# Can you survive a millonth of a second on the surface of the sun?

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Posted (edited)

Not for scifi...just curious if a human could survive this dressed in a plain t-shirt and jeans.

Race car drivers survive a 100g's or more for a brief momentary crash.

Sun's surface is 27 g's.

So if a person was teleported from Earth with same Earth speed they had from Earth to the surface of the sun and only stayed for a millonth of a second before teleporting back to Earth would they survive?

My guess? Maybe...but burned. Someone here already knows how much heat can do what in a millonth of a second to a human.

Edited by Spacescifi
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Posted (edited)
Quote

a millisecond on the surface of the sun

1 hour ago, Spacescifi said:

a millonth of a second

A millisecond is not a millionth of a second.

Edited by kerbiloid
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It is not a binary thing is person dead or alive. You can think execution in guillotine, which cut head very rapidly. What do you think is actually time of. Usually death is defined as stopping of EEG activity but brain cells stay alive couple of minutes without blood circulation. But of course the blade cause irreversible fatal injury in less than a second.

I guess that explosive evaporation of water at 5800 K ambient temperature would cause some kind of shock wave, which would proceed at about 1.5 km/s, which is speed of sound in water. Such a shockwave would clearly reach any point of human body and destroy all biological activities in less than a millisecond (1/1000 s).

If you mean a microsecond (1/1000000 s), shock wave would have proceeded just few millimeters and biological processes deep in the body would be unaffected.

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Excellent question!

For some values of "answered" because there the question was a nanosecond, not a microsecond. Here is the relationship between how long you stay there and how much thermal energy your body absorbs (assuming you are an adult human with a body surface area of around 1.7 m2):

• 1 nanosecond: 0.17 J
• 500 nanoseconds: 85 J
• 1 microsecond: 170 J
• 500 microseconds: 85 kJ
• 1 millisecond: 170 kJ
• 1/100 second: 1.7 MJ

That last bit is where things definitely get fun. 1.7 megajoules is the energy contained in about half a kilogram of TNT. So spending 1/100 of a second on the surface of the sun would be the equivalent of being painted with a thin layer of high explosive and then having it detonated. You will not have a good day.

But let's look at those lower numbers. Your brain can only react to things on the order of a millisecond, anyway, so whatever heating effects you have are going to be effectively spread out over a millisecond. This allows us to get an idea of the effective heat flux.

We assume you are somewhere cold and unpleasant and you are teleported to the surface of the sun and back for the specified period. What do you feel?

• 1 nanosecond. You experience an odd, dim flash of light in your eyes but you don't feel any warmer.
• 10 nanoseconds. The flash is bright. You experience the momentary sensation of being outside on a warm, sunny day but the feeling fades.
• 25 nanoseconds. You feel hot for a moment, like someone just opened a giant oven door right next to you but then closed it.
• 50 nanoseconds. You feel uncomfortably hot but it's not painful, just unpleasant, and the sensation fades quickly.
• 500 nanoseconds. The pain receptors are triggered and you recoil, but the pain is mild and doesn't last.
• 1 microsecond. A sudden burst of searing pain across your entire body. You find yourself screaming but look down at your body and see no burns or injuries.
• 10 microseconds. Once you stop screaming, you look down at your skin and find that you have a mild sunburn.
• 100 microseconds. Your entire body is sunburned and you will need hospitalization because of reduced skin respiration and other general injuries.
• 500 microseconds. Second-degree burns across your entire body. Very low chance of survival.
• 1 millisecond. You experience an odd, dim flash of light in your eyes but you don't feel anything else. Ever again.

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Posted (edited)

Assuming your surface area is 1.7 m^2,

and your specific heat is 3600 J kg-1 °K-1:

The sun's luminosity is 3.828×1026 J m-2 s-1 over its entire area. Dividing out so we get only the luminosity from the area of a human, the power you are exposed to is 1.07e8 W. So, assuming this power is being emitted from all points and in all directions around the edge of the photosphere,

Energy = Power * albedo * time

In 1 microsecond you will absorb 42.7 J.

Q = mcΔT

Your change in temperature is .0002 degrees.

In 1 millisecond you will absorb 42743 J. Your change in temperature will be .192 degrees.

Conclusion: An average person will not be killed by heat when visiting the edge of the Sun's photosphere for 1 microsecond. For 1 millisecond, the Sun would warm you to the core by .192 degrees Kelvin on average, which may not seem like much but your temperature would probably be much hotter near the surface, as you have to absorb all that heat through your skin. People with dark skin, dark hair, or darker spots on their skin might be burned visiting the Sun, even for short times on the microsecond scale. Other factors could hurt or kill you besides heat, like ionizing radiation.

Edited by cubinator
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Nice job with your math, guys! The amount of work done on these questions never fails to amaze.

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17 minutes ago, cubinator said:

The sun's luminosity is 3.828×1026 J m-2 s-1 over its entire area. Dividing out so we get only the luminosity from the area of a human, the power you are exposed to is 1.07e8 W. So, assuming this power is being emitted from all points and in all directions around the edge of the photosphere,

Doesn't this assumption suggest you will only be exposed to an area of the sun the size of a human, rather than the area of the sun that would actually be within light-millisecond-range of a person standing on it? The area of the sun that would radiate upon you would be larger than the area of yourself, wouldn't it?

