Pressure
Practice
practice problem 1
 Determine the mass of Earth's atmosphere using the value of the standard atmosphere.
 Determine the scale height of the atmosphere — the height the atmosphere would have if its density stayed constant instead of decreasing with altitude. (The scale height is a useful approximation for some calculations in atmospheric sciences.)
solution
Since pressure is defined as force divided by area, the force of the atmosphere can be found by multiplying pressure by area.
F = PA ⇐ P = F A The force of the atmosphere is due to gravity and is called weight (mg). The area it is pressing down on is the surface area of a sphere (4πr^{2}). Substitute these expressions.
mg = P(4πr^{2})
Solve for the goal of the problem.
m = P(4πr^{2}) g Put numbers in…
m = (101,325 Pa)(4π)(6.37 × 10^{6} m)^{2} 9.8 m/s^{2} Get answer out…
m = 5.27 × 10^{18} kg
Compare this to the mass of the whole Earth.
5.27 × 10^{18} kg ≲ 0.0001% 5.97 × 10^{24} kg The atmosphere is a tiny fraction of the Earth's total mass.
Once again, force is pressure times area because pressure is force divided by area.
F = PA ⇐ P = F A The force is still the weight of the Earth's atmosphere (mg), but this time don't do anything to the area term.
mg = PA
Solve for mass…
m = PA g …and then work on something else.
Since density is defined as the ratio of mass to volume, mass is the product of density and volume. (Remember, the symbol for density is the Greek letter rho not the Latin letter p.)
m = ρV ⇐ ρ = m V Treat the atmosphere as a very thin shell on top of the surface of the earth. Its volume would then be the surface area of the earth (A) times the thickness of the shell (h).
m = ρAh
Set the two mass equations equal to one another.
ρAh = PA g Solve for thickness and note how nicely area cancels out.
h = P ρg We're ready for numbers…
h = (101,325 Pa) (1.21 kg/m^{3})(9.8 m/s^{2}) …and an answer.
h = 8.5 km
Compare this to the radius of the whole Earth.
8.5^{} km ≳ 0.1% 6371 km Don't think about it as "How thick is the atmosphere" but rather "How thin is the atmosphere". A standard pack of copier or printer paper has 500 sheets. Unwrap two of them and stack one on top of the other. Remove one sheet of paper. That's the thickness of the Earth's atmosphere.
practice problem 2
solution
A lift pump works by reducing the pressure above of a column of water. The greatest difference possible would be atmospheric pressure at the bottom and vacuum at the top. Set this pressure difference equal to the pressure difference within the column of water and solve for height.







practice problem 3
In December I laid a common field gate on the ground, with some straw upon it, on which a white mare was cast on her right side, and in that posture bound fast to the Gate; she was fourteen hands and three inches high [150 cm], lean, tho' not to a great degree, and about ten or twelve years old. This and the abovementioned horse and mare were to have been killed, as being unfit for service….
Then laying bare the left carotid artery, I fixed to it towards the heart the brass pipe, and to that the windpipe of a goose; to the other end of which a glass tube was fixed, which was twelve feet nine inches long [388 cm]. The design of using the windpipe was by its pliancy to prevent the inconveniencies that might happen when the mare struggled; if the tube had been immediately fixed to the artery, without the intervention of this pliant pipe.
There had been lost before the tube was fixed to the artery, about seventy cubic inches of blood [1.15 L]. The blood rose in the tube in the same manner as in the case of the two former horses, till it reached to nine feet six inches height [290 cm]. I then took away the tube from the artery, and let out by measure sixty cubick inches of blood [0.98 L], and then immediately replaced the tube to see how high the blood would rise in it after each evacuation; this was repeated several times, till the mare expired….
solution
Use the formula for the gauge pressure in a uniform fluid, take the maximum height of the column of blood, and solve.
ΔP = ρgΔh ΔP = (1035 kg/m^{3})(9.8 m/s^{2})(3.88 m) ΔP = 39,354.84 Pa ΔP = 39 kPa 
Compared to the typical human values of 10 to 16 kPa for arterial blood pressure, this result seems reasonable. Horses are much bigger than people and thus need a generally higher arterial pressure to squeeze the blood to every distant nook and cranny. Also, blood pressure is generally higher when an animal is under stress. Slowly bleeding to death is definitely a stressful situation. This method of determining blood pressure is called invasive catheterization and is almost never used. Blood pressure is now routinely determined by much less deadly means.
practice problem 4
solution
In this problem we are given pressure difference, height difference, and density and are asked to find acceleration.







