# Is it possible for a planet oddly similar to the ones we have orbiting our sun to make a shield to protect an interstellar space ship?

The situation is bad, human kind have waited to the last minute (literally) with building the huge ship that is supposed to save man kind. The sun goes nova, nothing can stop that but the ship has to survive the shock wave.

So the idea is the following: if something blows up you hide in the bathtub, behind a stone or anything else that can shield the blast.

But since this is a ship holding millions of people and DNA for producing flora and fauna (think Titan AE) one would need a fairly large rock. Lucky for us we got 10 available in our solar system.

The question: is there a planet, similar to the ones in our solar system strong / dense / large enough to provide a shield for long enough to enter the subspace.

• Ship is estimated 3 km^3
• To achieve subspace the ship needs to get to quarter-c and it will take it 1 month to get to that speed.
• Assume the space ship is shielded against dangerous radiations since it has to enter subspace.
• Really quite relevant: what-if.xkcd.com/73 – Joe Bloggs Oct 20 '15 at 10:22
• If you have a spaceship that can achieve a non-trivial percentage of c, why do you need to hide behind a planet? If you can go that fast, just exit the solar system and find a new home. Also, you'll have millions of hears notice about when a star will go nova. That's plenty of time to plan. – Green Oct 20 '15 at 12:04
• @Green as i said, bad planning. – Magic-Mouse Oct 20 '15 at 12:16
• The sun will not go nova, suddenly or otherwise. Everyone go home. – JDługosz Feb 25 '16 at 12:27
• The neutrino radiation that is the bulk of the energy of the supernova will pass through a planet virtually unimpeded. There is no material that can protect against them. – John Meacham Feb 25 '16 at 12:38

NOTE: The following calculations are a Fermi Approximation only. They contain some small but inconsequential errors. The orders of magnitude stated are accurate.

In order for a planet to survive, the incoming energy from the supernova must be significantly less than the binding energy of the planet. If the energy delivered to a planet is higher than the binding energy then the planet blows up, à la Alderaan. A blown up planet has zero value as a shield for a piddly little $3 \, \text{km}$ spaceship.

Supernova energy output: $10^{44} \,\text{to}\, 2 \cdot 10^{44} \,\text{J}$.

Binding Energy of Earth: $2.41 \cdot 10^{32} \,\text{J}$

Binding Energy of Jupiter: $2.086 \cdot 10^{35} \,\text{J}$

Surface area of Jupiter exposed to blast: $\frac{61.42 \,\text{billion km}^2}{2} = 30.71 \,\text{billion km}^2$

Distance from Sun to Jupiter: $778.5 \,\text{million km}$

From the Inverse Square law, we lose 9 orders of magnitude in intensity between $0.05 \,\text{million km}$ from the supernova point source and $778.5 \,\text{million km}$ leaving us with

Intensity = $\frac{1}{(7.785 \cdot 10^{8} \, \text{km})^2}$

So, at Jupiter's distance from the Sun, we should expect an energy delivery of $10^{44} \,\text{J} \cdot \frac{1}{(7.785 \cdot 10^{8} \, \text{km})^2} = 1.6499955 \cdot 10^{26} \, \frac{\text{J}}{\text{km}^{2}}$

Jupiter, with an area of $1.61 \cdot 10^{10} \, \text{km}^{2}$ will receive $\frac{1.61 \cdot 10^{10} \, \text{km}^2}{7.62 \cdot 10^{18} \, \text{km}^2} = 2.11286089 \cdot 10^{-9}$

(where $7.62 \cdot 10^{18} \, \text{km}^{2}$ is the surface area of the spherical blast front with a radius of Jupiter's distance to the Sun)

for a total of $1.6499955 \cdot 10^{26} \, \frac{\text{J}}{\text{km}^{2}} \cdot 2.11286089 \cdot 10^{-9} = 3.4845888 \cdot 10^{17} \, \frac{\text{J}}{\text{km}^{2}}$

Back to the Orders of Magnitude table, energies in the $10^{17}$ range are equivalent to the Tsar Bomba rated at 50 megatons.

Roughly every square kilometer of Jupiter is getting hit with the largest nuclear weapon that man has ever made. (There we go, those are the mind blowing numbers I was expecting.)

While $3.4845888 \cdot 10^{17} \, \text{J}$ is significantly below the binding energy of Jupiter and Earth, getting hit with that much energy will do extremely unpredictable things to Jupiter's atmosphere and orbit. I don't have the math to figure out those kind of calculations but anywhere near a supernova is going to be a really uncomfortable place to live.

Possibly, if the ship bedded down deeply in Jupiter's atmosphere on the far side of star, it might survive. Maybe. If Jupiter's core gets a huge shove outward (which it very well may) then the ship may get pushed too far into Jupiter's gravity well and experience a hull failure because of the insane pressures. Or (I'm speculating and definitely don't have the math to prove it) that much energy will kick off a chain of thermonuclear reactions in Jupiter's core that will incinerate the ship.

And the procrastinators don't get on. Anyone who caused this kind of procrastination deserves a Darwin fate as they are too stupid to live.

