What would be the effect of a giant magical fireball burning in the ocean?

Question

Original prompt:

"What would be the effect of a giant magic fireball burning in the ocean?

Imagine there's a magical fireball that stays in one spot in the middle of the ocean and emits heat at a constant rate that is hot enough to evaporate any water on contact. It's visually sun-like but doesn't do all that fusion crap, it just sits there perpetually emitting its fiery hotness."

First edit with new specifications:

• Solid, non-buoyant, fiery magical sphere
• 10km in diameter
• Temperature of consistent and perpetual 1250K/976C
• DOES NOT necessarily evaporate/vaporize water on contact (but does boil)
• Will not sink into the Earth
• Sits on floor of South China Sea (5km deep, top half of sphere exposed)

• 55 mi from Scarborough Shoal
• 130 mi WSW to next closest (Unnamed) Island
• 136 mi from Truro Shoal
• 185 mi from closest coastline of Philippines
• 245 mi from Manila, Philippines
• 335 mi from Puerto Princesa, Philippines
• 545 mi from Qui Nhon, Vietnam

Latitude and longitude from Google are 14°24'09.6"N 117°20'05.9"E

I'm using a real-world example location just so all the data is already there and I don't have to invent a bunch of it. This location in the South China Sea is perfect because it's very similar to where the sphere would be on my fictional planet. There's an interdimensional rift on the opposite side of my fictional planet, so assume the same for this scenario, and assume it absorbs the excess heat generated by the sphere. Iceland would be a comparable real-world location for said rift. For those who've commented about these elements of my planet creating a smaller habitable region for people to live, you've got the right idea about where I'm going with this.

I expect the area close to the sphere (within 5km) to see some pretty extreme water boiling effects, like what is seen in this video right when the 1000C kettlebell makes contact with the water, only continuous and on a much bigger scale:

https://www.youtube.com/watch?v=qzxrnZfXo4E&t=1m10s

For the area between 5km and 200km, I expect there to be dangerous storms on the scale of hurricanes. More than one person has made such suggestions. The math can be found in the top responses.

To wrap this up, I'd like to limit myself to the environmental/weather questions as before, but with the South China Sea location as a reference point. What would be the effect of this sphere on the specified neighboring locations at their various distances? How would it interact with the air and ocean currents that flow primarily northeast and southwest through the South China Sea?

If there's not much to add/change in the "within 200km" range, then I'd still be curious about the effects of the sphere beyond 200km. I'd guess that once the storms from the sphere hit land (especially higher elevation), then they'll start slowing down and dissipating. Is this accurate? The furthest location I chose for reference is Qui Nhon, Vietnam, but if there are potentially-affected locations further away, feel free to include them in your response.

I hope this is an improvement from before and worthy of a revisit from everyone. Please let me know of any errors. Thank you!

Answer 1

Let's say your magical sphere has radius $r$ of 10km (so just poking up into the outer atmosphere) and is at a temperature $T$ of 1,250K (so glowing a nice warm yellow). The total radiative heat flux from the sphere is given by:

$$ Q = \sigma T^{4}. 4\pi r^2 \approx 1.7 \times 10^{14} \mathrm{W} $$

Where $\sigma$ is the Steffan-Boltzman constant. A proportion of this energy, however, will be lost out to space and so will not affect the biosphere. Let's say that the energy absorbed by the rest of the planet is $10^{14} \mathrm{W}$, or about 100 terawatts.

This is not actually that much energy, on a planetary scale. The solar constant (the measure of solar power incident at the Earth's surface) is about 1.3kW per square metre, so this is equivalent to doubling the solar energy deposited over a circular area just over 200km in radius, which is probably about the size of the 'chaos zone' around your artefact in any case.

(Note that because this equation is quartic in the temperature of the sphere, you can go from innocuous to world-destroying very quickly. Raise the temperature to 2,500 Kelvin, and your power goes up sixteen-fold, and the size of your chaos zone increases to 1,600km diameter. Go up to 8,000 kelvin, and the power input is equivalent to doubling the solar power across the whole earth; this would almost certainly cause a Venus-like biosphere destruction. At 100,000 Kelvin, you will deposit enough energy to exceed the gravitational binding energy of the Earth within a millennium; I'm not sure exactly how the disaster would unfold in this case, but it's sure to be pretty bad. But If you keep the temperature 'reasonable', you can maintain a stable biosphere.)