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And now we should calculate the proton and alpha dose (let it be k~20 Sv/Gy).
Probably, gamma and X-ray, too.

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6 minutes ago, Codraroll said:

Doesn't this assumption suggest you will only be exposed to an area of the sun the size of a human, rather than the area of the sun that would actually be within light-millisecond-range of a person standing on it? The area of the sun that would radiate upon you would be larger than the area of yourself, wouldn't it?

In this situation you are functioning as a heat engine. One of the consequences of the second law of thermodynamics is that radiative heat transfer is reversible and so you cannot raise the temperature of your target above the temperature of a source, no matter how much focusing you do.

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Posted (edited)
23 minutes ago, Codraroll said:

Doesn't this assumption suggest you will only be exposed to an area of the sun the size of a human, rather than the area of the sun that would actually be within light-millisecond-range of a person standing on it? The area of the sun that would radiate upon you would be larger than the area of yourself, wouldn't it?

Yeah, I thought about that, but I think the luminosity figure takes that into account, since it basically says "here's how much power per unit area is coming out" no matter if you're in a diffuse cloud of hot gas or in contact with a hot stove pan. If you were to use the figure for that larger area, you'd be including a bunch of photons that actually weren't going to contact your body.

Edited by cubinator
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Posted (edited)
1 hour ago, sevenperforce said:

Excellent question!

For some values of "answered" because there the question was a nanosecond, not a microsecond. Here is the relationship between how long you stay there and how much thermal energy your body absorbs (assuming you are an adult human with a body surface area of around 1.7 m2):

• 1 nanosecond: 0.17 J
• 500 nanoseconds: 85 J
• 1 microsecond: 170 J
• 500 microseconds: 85 kJ
• 1 millisecond: 170 kJ
• 1/100 second: 1.7 MJ

That last bit is where things definitely get fun. 1.7 megajoules is the energy contained in about half a kilogram of TNT. So spending 1/100 of a second on the surface of the sun would be the equivalent of being painted with a thin layer of high explosive and then having it detonated. You will not have a good day.

But let's look at those lower numbers. Your brain can only react to things on the order of a millisecond, anyway, so whatever heating effects you have are going to be effectively spread out over a millisecond. This allows us to get an idea of the effective heat flux.

We assume you are somewhere cold and unpleasant and you are teleported to the surface of the sun and back for the specified period. What do you feel?

• 1 nanosecond. You experience an odd, dim flash of light in your eyes but you don't feel any warmer.
• 10 nanoseconds. The flash is bright. You experience the momentary sensation of being outside on a warm, sunny day but the feeling fades.
• 25 nanoseconds. You feel hot for a moment, like someone just opened a giant oven door right next to you but then closed it.
• 50 nanoseconds. You feel uncomfortably hot but it's not painful, just unpleasant, and the sensation fades quickly.
• 500 nanoseconds. The pain receptors are triggered and you recoil, but the pain is mild and doesn't last.
• 1 microsecond. A sudden burst of searing pain across your entire body. You find yourself screaming but look down at your body and see no burns or injuries.
• 10 microseconds. Once you stop screaming, you look down at your skin and find that you have a mild sunburn.
• 100 microseconds. Your entire body is sunburned and you will need hospitalization because of reduced skin respiration and other general injuries.
• 500 microseconds. Second-degree burns across your entire body. Very low chance of survival.
• 1 millisecond. You experience an odd, dim flash of light in your eyes but you don't feel anything else. Ever again.

Thanks.

I like to use this as a trick question with average folks:

"Which is would you choose if you must? Teleportation to the moon's surface from Earth's surface for ten minutes in a space suit? Or teleportation to the surface of the sun for a nanosecond and return back to earth?

Everyone picks the moon and dies. Speed difference kills but most are unaware of that LOL.

Edited by Spacescifi
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3 minutes ago, Spacescifi said:

"Which is would you choose if you must? Teleportation to the moon's surface from Earth's surface for ten minutes in a space suit? Or teleportation to the surface of the sun for a nanosecond and return back to earth?

Everyone picks the moon and dies. Speed difference kills but most are unaware of that LOL.

Doesn't that require you to bake in (heh, no pun intended) the assumption that your speed at Earth is not matched with your speed on the surface of the moon?

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Posted (edited)

There's always an XKCD for every occasion:

Edited by Codraroll
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1 hour ago, cubinator said:

Energy = Power * albedo * time

In 1 microsecond you will absorb 42.7 J.

Q = mcΔT

Your change in temperature is .0002 degrees.

In 1 millisecond you will absorb 42743 J. Your change in temperature will be .192 degrees.

Our numbers are different because we came at the equation from two different angles, but they are at the same order of magnitude so we are in the right place.

I think the issue here is heat conduction rate. Your body is basically water, but you don't have any convective heating going on, so you are limited by the direct heat conduction rate. And water is an insulator. So the "change in temperature" doesn't even come close to reaching your entire body; it's limited to the outer layer of your skin.

