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.

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

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.

Highly related, or maybe even kinda duplicate: https://worldbuilding.stackexchange.com/q/100423/809 — Mołot, Apr 24 '18 at 21:41
@Mołot, not a duplicate, but dang close. Only one of the answers goes into the after effect of the blast and it doesn't touch at all on how the gravity change (if there is one) affects the combatant ship. I might be able to use the energy calculation to estimate impact on the shield, but it's the energy of the beam, not of the impacting planetary mass. I'm grateful you pointed it out, though. It has useful info. — JBH, Apr 24 '18 at 22:03
@JBH "disperses" assumes that there are no chunks, right? And it can't disperse all the way down to 0% unless you stipulate that Death Ray should leave a "donut hole" in a place where the planet had been. — Alexander, Apr 24 '18 at 22:08
@Alexander, That's true from the perspective of the center of the planet, but not true from a god's-eye view above the planet. If you're standing far enough away, then as the mass of the planet disperses, the density of mass seen by a single object "above" the mass also decreases. You bring up the point that an unstated condition (because it's variable) is whether or not the ship's final safe distance would even allow it to see a density of 0%. For example, Mark's answer would leave the density in the volume before the Abhorrent at 0.8%, (wait, that's wrong, just a sec...) — JBH, Apr 24 '18 at 22:12
@JBH hmm, if I follow Mark's calculations correctly, dispersing a planet over a million miles range should take way longer than 5 minutes. — Alexander, Apr 24 '18 at 22:30
@Alexander :-) Probably. Note in footnote #4, "...as impossible as it may be..." I was expecting the scenario to require more suspension of disbelief than I'd hoped for. But it didn't stop me from hoping. — JBH, Apr 24 '18 at 22:36
Any well designed space battleship armed with planet-busting weapons should immediately activate its FTL drive and retreat from explosion in safety. Any radiated energy will be moving at lightspeed. Debris at less than lightspeed. Escape at superluminal velocity provides a much better safety margin than relying on crude, old-fashioned force-shields. Also, you can attack, firing your Death Ray, at closer distances before departing. Or is that idea abhorrent? — a4android, Apr 25 '18 at 04:06
I'm not sure about the physics in your note "0.333%/second or a loss of 0.184g/cm3 per second.". That would be an extremely heavy planet, wouldn't it, exactly 10 times as dense as the Earth. — Mr Lister, Apr 25 '18 at 06:54
@MrLister, Yup... missed the decimal point. Didn't seem to stop people from getting the gist of my question, though. — JBH, Apr 25 '18 at 08:59
As a realism check, if your weapons process that much energy, those shields are roughly the equivalent of nuclear shell firing tanks with paper lantern armor. But, you know, more extreme than that. Calling that "armor" or "defense screens" seems wrong; I mean, it will slow down insects, but not your enemies actual serious weapons. — Yakk, Apr 25 '18 at 18:31
In regards to footnote #4, it's not clear how the death ray "fires". Is it a brief firing, like a projectile, with energy being added to the target in an instant or otherwise extremely brief period of time? Or is it a stream that is constantly adding more and more energy into it's target over a given length of time? This could affect answers since having a stream-type of firing mechanism would potentially add some "safety" since an active firing stream would also be impacting debris that would otherwise fly towards your ship. — Ellesedil, Apr 25 '18 at 18:56
@Ellesedil, unless that stream was measured in at least hundreds of miles in diameter, it won't have a measureable impact due to additional destruction of material in the path of the beam. But, for all intents and purposes, assume it's pencil thin. — JBH, Apr 25 '18 at 19:28
I'm disappointed footnote 1 doesn't have a footnote-footnote explaining the meaning of "JBH" — TemporalWolf, Apr 25 '18 at 22:11
Not an answer by itself, but a very useful set of data on the energy required to destroy an Earth-like planet using a variety of methods can be found in the one of the appendices of How to Destroy the Earth. The "Cooked in a solar oven" and "Blown up/deconstructed/smashed to pieces/overspun/etc." sections are the most likely to be relevant to your death ray approach. — Ray, Apr 27 '18 at 00:37
How fast can your ship accelerate? You'd have a great view for a longer time if you were able to shoot from a front-row-seat distance, then get out of the way ever-so-slightly ahead of the blast. — Dan, Apr 30 '18 at 03:58

