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# Pressure Inside Shuttle Infinite?

cr3ativ3
Ok, I did a quick search of this on google maybe I wasn't searching properly but I couldn't get any matches to my problem.

I probably have no idea what I am talking about just bear with me point out how I am wrong cause this is bothering me.

So me and my friend today when I go on my daily 30km bike ride we were talking about why the space shuttle doesn't just explode in space because of the pressure. This is what I was getting at, there is no pressure in space right? There is an atmosphere in the shuttle so shouldn't the pressure in the shuttle be infinitely powerful compared to the pressure outside in which case the shuttle would explode because only an infinitely strong force could contain the atmosphere which the stucture of the shuttle isn't infinitely strong.

My friend was saying that there would have to be a hole in the shuttle for the pressure to have any effect on it. If so why does a balloon pop when you let it go higher in the atmosphere to the point at which the pressure on the inside is to powerful for the pressure outside.

Sorry, the problem has just been bugging me

Thanks to whoever can answer this.
ocalhoun
The pressure in the shuttle is infinitely higher than outside it, because anything is infinitely higher than 0. However, the pressure inside the shuttle is certainly not infinite, when compared to anything other than 0.

The outside pressure is not the important factor, though.

All the hull of the shuttle has to hold in is one atmosphere of pressure (The air pressure of sea level on Earth (Actually, I suspect that it is a little lower than that after takeoff, as reducing the air pressure inside would reduce the weight, and enable them to use weaker materials for the hull)).
As long as the hull is strong enough to hold in that pressure, it doesn't matter what the outside pressure is (as long as the outside pressure is less than the inside pressure).

Remember, it's not the vacuum trying to pull the shuttle apart, it's the pressure inside it trying to push it apart. The opposite thing happens if you remove some of the air from a can or plastic bottle (either by sucking it out or by reducing the temperature rapidly). The can or bottle will be crushed. It will not be crushed by the vacuum within it, though. It is crushed by the air pressure above it.
cr3ativ3
Thanks, I am still a little confused but it's probably just me like I understand what your saying I just don't understand why it doesn't get pushed apart if the pressure in side the shuttle is infinitely larger then outside the structure. Because then wouldn't you need a structure that could hold infinitely?

Hey, also helium it's lighter then air does that mean it has less pressure if so would a balloon that you let go would pop farther up than a balloon filled with "1" atmosphere at the same altitude?

Thanks, for bearing in with me on this.
Gagnar The Unruly
Any numeber is an infinite number of times greater than zero, but not infinitely more than zero. For example, 1 is only 1 more than 0, even though 0 times infinity is less than 1. 3 is 3 more than zero, so it's bigger than one is, with respect to 0. Ocalhoun is right that in space, all that matters is the 1 atmosphere of pressure that needs to be contained by the ships hull. If the ship had to maintain 100 atmospheres, it would have to be (at minimum) 100 times stronger. But what really matters is the pressure difference. If the absolute pressure in space were 0.5 atmospheres, a ship at 1 atmosphere would have to hold in 0.5 atmospheres of the net pressure difference. Likewise, if space had a pressure of 2 atmospheres, then the ship would have to hold off 1 atmosphere of external pressure. If there was barely any difference, then it wouldn't matter too much because the net pressure on the hull would be very low.
Indi
 ocalhoun wrote: The outside pressure is not the important factor, though. All the hull of the shuttle has to hold in is one atmosphere of pressure (The air pressure of sea level on Earth (Actually, I suspect that it is a little lower than that after takeoff, as reducing the air pressure inside would reduce the weight, and enable them to use weaker materials for the hull)). As long as the hull is strong enough to hold in that pressure, it doesn't matter what the outside pressure is (as long as the outside pressure is less than the inside pressure).

No no no. >_< That's all wrong.

The outside pressure is very much a factor in whether the shuttle pops or not. The stress on the hull is very different when the shuttle is on the ground (and sealed up) versus whether or not it is in space. In fact, on the ground, it's very likely zero.

To determine the stress on the hull, you need to consider both the pressure inside and the pressure outside. In actuality, what you're concerned about is the pressure differential (p = abs(PINSIDE - POUTSIDE)).

Once you determine the pressure differential, you can figure out whether the shuttle will explode or not. For simplicity, you can model the shuttle as a cylinder, and use these equations:
σH × t = p × R
σL × t = p × R ÷ 2
where:
R = the radius of the shuttle's body
p = the pressure differential
t = the thickness of the shuttle wall
σH = hoop stress
σL = longitudinal stress

If you go beyond the hoop stress, the shuttle will burst outwards like an overcooked sausage. If you go beyond the longitudinal stress, the shuttle will tear apart with the front end going one way and the back the other.

