# Re-Entry Heating and G's math (Help I never went to college)

#### Char0n

##### New member
Looking for equations or graphs related to re-entry heat. Assuming a human-shaped object weighing 350 pounds. Starting with a 0 vertical and horizontal velocity @100km. Looking for the G's you would pull on deceleration given the same problem.

#### Urwumpe

##### Not funny anymore
Donator
Looking for equations or graphs related to re-entry heat. Assuming a human-shaped object weighing 350 pounds. Starting with a 0 vertical and horizontal velocity @100km. Looking for the G's you would pull on deceleration given the same problem.

There are some approximations, but generally, simple equations don't exist, only graphs. For example, there is a graph of the destruction altitude of a solid sphere of metal reentering at orbital velocity.

In your case, you would need the approximation for eccentricity just below 1 and 100 km apogee during a ballistic reentry.

Sorry, bad example for equations, since these assume that the drag force is much stronger than gravity and this the effects of gravity can be ignored. But this is not the case for your scenario. But it could ,aybe be adapted to give you an approximation, by calculating the terminal velocity first of your "human shaped object" (You don't want to treat objects like humans, do you?) and then use this as your initial ballistic reentry velocity $v_e$.

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#### Char0n

##### New member
There are some approximations, but generally, simple equations don't exist, only graphs. For example, there is a graph of the destruction altitude of a solid sphere of metal reentering at orbital velocity.

In your case, you would need the approximation for eccentricity just below 1 and 100 km apogee during a ballistic reentry.

Sorry, bad example for equations, since these assume that the drag force is much stronger than gravity and this the effects of gravity can be ignored. But this is not the case for your scenario. But it could ,aybe be adapted to give you an approximation, by calculating the terminal velocity first of your "human shaped object" (You don't want to treat objects like humans, do you?) and then use this as your initial ballistic reentry velocity $v_e$.
When looking up articles about theses problems there was a lot of mention of compressible and incompressible flow how does that relate to re-entry heating?

#### Urwumpe

##### Not funny anymore
Donator
When looking up articles about theses problems there was a lot of mention of compressible and incompressible flow how does that relate to re-entry heating?

Thats first of all a matter of aerodynamics. If you move at slow speeds, the air can easily move out of your way without density increases, you have incompressible flow. The closer you get to the speed of sound, the less the air can move out of your way and gets compressed, density increases, shockwaves form. You have compressible flow. Now, remember the air pump: If you compress the air, temperature increases, too. Behind the shockwaves, velocity changes, resulting in only small parts of your vehicle actually experiencing supersonic flow. Once your spacecraft travels at a significant multiple of the speed of sound (hypersonic speeds), another phenomena happens: The shockwaves become so densily compressed to the spacecraft hull, that large parts of its surface experiences supersonic flow. At this point you start the transition to hypervelocity flow: The speeds at which the speed of sound is completely irrelevant again, because the air behaves no longer as a gas, but more like individual air molecules hitting your hull. Like tiny balls, just smaller. Compressibility also stops to matter much, because the air is already compressed that dense, that any further increase in density results in much larger heating. That is the speed where most of the reentry happens.

If you need even more information, see here, it pretty much describes what engineers thought before flight and what reality happened and talks about many different phenomena that happen beyond Mach 10:

#### Char0n

##### New member
Thanks this definitely helps.

#### Linguofreak

##### Well-known member
When you ask for equations for a problem like this, the difference between asking someone who went to college and asking someone who didn't go to college is that the person who didn't go to college (or the person with a humanities degree) will say "No clue", while the person who did go to college will say "No clue, but there's a PhD and a Nobel Prize waiting for you if you can find an equation". If you're lucky, the person who went to college might be able to provide a link to research results where someone has had a computer (or a datacenter full of computers, depending on the difficulty of the problem) solve many equations step-by-step to get a result that's a close approximation to what you'd get in reality.

#### Urwumpe

##### Not funny anymore
Donator
And sometimes, you need to figure out, what the problem actually is. Is it about understanding? Is really a more accurate result needed than just letting orbiter run under these conditions? Or is it even about HOW to explain this problem to Orbiter?

#### n72.75

##### Move slow and break things
Orbiter Contributor
Tutorial Publisher
Donator
The equation and concept for stagnation temperature is probably worth bringing up here, but it's worth noting that most of the fluid and indeed the vessel will be well below the stagnation temperature, at orbital velocities this is around 30000 K. https://en.m.wikipedia.org/wiki/Stagnation_temperature

You also need high density, and high temperature, and high proximity with the vessel in order to actually cause heating.

Note the example below:

The math required to calculate something like that isn't necessarily a super complicated equation, it's more that it's a bunch of moderately complicated equations that need to be solved millions of times. For all of the time we've been building reentry vehicles this has been calculated by: rough guesses, intense computer simulation, or physically modelling and calculating.

In the example you gave of someone falling from 100km at 350 lbs the best way to approach that is probably by looking at potential energy.

PE = m x g x h

Mass times gravity (acceleration) times height.

159 kg x 9.81 m/s/s x 100000 m

Is appropriately 1.55 MJ (megajoules)

On descent you would need to disipate that energy somehow. If you assume that a person is mostly water and use the specific heat of c = 4.18J(gK) and the formula:

Delta T = E/ (m * c)

Your temperature rise will only be a few degrees
on average. The local heating from shock waves could be much higher but as long as you have proper gear (good reentry shoes?) It's probably survivable with the right suit design.

edit: See below.

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#### Urwumpe

##### Not funny anymore
Donator
Actually, its 155 MJ (just check the orders of magnitude) and if that energy would not dissipate different, the poor astronaut would heat up by 234 K. Enough to bake a good Pizza Astronaut.

Luckily, that does not happen, but I would expect a peak temperature difference of ~100K, since most of the decelleration and heating happens over a very short altitude range around 40 km altitude. With a good space suits, no problem.

#### n72.75

##### Move slow and break things
Orbiter Contributor
Tutorial Publisher
Donator
Actually, its 155 MJ (just check the orders of magnitude)

I knew it was too early to do math...

More coffee should've been consumed first.

Thank you.

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