Mark Olson · Answer 1 · 2018-04-25T01:42:18.733

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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.

edited Apr 25 '18 at 01:42

answered Apr 24 '18 at 22:00

Mark Olson

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the one true source of truth ... I'll finish reading your answer when I'm done laughing. It'll take a minute... – JBH Apr 24 '18 at 22:04
Great analysis of the energy impact. But would there be a gravitational effect at 1M miles out? Would the ship lurch backward, or is the force of gravity that far away so weak that it didn't need engines burning to hold it in place? (I should probably mention that in my question. Sorry.) – JBH Apr 24 '18 at 22:06
Comment added about the gravitational effect. – Mark Olson Apr 24 '18 at 22:10
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"And friends don't let friends blow up planets." Vader and Tarkin: not friends. – kingledion Apr 25 '18 at 14:08
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I just joined this community and am absolutely floored by this answer. – Troy Bryant Apr 25 '18 at 14:27
How are you calculating the slowing of the ejecta? What's the mechanism? – Samuel Apr 25 '18 at 16:56
The ejecta slow due to gravity. As explained above, the fact that the Earth is rapidly disassembling doesn't have much impact on the Earth's gravity field outside the sphere of expanding ejecta. – Mark Olson Apr 25 '18 at 16:59
The ejecta could also slow depending on the firing mechanism of the death ray. If the death ray has to fire a stream of energy for a decent amount of time to properly destroy the planet, then said death ray will also be impacting ejecta that would otherwise want to fly towards the ship. – Ellesedil Apr 25 '18 at 19:01
@Ellesedil Since we know nothing whatsoever about the death ray, once you get into its internal details, you can get any answer you want. This lack elegance. – Mark Olson Apr 25 '18 at 19:19
While it's true that the gravitational attraction to a hollow sphere of matter is the same as if it were a point, after the expanding sphere passes the hovering ship (and the ship is inside the sphere) the gravitational attraction is exactly zero at all positions inside the sphere.(I clearly remember doing the integral calculus claculation in high school physics!) – Bohemian Apr 25 '18 at 20:39
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Yes. I noted that in my answer. – Mark Olson Apr 25 '18 at 20:40
Assuming the mass is sprayed uniformly angularly, only about 1 part in 10^(17) of its mass flies through the 1 sq. mi. of the ship at 100 Mmi. (And now I feel dirty using SI prefixes with non-SI units.) This is only about 10 Mg spread over 25 Msq.in. Depending on what the ship's made of, it may not need shields... Although there may be a loud, Earth-shattering kaboom -- yes transmitted through space -- by dust -- lots and lots of dust. Note that 1 g km/s = 1 kg m/s... – Eric Towers Apr 26 '18 at 03:34
@MarkOlson Wouldn't there be some gravitational effect from the weapon itself? IIRC energy and mass are directly related so to pump that much energy into something would require an enormous amount of mass too, be it a huge kinetic warhead or a relativistic beam. As in, weird as it sounds, a super-powerful laser beam exerts a non-trivial amount of gravitational force around it, virtually the same as if all that energy were physical mass instead. – thanby Apr 26 '18 at 16:44
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@thanby: Probably, but since the weapon is for all practical purposes magical, how can we possibly make an estimate? Similarly, since all real weapons are less than 100%-efficient, a real world-killer would not be a good thing be near when it is fired. But other than saying "use a long string to pull the trigger" what can we say, since we know nothing at all about the weapon? – Mark Olson Apr 26 '18 at 16:52

Willk · Accepted Answer · 2018-04-24T22:56:38.343

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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.

