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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)?


1Representing the Universal JBH Empire dontchaknow.

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

3Only wusses use buttons.

4And 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.

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    $\begingroup$ @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. $\endgroup$ – JBH Apr 24 '18 at 22:03
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    $\begingroup$ @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. $\endgroup$ – Alexander Apr 24 '18 at 22:08
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    $\begingroup$ 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? $\endgroup$ – a4android Apr 25 '18 at 4:06
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    $\begingroup$ 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. $\endgroup$ – Mr Lister Apr 25 '18 at 6:54
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    $\begingroup$ @MrLister, Yup... missed the decimal point. Didn't seem to stop people from getting the gist of my question, though. $\endgroup$ – JBH Apr 25 '18 at 8:59
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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.

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    $\begingroup$ 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. $\endgroup$ – JBH Apr 24 '18 at 22:44
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    $\begingroup$ Now we need a question asking if the Moon would be enough of a shield. $\endgroup$ – Mr.Mindor Apr 25 '18 at 13:55
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    $\begingroup$ 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. $\endgroup$ – xDaizu Apr 25 '18 at 16:30
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    $\begingroup$ 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. $\endgroup$ – jvriesem Apr 25 '18 at 17:39
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    $\begingroup$ 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.) $\endgroup$ – jvriesem Apr 25 '18 at 18:13
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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.

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    $\begingroup$ the one true source of truth ... I'll finish reading your answer when I'm done laughing. It'll take a minute... $\endgroup$ – JBH Apr 24 '18 at 22:04
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    $\begingroup$ "And friends don't let friends blow up planets." Vader and Tarkin: not friends. $\endgroup$ – kingledion Apr 25 '18 at 14:08
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    $\begingroup$ I just joined this community and am absolutely floored by this answer. $\endgroup$ – Troy Bryant Apr 25 '18 at 14:27
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    $\begingroup$ Yes. I noted that in my answer. $\endgroup$ – Mark Olson Apr 25 '18 at 20:40
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    $\begingroup$ @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? $\endgroup$ – Mark Olson Apr 26 '18 at 16:52
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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.

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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.

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    $\begingroup$ Great name for that quantity of energy. $\endgroup$ – Samuel Apr 25 '18 at 16:52
  • $\begingroup$ 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. $\endgroup$ – jvriesem Apr 25 '18 at 18:25
  • $\begingroup$ @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). $\endgroup$ – LSerni Apr 25 '18 at 19:31
  • $\begingroup$ @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. $\endgroup$ – jvriesem Apr 25 '18 at 20:51
  • $\begingroup$ 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. $\endgroup$ – KalleMP Apr 26 '18 at 18:38
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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.

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  • $\begingroup$ 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! $\endgroup$ – JBH Apr 26 '18 at 0:25
  • $\begingroup$ +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. $\endgroup$ – Daniel Apr 26 '18 at 8:33
  • $\begingroup$ @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. $\endgroup$ – bta Apr 26 '18 at 16:54
  • $\begingroup$ @bta Wisdom, from a megalomaniac? Bah! Wisdom? We don't need no stinking wisdom! $\endgroup$ – JBH Apr 26 '18 at 17:40
  • $\begingroup$ 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. $\endgroup$ – Vladislavs Dovgalecs Apr 26 '18 at 17:44

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