The non-convective thermal conductivity of water is about 0.606 W/(m*K). The sun is 5800 K and your body temperature is 310 K. Unfortunately I am blanking on how to effectively drag a "heat transfer wavefront" speed out of that.

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Wouldn't you get a few ablative effects as well, or is the time frame too short for that? I.e. your outer skin cells cooking to a crisp, but the heat not penetrating particularly deeply anywhere before you're safe and sound back on Earth.

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Posted (edited)
4 hours ago, sevenperforce said:

Doesn't that require you to bake in (heh, no pun intended) the assumption that your speed at Earth is not matched with your speed on the surface of the moon?

Nope...would probably give away the gambit. Though actually I probably should, since most average joes and janes don't know the moon moves a kilometer per second. Might give them a fighting chance.

To put things in perspective, putting your foot on the road outside while driving is a bad idea...the moon is like that...on steroids.

Anyways...at least they learn somethong about space in the offing. That speed is always relative to something else, no matter how fast you are going.

16 minutes ago, Codraroll said:

Wouldn't you get a few ablative effects as well, or is the time frame too short for that? I.e. your outer skin cells cooking to a crisp, but the heat not penetrating particularly deeply anywhere before you're safe and sound back on Earth.

Yeah I think the one thing no one calculated for is the solar plasma surrounding the sun.

That would actually apply some conductive heat, not just radiation

In other words,no matter the skin color, at best you get a tan, mid to worst you get crispy skin on one side of your body and depending on what stuff you put in your hair...it may be on fire or at least smoking.

Edited by Spacescifi
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Posted (edited)
1 hour ago, Spacescifi said:

Nope...would probably give away the gambit. Though actually I probably should, since most average joes and janes don't know the moon moves a kilometer per second. Might give them a fighting chance.

Of course you should. It's a conditional premise that is not evident from the phrasing of the question. To reiterate the XKCD above, communicating poorly and acting smug when you're not misunderstood is not cleverness. It's not "a gambit". It's being an annoying prickwad.

Edited by Codraroll
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103 m/s  * 10-6 s = 1 mm

Why at all mention the speed?

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I have to agree. Winning a game by not explaining the rules is ... let's say ... a bit childish.

That said, being teleported (in the above sense)  to the surface of the moon for 10 minutes might actually be quite survivable, depending on where you land exactly. In the centre of the trailing hemisphere for example you would be catapulted (more or less) straight upward which would easily give you twenty minutes before having to worry about the moon's surface again. Of course, the teleport back would probably kill you then.

Regarding the teleport to the sun, it's probably less the difference in relative speeds that wouldn't kill you (which is far more severe in the sun's case) but the fact that the sun's surface is not exactly a surface. Still, being bombarded by photospheric plasma at 30km/s is probably not exactly healthy.

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Posted (edited)

Magnetic fields.

Can we feel safe when the solar magnetosphere produces planetary-sized plasma jets?
Won't the body be ionized? Won't the alcali and HCO3 ions tear it apart? Even the H and OH ones?

Edited by kerbiloid
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Posted (edited)

The swept distance over a microsecond at 30km/s is 3cm. The average human is ~500cm² frontal area so you'd sweep out 1.5 litres of stellar corona. As a diffuse plasma it would reasonably be expected to penetrate skin rather than be deflected.

The density of the Corona at the edge of the photosphere is ~0.0000002g/cm3.

The 1.5 litres contain 0.3g of stellar material, 9kgm/s of momentum, 135kJ of kinetic energy, and approximately 15kJ of thermal energy.

This would all be transferred to the unfortunate traveller in addition to the radiative effects other people have outlined. A human is about 20cm thick and weighs 60kg, so the wind-facing 3cm would account for about 9kg. Those 9kg would be kicked backwards at 1m/s and heated by ~4 Kelvin near-instantaneously.

It wouldn't be a fun kind of 1m/s either. More like hitting a less-friendly concrete wall. Could come with severe deceleration injuries.

Edited by RCgothic
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Glad someone else did the math, although I think your calculation is off by two magnitudes. Using your numbers, the amount of stellar material swept up would be only 0.3 mg.  Also, the human frontal area is closer to 5000 square centimeters rather than 500. This would give you 1.35 kJ of kinetic energy which is close to the 4 kJ I got.

Still, quite a significant amount compared to thermal energies involved.

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Posted (edited)
1 hour ago, Piscator said:

Glad someone else did the math, although I think your calculation is off by two magnitudes. Using your numbers, the amount of stellar material swept up would be only 0.3 mg.  Also, the human frontal area is closer to 5000 square centimeters rather than 500. This would give you 1.35 kJ of kinetic energy which is close to the 4 kJ I got.

Still, quite a significant amount compared to thermal energies involved.

I'm having an off day. Think I misplaced a kg for g somewhere, and 1k cm2 to a m2 instead of 10k cm2.

Thanks for the corrections. Think a 1cm/s kick would be more noticeable rather than seriously injurious.

Edited by RCgothic
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I dunno about you guys, but I can

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4 hours ago, NFUN said:

I dunno about you guys, but I can

Me too Provided I'm teleported there at night, when - as everyone knows, Sun is turned off

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