edited Apr 24 '18 at 22:56

answered Apr 24 '18 at 22:38

Willk

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That sounds like fun. Can I watch, too? – Mark Olson Apr 24 '18 at 22:41
I've said it before, Will... you're my hero! What would be the gravitational effect on the ship as such a close distance? Anything? – JBH Apr 24 '18 at 22:42
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A little flash of light, a great disturbance in the force; bah. I'm going to be laughing at that for a week. I won't be able to drink fluids in all that time. – JBH Apr 24 '18 at 22:44
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Now we need a question asking if the Moon would be enough of a shield. – Mr.Mindor Apr 25 '18 at 13:55
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After taking the shot you have a luxurious 1.3 seconds to coast into the shadow of the moon. I'd think you'd have more time. If you fire a projectile at light speed, wouldn't you have at least 2.6 seconds (double deluxe!)?. And certainly, the chunks of magma would take a few more seconds to arrive. Maybe you can spare 4-5 seconds to make a couple of pictures for Insta-death-a-gram. What would actually be the first hazard to reach the ship? A burst of energy? Would it move at light speed? Honest question. – xDaizu Apr 25 '18 at 16:30
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Planetary scientist here: Actually, hiding behind the moon could backfire. As planetary debris flies outward from the Earth, the Moon's gravity will bend the path of some of the outward-travelling debris towards the Moon. For debris traveling slow enough, the Moon's gravity will be able to bend the path enough that debris will impact on the back of the Moon, meaning there's no safe place to hide. The Moon will have less time to bend faster-moving debris, so if the debris is moving fast enough, it's possible there will be a "cone of safety" in which one could hide. – jvriesem Apr 25 '18 at 17:39
@jvriesem So you have to hide right behind the moon, not far back enough for the bending to matter. – Monty Harder Apr 25 '18 at 17:43
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So what would happen? The Moon's escape speed at the surface is ~2.4 km/s (1.5 mi/sec). Although if we assume the debris is coming at the Moon at 5 mi/sec when it leaves Earth, it'll slow down very rapidly by the time it gets to the Moon's distance from Earth. If it slows down to 2.4 km/s, then there's no safe place to hide. The faster it is above 2.4 km/s by the time it got near the Moon, the larger the safety zone. (This is me being a little inexact with numbers. This is an easy problem to solve analytically...it just takes more time and diagrams to explain than this comment has room for.) – jvriesem Apr 25 '18 at 18:13
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@MontyHarder: Probably, assuming Earth-debris leaves Earth at 5 km/s, but it depends on how fast the debris is going. If the debris is moving in the right range of speeds (not slow enough that debris hits only the "front" of the moon, and not fast enough that the debris doesn't impact the Moon), then the back of the moon will get pelted, leaving no safe place to hide. What complicates this is quantifying how much self-gravity will slow the debris as it leaves Earth-That-Was. Again: analytic, but it requires an assumption about the initial debris speeds (5 mi/sec from MarkOlson's answer). – jvriesem Apr 25 '18 at 18:19
And any debris that doesn't get launched at full speed will eventually come back towards the largest gravity well around - the moon, and I doubt there would be a safe place when material is coming in from every side. – bendl Apr 25 '18 at 19:57
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Why wouldn't you be core sampling the moon to clear the path to the planet? What kind of cheap Death Ray are you using here? – Eric Towers Apr 26 '18 at 03:36
@jvriesem: Are you accounting for the Moon's gravity accelerating debris on its way from Earth to Moon, counterbalancing some of the pull of the (less and less dense) Earth-cloud? Also, the ship has powerful shields, and doesn't need to completely avoid getting hit, especially by debris that's slowed down a lot. (@JBH, the shields have some recharging power, right? So being pelted over a longer duration while hiding behind the moon can give us a much larger energy budget than 100TJ over the course of a couple hours? Or is the recharge time very very long or uses up consumables?) – Peter Cordes Apr 26 '18 at 04:32
Hiding behind the moon to avoid the initial blast seems reasonable but why stay there riding out the entire event? We came from a different system. Unless we're on some insanely slow generational ship that prioritized death ray research over propulsion we should have something that can move us outta here before the moon gets all sticky. – candied_orange Apr 26 '18 at 16:50
@jvriesem - the cone idea is neat. I did not realize it would be a cone. I can envision a diagram with colored lines / different sized safe cones / impact zones depending on incoming speed of earth fragments. Please, if it is in your power to make such a diagram, edit this idea and attach the diagram. – Willk Apr 26 '18 at 17:47
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@Willk: It's not quite a cone—more like a cone with slanted walls that bend inward, making the cone shorter than it would have been had the walls continued at the slope they had at the base of the cone. I'm pretty busy in the next week or so, but ping me if you come back to this. I definitely should be able to do this. It sounds like a fun problem! – jvriesem Apr 26 '18 at 17:49
The moon is a great place to ride out the storm. The ship has respectable shields and any debris that is slow enough to land on the moon in the first wave is easily stopped by a good shield. The impacts will reduce and then build up with time and have increasing velocity as more distant fragments start to fall towards the moon, time to leave. – KalleMP Apr 26 '18 at 18:27