What would this sort of energy look like? Estimates of the power of hurricanes are on the order of $10^{14}$ to $10^{15}$ Watts, so at the mid-range the effects would probably manifest as a large hurricane surrounding the artefact, although the wind and water flows near the centre would be much more confused, with similarities to nuclear mushroom clouds. Modelling the behaviour very close to the artefact ('very close' here probably meaning up to a few kilometers) would be very computationally challenging.

On a planetary scale, this energy input is on the same order as the 3.2TW of greenhouse-gas-induced heating that's currently causing our climate woes, so the sudden appearance of such an artefact would put the Earth on the same sort of global warming course that we're currently facing. To be honest, the fact that we're doing twice as much damage to the biosphere with cars and power stations, as a city-sized white-hot alien artefact, kind of makes me want to move in with the aliens...

Answer 2

There is no scenario in which the biosphere survives long. You have, at best, a few centuries. Say the fireball is as cool as possible while still being a fireball; 100 degrees celsius. All the oceans will continually drain towards the fireball and will boil on contact, as you said. This is bad news for your biosphere, because that's a huge amount of water vapor entering the atmosphere, and water vapor is a major greenhouse gas. The hotter the ball is, the faster this greenhouse effect ramps up.

If the ball is hotter than about 1400C, you will start melting the ocean floor. This is also very bad for the biosphere. Most mass extinctions have coincided with massive volcanic eruptions, and that's basically what you have here. Molten rock gives off some pretty nasty fumes, which further alter the climate. If the ball is effected by gravity, it may sink into the mantle. This is the best case scenario I'd think. Eventually it would sink to a layer as hot as it was, and would stop causing chaos. The biosphere will probably be wrecked before it settles, but it might be able to recover. If it doesn't sink, you have a permanent volcanic eruption, which as I said above, is really bad news. Most life will be dead within a few years.

The Earth's core is 5400C. If the ball is hotter than this, even if it sinks, it never stops causing problems. If it doesn't sink, just know that 5400C is ridiculously hot. Iron boils at 2750C. This is like thousands of atomic bombs going off constantly in one place. Weather will be extreme immediately. We're talking winds many times faster than the fastest hurricane. The rain will be extreme as well, and it won't just be water. All that rock boiling will come back down, and it'll be super-heated. Lava rain basically. All macroscopic life will probably be dead within the day. Maybe people in hardened air-tight bunkers could survive a little longer, but the extreme earthquakes will get them soon enough. If the ball does sink, the Earth's plate tectonics will go crazy as it sinks. Massive earthquakes and volcanic eruptions everywhere. Mass extinction, just like in the 1400C example. Once it reaches the core, it'll keep sinking, and will disrupt the Earth's magnetic field. No magnetic field means no UV protection. Earth's surface will be sterilized in months, maybe a year. Life around deep ocean vents may survive, if the rampant volcanism didn't already get them.

Somewhere around 10 million C, the ball is as hot as the surface of the sun, and the Earth will be vaporized. Nothing survives.

Answer 3

You are basically cooking your planet on a stove.

This magic fireball is an infinite source of heat, so it will sit there continually pouring heat into the planet's system. Water near it will be heated into steam, but the coolness of the water will not cool the fireball at all. This will cause the overall temperature to rise and rise until everything on the planet eventually matches the heat level of the fireball itself. Eventually the ocean will boil and steam in the atmosphere will render all life extinct. You will end up with a giant ball of ash (or maybe lava). And a 900km fireball is easily large enough to make this happen in short order.

Answer 4

The fundamental problem is that there won't be an ocean left. And evaporating the ocean will destroy the biosphere from the heat.

The bottom of the ocean is 1000 bar of pressure. To stop the ocean from flowing in, you need 1000 bar of steam, which requires near star-core scale temperatures (400,000 K). And then you have a star on your planet, which means you don't have a planet for long.

If you simply evaporate the incoming water, well then the water flows out into the artifact at a rate equal to how fast it would be moving if you dropped water in a vacuum at the top of the ocean.

