Practice
practice problem 1
Determine the mass of the Earth's atmosphere.
solution
Since pressure is force divided by area, the force of the atmosphere pressing on the surface of the Earth can be found by multiplying standard atmospheric pressure by the surface area of the Earth. Given that the force of an object's weight is its mass times the acceleration due to gravity, the mass of the Earth's atmosphere is the force it exerts divided by the acceleration due to gravity. Or symbolically…
F = 
PA = P(4πr^{2}) = W = mg 




m = 
(101,325 Pa)(4π)(6.37 × 10^{6} m)^{2} 
9.8 m/s^{2} 


practice problem 2
Determine the maximum height that a lift pump can raise water from a well.
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.


h = 
(101,325 Pa) 
(1,000 kg/m^{3})(9.80 m/s^{2}) 


practice problem 3
The first measurements of blood pressure were made in 1726 by the English botanist, physiologist, and clergyman, Stephen Hales. Hales performed several experments on horses deemed "unfit for service". You must recall that at the time horses were primarily used as working animals. Those that were seriously injured, chronically ill, or otherwize unable to perform their duties were routinely slaughtered and eaten. Read Hales' description of one such experiment, then determine the blood pressure of his poor, unfortunate horse.
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….
Stephen Hales
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
When the human body is accelerated vertically, blood pressure in the brain will drop. Determine the maximum vertical acceleration that a human can withstand before losing consciousness; that is, determine the acceleration that would reduce the blood pressure in the brain to zero. Assume a typical systolic pressure of 16 kPa and that the base of the brain is 20 cm above the top of the heart.
solution
In this problem we are given pressure difference, height difference, and density and are asked to find acceleration.


g′ = 
(16,000 Pa) 
(1035 kg/m^{3})(0.20 m) 

g′ = 
77.294… m/s^{2} ≈ 8 g 



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 very 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
Astronomical Pressures.
 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…
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…
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 = 
⌠ ⌡ 
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…
at the center where r = 0.
 Let's do it.
 For the Earth…

P_{0} = 
3(6.67 × 10^{−11} Nm^{2}/kg^{2})(5.98 × 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 2300 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} = 
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?