Samuel · Answer 3 · 2018-04-25T16:50:35.883

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.

LSerni · Answer 4 · 2018-04-25T17:21:52.047

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.

Regarding the first sentence of this answer: all things emit light in the form of blackbody radiation. Whatever causes the planet to explode in the first place will require a tremendous amount of heating as it breaks chemical bonds—just from frictional heating and deformation alone. This energy will be in the form of heating. Hot things glow, and very hot things shine. It's not at all unreasonable to assume that something delivering enough energy to make a planet explode will also cause enough heating to make it shine. If NOTHING else, the magma will be glowing. — jvriesem, Apr 25 '18 at 18:25
@jvriesem absolutely true, but I wasn't talking of "emitting light" - I was interpreting the "exploding in a ball of light" requirement (that's perhaps made clearer in the second paragraph). — LSerni, Apr 25 '18 at 19:31
@LSerni: Totally! :-) My contribution was just that the conversion to energy (from your first sentence) could be by deformation and friction, and that this heating would probably be pretty significant. — jvriesem, Apr 25 '18 at 20:51
Perhaps sunlight can be handled with solar collectors that power radiators on the side facing away from the sun. Or the ship might make a point of sitting in a moon or planets shadow when close to a star. I will bet that this radiant heat load has been ignored by many a SciFi writer having large ships near stars. — KalleMP, Apr 26 '18 at 18:38

score 7 · Answer 5 · answered Apr 26 '18 at 00:09

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.

Ooooh... this is a very good point! The calculation of plasma ejecta impacting the screens is much more complicated (now we really do need to know about that dang Death Ray), but the ship could be close enough to feel the gravimetric displacement. Cheers! — JBH, Apr 26 '18 at 00:25
+1 this, while the answer of @Mark Olson is impressive, I think this would only apply if you planted a doomsday device right in the core. With your death ray, I´d imagine a ring-like acceleration of the debris perpendicular to the beam. At least that would be the implementation I´d look for when designing planet-destroying weapons for maximum visual effect. — Daniel, Apr 26 '18 at 08:33
@JBH Don't forget asteroid form belts based on the gravitational interactions between nearby planets. As the planet remnants drift away and scatter, any neighboring asteroid belts will no longer be in gravitational equilibrium and may eject asteroids in unpredictable directions. You'd be wise to move far away as quickly as possible after taking the shot. — bta, Apr 26 '18 at 16:54
@bta Wisdom, from a megalomaniac? Bah! Wisdom? We don't need no stinking wisdom! — JBH, Apr 26 '18 at 17:40
Actually, for Sun to clean up most of the mess (capture the debris), you must shoot direction that is opposite to its orbit around this Sun. — Vladislavs Dovgalecs, Apr 26 '18 at 17:44
Hmm, if you plan to trash the planet with a kinetic weapon your ship will have planet sized recoil to contend with, not likely to be practical unless you use reaction powered projectiles. The projectile mass/es would also have to be huge unless the velocity was creeping up on light speed. — KalleMP, Apr 26 '18 at 18:30

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

5 Answers5