The value g′ is the apparent acceleration due to gravity. Since gravity is already pulling us down with 1 g, the absolute acceleration that a human could withstand is on the order 7 g. Since the height difference was measured from the bottom of the brain, 7 g would be the acceleration at which the brain was entirely emptied of blood. The actual acceleration that would induce unconsciousness would be somewhat lower and would be preceded by a period of greyout and then blackout as the visual cortex was drained of blood. With training and special clothing, it is possible to remain conscious at accelerations greater than what we just calculated. At the 2002 Ilopango Airshow in El Salvador, aerobatic pilot Greg Poe pulled a maximum of 11.4 g apparent acceleration for a second or two at the start of a rapid ascent. This is the current record for a civilian pilot and may be an overall record. Since most air forces keep this kind of information classified, we can't be sure.
practice problem 5
 Derive an expression for the pressure in a spherical, astronomical body with uniform density.
 Use this formula to estimate the pressure at the center of…
 the earth
 the sun
solution
We simplify things a bit by assuming a constant density…
ρ = m = 3m V 4πr^{3} but we can't do the same for gravity. On astronomical scales, gravity varies considerably. This means it's time to reach for a calculusbased solution. We'll begin by determining just how gravity varies. Start with Newton's law of universal gravitation…
g(r) = Gm(r) r^{2} On the surface of the Earth we'd use the whole mass of the Earth in this equation, but inside the Earth we use only the fraction that's at a greater depth; that is, at a distance r from the center of the Earth smaller than the the radius R of the whole earth. The mass of this portion can be found by multiplying density and volume…
m(r) = ρV(r) = 3m 4πr^{3} = mr^{3} 4πR^{3} 3 R^{3} which makes sense. Mass varies as the cube of length, so the fraction should be some sort of ratio of the cubes. Now, substitute and simplify.
g(r) = G mr^{3} = Gmr r^{2} R^{3} R^{3} Pressure in a fluid (yes, I know the Earth is mostly solid, but the equation works) is the weight of the fluid above a surface divided by the area of the surface. The surface can have any area and, through the magic of algebra, disappears from the equation so that we are left with the product of density (ρ), gravity (g), and height (h in swimming pools and blood vessels, r in astronomical situations like this). Now for the calculus. You can't assign a value for gravity in this situation. It varies from 9.8 m/s^{2} on the surface to zero at the center. We can reduce the amount of variation if we examine just a part of this total distance (Δr). We can reduce it even more if we examine an even smaller part. And we can reduce the variation to nothing if we examine an infinitesimal part (dr). Now the product of density (ρ), gravity (g), and height (dr) works again. All we have to do is add up the contributions to the pressure made by the infinite number of infinitesimal parts from the surface of the Earth down to its center. The process of adding infinitesimals is called integration.
R P = ⌠
⌡ρg(r)dr r R P = ⌠
⌡3m Gmr dr 4πR^{3} R^{3} r R P = 3Gm^{2} ⎡
⎣r^{2} ⎤
⎦4πR^{6} 2 r and here's our equation…
P = 3Gm^{2} (R^{2} − r^{2}) 8πR^{6} which reduces to…
P_{0} = 3Gm^{2} 8πR^{4} at the center where r = 0.
Let's do it.
For the Earth…
P_{0} = 3Gm^{2} 8πR^{4} P_{0} = 3(6.67 × 10^{−11} Nm^{2}/kg^{2}) (5.97 × 10^{24} kg)^{2} 8π(6.34 × 10^{6} m)^{4} P_{0} = 1.7 × 10^{11} Pa = 170 GPa P_{0} = 1.7 million atmospheres The actual value is closer to 360 GPa or about twice the value calculated above, which is annoyingly big, but at least we got the right order of magnitude. To do this correctly, we'd have to account for variations in density with depth. The density of the Earth starts at about 2,300 kg/m^{3} at the crust, increases (nonuniformly) with depth in the mantle, jumps drastically at the outer core where it nearly doubles, and keeps increasing (nonuniformly) hitting a maximum of 12,580 kg/m^{3} at the center.
For the sun…
P_{0} = 3Gm^{2} 8πR^{4} P_{0} = 3(6.67 × 10^{−11} Nm^{2}/kg^{2}) (1.99 × 10^{30} kg)^{2} 8π(6.96 × 10^{8} m)^{4} P_{0} = 1.3 × 10^{14} Pa = 130 TPa P_{0} = 1.3 billion atmospheres The actual value is on the order of 100 to 300 billion atmospheres. I wonder what's going on here? Probably something to do with density variation again. The density of the sun is a whispy 0.0002 kg/m^{3} on the visible surface and increases exponentially until it maxes out at 150,000 kg/m^{3} in the core.