to the limitations of its defensive shield, the Abhorrent needs to be able to perform a FTL escape (or, as Willk suggested, hide behind the Moon) in order to survive.

However, the energy release might be enough to also wipe way the Moon at a piddling 400,000 km distance. The energy from a 0,005 foes explosion is enough, if emitted as high-energy gamma rays, to photodisintegrate the Moon and presumably blast the Abhorrent, a few milliseconds after the gamma leak has crashed its shield and thoroughly sterilized it.

It turns out that I was wrong in fearing neutrinos, though. If that same energy went into neutrinos, the safe distance would be 2.3/200² AU, or 8625 km (I took Randall Munroe's 2.3 AU and just compensated for the explosion being two hundred times weaker than a supernova). At a distance forty-four times greater, behind the Moon, the neutrino radiation dose would be only 2.6 milliSievert, not enough to cause damage, as the planet would softly and suddenly vanish away.

Answer 5

The answers given so far use a model based on your classic sci-fi exploding planet. The whole thing blows apart from the center outward, as if someone replaced the core with dynamite. This is probably an oversimplification, though, and one that might not be working in your favor.

If the planet is destroyed by a weapon fired from space, it probably won't result in a nice, radially-symmetric explosion (full disclosure: I've never tried this personally). The weapon will impact the planet at a point on the surface, and the explosion will travel outwards from that point in a cone-like shape away from the attacker. For an example of what I mean, look for one of the many videos online of slow-motion footage of someone shooting a watermelon or pumpkin with a rifle. Matter gets ejected in all directions, but the vast majority of it travels in the same general direction as the bullet.

The kinetic energy your weapon gives to the planet should keep most of the remnants moving away from you. You should be able to remain much closer to the planet than the "symmetric core explosion" answers suggest. For best results, fire when the planet is directly in line between you and the sun. The sun's gravity can capture much of the debris and clean up part of your mess.

The downside is that this can have more severe gravitational effects. An asymmetrical explosion means the center of mass will drift away from the shooter, and you're almost certainly going to be close enough to feel the planet's gravity. Beware of natural satellites that might be yanked off-course and sent in your direction. You probably want to start accelerating away from the target immediately after firing. If you really want to watch the carnage, leave a disposable camera probe behind to film it.