In either case, you do need to take the outside pressure into account. As a matter of fact, assuming the pressure inside the shuttle is 1 atm:
 Code: Location      Outside pressure    Result ----------------------------------------------------------------------------- In orbit            0 atm         Straining outwards like an inflated balloon Under water         2 atm         Straining inwards threatening to be crushed On the ground       1 atm         No strain at all

If you were to create a large enough negative absolute pressure outside the shuttle (which you can't do hydro-statically, but is possible by other means), the shuttle would explode, regardless of how strong the hull is or what the internal pressure is.

 cr3ativ3 wrote: Hey, also helium it's lighter then air does that mean it has less pressure if so would a balloon that you let go would pop farther up than a balloon filled with "1" atmosphere at the same altitude?

No, the molar mass of the fluid has no effect on that kind of thing. No matter what the fluid in the balloon is, whether it's helium, air, carbon dioxide or even water, as long as you use the same pressure and the same balloon, it will pop in the same place - when the pressure differential between whatever is inside and whatever is outside exceeds the stress tolerance of the balloon's material.

The pressure difference is the only thing that matters in that case.
dwinton
 cr3ativ3 wrote: So me and my friend today when I go on my daily 30km bike ride we were talking about why the space shuttle doesn't just explode in space because of the pressure. This is what I was getting at, there is no pressure in space right? There is an atmosphere in the shuttle so shouldn't the pressure in the shuttle be infinitely powerful compared to the pressure outside in which case the shuttle would explode because only an infinitely strong force could contain the atmosphere which the stucture of the shuttle isn't infinitely strong.

Who says the pressure is infinite? Pressure is created by molecules bouncing off the walls of the container. The pressure is limited (though still quite large). That is why they use special metal for the space craft rather than a large balloon.
benjmd
Hmm, Indi definitely offered the fancy explanation, but hopefully I can offer something more basic.

If you've ever had a physics course, you know that forces are expressed as vectors -- that is, they have a magnitude and a direction. The net force on something is the sum of those magnitudes and directions (5 Newtons in one direction and 6 Newtons in the opposite direction result in a net of 1 Newton in the latter direction). Physical forces are discrete values in a 3-dimensional space. Vectors are added and subtracted from each other; they are not multiplied or divided in this case.

A simple trick to understanding this is simply to look at units in the factor-label method.

Force (F) = Pressure (P) * Area over which pressure is applied (A)

Think about a tire on a car. On my car, the tire has a 14" radius, an 8" width, and a side wall about 5". I estimate the surface area of the inside of the tire at about 1400 square inches (call this A). The pressure in my tires is 35 pounds/square inch (call this P1). Atmospheric pressure is 14.7 pounds/square inch (call this P2).

If we want to determine the pressure on the tire material, our vectors would say that there is a pressure pointing "out" which is +35 pounds/sq inch (psi) and a pressure pointing "in" which is -14.7 psi. Because 14.7 is pointing the opposite direction of the 35, I have given the two values opposite signs and arbitrary designated "out" as the positive direction because I know that the pressure in the tire has to have a net force outward to allow it to hold up the car. The net addition of the pressures is 35psi + -14.7psi = 20.3psi.

F = 20.3 Pounds/Square Inch * 1420 Square Inches = 28,826 Pounds. The pound is a unit of force in the American weights and measures system.

If we used the ratio of the inside and outside pressure, our "net pressure" would be calculated as:
(35 pounds/sq inch) / (14.7 pounds/sq inch) = 2.38. Note that this no longer has units, as when you divide pounds/sq inch by pounds/sq inch the units cancel out! Plugging this in to the force equation:

F = 2.38 * 1420 Square Inches = 3381 Square Inches.... The force is in units of area!! This surely is not correct. Therefore, we must calculate the difference in the pressures rather than the ratio of the pressures to figure out the force on the surface separating the pressures.
cr3ativ3
Ok thanks everyone, by the time I am threw high school I probably would even wonder this, I am going to high school next year so that's why I didn't really know, just wanted to know before some physics course in highschool .

Anyways I understand now I was just confused thinking if the pressure is absolutely 0 outside then any pressure on the inside must be infinitely strong. But I now know that's not true.

Thanks everyone for your help with my problem!
qsmith
I think the answer to question is simpler than all of the current explanations. The original confusion is that you are using the wrong way of comparing. What we are talking about is a differential (i.e for difference you subtract) not a ratio (for ratio you divide)
so the pressure that needs to be contained is 1atm(inside)-0atm(otside)=1 atmosphere

There is no infinite difference
qsmith
Oh, and about that balloon popping - It may not actually pop as it goes up, provided that it was not inflated to its maximum streching capacity - As the balloon rises, the outer pressure drops, the differential gets bigger, and the balloon will just get bigger. one of two things will happen - it will grow to the point where the weight of the latex is balanced out and it will just float there, or a weak point in the ballon will tear and it will pop.