As a simple model, I'll assume a 3 km tall cylinder 1 km in radius. This has a surface area of 3 km * 2 * pi * 1 km, or ~20 km^2.

It has a diameter of 6 km. The rate at which water inflows is roughly $\sqrt{ h * 20 m/s^2 }$, so we want $1 g/cm^2 * 6 km * \int_{h=0}^{3 km} \sqrt{ h * 20 m/s^2 } dh$

Using wolfram alpha we get a flow of $2.939×10^{12} \frac{kg}{s}$. Evaporating that is 40.65 kJ/mol -- water is 18.0153 g/mol, so this is $\frac{40.65 \frac{kJ}{mol} }{ 18.0153 \frac{g}{mol} } * 2.939*10^{12} \frac{kg}{s}$

Or $6.6316048 × 10^{18} W$.

This exceeds the heat provided to the Earth by the Sun. So bye-bye biosphere.

Answer 5

The above answers, particularly Stephen's first one which sets out the total power of the object and shows that it can be achieved while maintaining a habitable planetary system, cover most of what you ask, but you do need to consider the location of the object with regard to latitude and relative position of continents. Atmospheres and oceans will convect and create patterns of moving air - think jet stream winds, Atlantic gulf stream etc. On an ideal planet rotating about its poles and heated from a central star, with a uniform atmosphere / ocean, you will get jet streams (or ocean streams) in the form of belts and zones rotating in opposite directions around the planet, with cyclonic and anti-cyclonic circulation at the boundaries between the belts and zones. Your object will inject a large amount of heat and moisture into the oceanic and atmospheric circulation at its location. If that is at a mid latitude with an uninterrupted flow around the planet (think the Southern ocean but a bit further up) then you would likely see a hurricane-scale perpetual storm around its location, but also a much more powerful water vortex arising from the water circulation around the object: that's something we don't have an equivalent of on Earth, but with the rate of water uptake from evaporation on contact with the object it would kind of be like a reverse plughole, with water taken up into the atmosphere rather than down the plughole. Obviously there would be intense rain in the vicinity too, on the scale of the atmospheric storms. On the other hand, if plonked in the middle of the Earth's arctic ocean with surrounding continental masses there would be very little rotational effect but the object would still cause large scale dumping of water vapour over the surrounding few hundred miles - in this case you could potentially generate a ring of ice mountains around the pole.

The other thing to consider is the mass of this object. Even though at 1250K it won't melt through to the mantle, if it's a sphere then all of its mass will act on a small contact patch at the bottom of the object: at 8-10km diameter as discussed in other answers, if the object were made of something like rock or metal the pressure would be colossal and it would sink part-way into the planetary crust anyway, like if you put a big ball-bearing on soft mud. At that mass it would definitely have a measurable effect on the direction of the gravitational field in the vicinity (google Schiehallion gravity experiment for an idea of what I mean).

On the other hand, it's magic so maybe its weight is negligible?

Answer 6

Hm... It does sound to me like you will just keep on putting Energy into your planet, which will make the whole system gradually heat up indefinitely even if it is a small ( < 10km) Ball of 600-2500°C. I don't have that much knowledge about the details as some of the others have, but how about this:

Try to have it as cool as possible (600-800°C is deep red while ~2500°C is glowing white) so it doesn't have any global consequences. Have the ball somehow float at the same height so it doesn't fall down and melt through the earth's mantle. Also create another ball that absorbs heat and sends it to the Fireball on the other side of the earth so you don't have to deal with the whole planet heating up for millenia and keep the law of energy conservation. That way you get a nice pole of extreme heat and a pole of extreme cold, which should also be a nice point for the plot.

Answer 7

Well it will be loud.

No one has mentioned so far is how loud this thing is going to be. The second you start boiling water in a continuous stream deeper than about an atmosphere (10.2 m) you need to be aware that you will be creating a choked flow scenario. That means water reaching the surface at supersonic velocities. Any supersonic boom will have a volume of 191 dB, and given the size of your fireball in comparison to the thicker parts of the atmosphere, this sound will dissipate as a line source rather than a point source. Line sources decrease with the square of distance rather than the cube. To get to the threshold of pain at 130 dB you will need to be 335 km away from the center of the ball, the continuous hearing damage limit of 85 dB will require you to be about 4500 km away.