• Cool, you had the math to tell me what I thought was correct. Get the hell out of Dodge. – bowlturner Oct 20 '15 at 14:30
• Yeah, I did a supernova calc for our planets a while ago and found that they all evaporate. Which makes sense. Astronomers think they found a small planet orbiting a neutron star (supernova remnant). They think that small planet is the remnant core of a stellar companion. If only the corpse of a star can barely survive a supernova blast - all the planets are toast. – Jim2B Oct 20 '15 at 14:49
• "And the procrastinators don't get on. Anyone who caused this kind of procrastination deserves a Darwin fate as they are too stupid to live." - But yet - we still have people denying that global warming is a problem. – Magic-Mouse Oct 20 '15 at 14:53
• Supernova output energy right at the beginning there kind of looks like 1 J to 2E44 J; which is a funny-looking range. – Williham Totland Oct 20 '15 at 14:57
• @Magic-Mouse I thought of the same thing. – Green Oct 20 '15 at 15:07

Well according to Wiki the material/shockwave of a supernova explosion travels out from the sun at approximately 1/10th the speed of light. So in theory, as long as the ship can reach .1c before the wave hits it will be able to outrun the wave. If they have enough warning to even try and get behind a planet they should really just focus all their energy on reaching .1c as fast as possible.

Pluto is about 13 light hours from the sun and Saturn is about 80 minutes.

So traveling at .1c the shockwave would take about 12-13 'hours' to reach it. There will be a 'small' shadow on the back side of the planet but because of diffusion the farther away from the planet you get the more material will 'wrap' around the planet and fill in that hole, eventually filling it up back up.

So really the only way is to start speeding away from the sun as soon as able, try to get to get to .1c before the wave catches up with you. Being in Jupiter's or Saturn's shadow might help but not enough to likely save the ship. Think of it like an eclipse. It only works when the item being eclipsed is close enough and positioned right. If the moon was significantly smaller our eclipse would be more like a dimming of the sun. So the farther beyond the 'protective' planet you go the less protection it will provide. So reach .1c before the wave reaches you!

Though it does appear that being in the center of a shadow of a planet far enough away when the wave goes by could protect a ship, but it would still be a close thing. The biggest thing a planet could do is protect you from the x-ray/gamma ray blast which would likely do a lot of damage all on it's own, might need a planet anyway just to survive that!

• According to @burgis calculations it will take 12 days to reach 0.1c and the blast of the supernova would last approximately 2 minutes. – Magic-Mouse Oct 20 '15 at 13:50
• @Magic-Mouse Well, starting points, distances to possible safe planets to hide behind, actual time until super nova etc. all go into it what is the best solution. It's not like you can predict to within days when a supernova is going to go off. basically if you are that worried, as soon as a ship is loaded to capacity it should push off and start accelerating away as quickly as possible, no looking back. – bowlturner Oct 20 '15 at 14:03
• Added more. Forgot about the Gamma rays, They will kill most everything in the solar system by themselves.... I think you need 30 YL distance for even a planet to be reasonably safe. – bowlturner Oct 20 '15 at 14:09
• Ill add to the description, but lets assume that a space ship that can enter subspace, is sufficiently shielded against various types of radiation. – Magic-Mouse Oct 20 '15 at 14:14
• Gamma rays can be blocked, To reduce typical gamma rays by a factor of a billion, thicknesses of stop-gamma shield need to be about 13.8 feet of water, about 6.6 feet of concrete, or about 1.3 feet of lead. So, if you're ship has a large amount of water between the emitter and crew, which can be used for drinking, air, and propulsion, then the population should be reasonably safe from the initial gamma burst. Being on the far side of a planet or moon would also shield the ship easily. – AndyD273 Oct 20 '15 at 14:18

My initial thought at seeing this question was "why don't they just hide behind the planet until the shockwave has passed by?". Jupiter is the largest planet so by definition has the largest mass buffer. I don't know how long the planet would survive. I admit this doesn't answer your question but gives you an option.

I then thought about maybe the ship could fly a spiral course inside the penumbral cone until they hit the "88mph" to jump into subspace, I don't know if there is enough volume in the cone for you to execute the manuover. I found this NASA paper on calculating the penumbral cone, unfortunately the physics calculations are a bit beyond me.

I can do the maths to work out the distance required to hit 0.25 c though.

299,792,458 / 4 = 74,948,114.5 m/s
30 days = 2,628,000 seconds
74,948,114.5 / 2,628,000 = 28.51 m/s^2

0.5 x (28.51) x (2,628,000)^2 = 98,481,822,453,000 metres


Acceleration and distance formulas from here.

That is about 658.31 AU. 1 AU is the distance of the earth to sun.

• You might have a point there... however I still don't think you will have the room.I'll check my maths and come back to you. – Burgi Oct 20 '15 at 12:52
• Thats only 1/411th of the distance to proxima centuri. – Magic-Mouse Oct 20 '15 at 13:39
• The point is that it is well outside the penumbral cone unless you do some crazy spiral trajectory. – Burgi Oct 20 '15 at 13:47
• – Burgi Oct 20 '15 at 14:09