How did theorists determine that the atmosphere attenuates enough to support unpowered orbits?

Before getting rockets into space were scientists certain that the density of the earth's atmosphere would decrease with height enough to permit low earth orbits? If so how was that determined?

(If the atmosphere does not become negligible then any attempt to orbit would require significant ongoing propulsion to overcome the atmospheric drag. I suppose one could deduce from observing natural satellites that at some distance from the surface that happens, but prior to getting an observable artificial satellite in orbit I assume the only observable natural earth satellite was the moon, and there is quite a distance between that and low-earth orbit!)

• sky-lights.org/2021/06/14/qa-how-we-knew-space-was-a-vacuum Mar 6 at 18:39
• Every meteoroid would be another natural data point.
– mdc
Mar 6 at 22:03
• Before V-2 there has already been canons that could shoot into orbital altitudes. The ballistics of those would make it clear that there's almost no drag at high altitude. Mar 7 at 8:19
• @MichaelCoon what information about upper atmosphere could be deduced from earth-observable meteroids? (Other than the previous conclusion that at some distance between earth and moon the atmosphere becomes negligible.) Mar 7 at 16:18
• Additional context: By the time rockets and Sputniks were attempting to orbit, the exponential fall-off in atmospheric density should have been well-known for 100 years. But i guess then you are asking how did they know. Mar 7 at 17:53

If 200 miles up counts as a low earth orbit height, then the appreciation is as old as Newton's Principia, that there was no atmosphere left there, and nothing like atmospheric drag to oppose motion.

In book 2 of the Principia, Newton made it a point to study the effects of resisting media on motion. Applying Book 2 Prop.22 (scholium) (https://books.google.com/books?id=6EqxPav3vIsC&pg=PA73 up to PA76), he worked out at what rate the combination of gravitational and (de)compressive effects should cause the density of the earth's atmosphere to decrease with distance above the earth's surface (see esp. Book 3 Prop. 10, (https://books.google.com/books?id=6EqxPav3vIsC&pg=PA230).

The figures in his calculations were off by today's standards, because they did not include temperature effects -- in turn because in Newton's time there was not yet any science or technique of thermometry, and he had no means of knowing about the temperature component of the gas laws.

Nevertheless, he calculated that at an altitude of even 200 miles up, the air should be 'rarer' (i.e. less dense) than at the earth's surface in a ratio of roughly 75 x 10^12 to 1. This is effectively a vacuum (and the conclusion answers a quantitative point left open by the discovery that the height of mercury in a barometer stands lower when the instrument is carried to higher altitudes).

In Book 3 Prop.10 Newton went on to conclude that there should be no atmospheric drag on the motion of Jupiter, and in the General Scholium at the end of the book he afterwards wrote: (https://books.google.com/books?id=6EqxPav3vIsC&pg=PA388) (translation slightly amended here, closer to the Latin original):

"Bodies, projected in our air, suffer no resistance but from the air.

Withdraw the air, as is done in Mr. Boyle's vacuum, and the resistance ceases. For in this void a bit of fine down and a piece of solid gold descend with equal velocity.

And the same reasoning applies to the celestial spaces which are above the Earth's atmosphere".

The scale height is a useful way to describe how the atmospheric pressure decreases with altitude. From Wikipedia,

For planetary atmospheres, scale height is the increase in altitude for which the atmospheric pressure decreases by a factor of $$e$$. The scale height remains constant for a particular temperature. It can be calculated by $$H = \frac{RT}{Mg}$$ where $$R$$ is the gas constant, $$T$$ is the temperature in Kelvin, $$M$$ is the mass of 1 mole of atmospheric particles, and $$g$$ is the gravitational acceleration.

That formula assumes that $$T$$ and $$g$$ are constant. With a bit of calculus, we can produce a formula that handles the variation in $$g$$ with altitude, but that's not really necessary, since $$g$$ at the Kármán line isn't much lower than what it is at sea level. The variations due to $$T$$ are much more significant, and it's not easy to measure those from the ground, even with modern technology, but using weather balloons we did have a fairly good idea how the atmospheric temperature varies before the space era began.

Here are some typical scale height values for Earth, courtesy of Wikipedia:

Temp (K) Height (m)
290 8500
273 8000
260 7610
210 6000

Eg, if the scale height is $$H = 8500$$ m, then $$6H=51$$ km, and $$e^{-6}\approx0.00248$$, so the air pressure at $$51$$ km is around $$\frac14$$% of the sea level pressure.

Curiously, that Wikipedia article gives no historical info regarding when we first learned about scale height. However, the article on the barometer does mention that

However, Pascal went even further to test the mechanical theory. If, as suspected by mechanical philosophers like Torricelli and Pascal, air had weight, the pressure would be less at higher altitudes. Therefore, Pascal wrote to his brother-in-law, Florin Perier, who lived near a mountain called the Puy de Dôme, asking him to perform a crucial experiment.

Perier was to take a barometer up the Puy de Dôme and make measurements along the way of the height of the column of mercury. He was then to compare it to measurements taken at the foot of the mountain to see if those measurements taken higher up were in fact smaller. In September 1648, Perier carefully and meticulously carried out the experiment, and found that Pascal's predictions had been correct. The column of mercury stood lower as the barometer was carried to a higher altitude.

In the subsequent years, variations on this experiment have been performed, measuring temperature as well as barometric pressure, and carefully recording the time of measurement.

By the dawn of the space era, sensitive barometric pressure measurements were the standard way for aviators to determine their altitude, so estimating the air pressure at orbital altitudes was a simple exercise for any aeronautical navigator. See Altimeter for details.

The change in temperature with altitude is generally known as the lapse rate. From Wikipedia,

The lapse rate is the rate at which an atmospheric variable, normally temperature in Earth's atmosphere, falls with altitude.

[...]

The environmental lapse rate (ELR), is the rate of decrease of temperature with altitude in the stationary atmosphere at a given time and location. As an average, the International Civil Aviation Organization (ICAO) defines an international standard atmosphere (ISA) with a temperature lapse rate of $$6.50$$ ºC/km from sea level to $$11$$ km. From $$11$$ km up to $$20$$ km the constant temperature is −56.5 °C, which is the lowest assumed temperature in the ISA. The standard atmosphere contains no moisture.

Unlike the idealized ISA, the temperature of the actual atmosphere does not always fall at a uniform rate with height. For example, there can be an inversion layer in which the temperature increases with altitude.

The behaviour of dry air is relatively simple to model. Moist air is much more complicated because water vapour is not well-modelled as an ideal gas. Water has a substantial latent heat of vaporisation, so it has a big effect on the air temperature when it vaporises and condenses.

No answer has yet mentioned the concept of hydrodynamic equilibrium, which dates back to Newton. The Earth's atmosphere has been more or less stable for billions of years. It is not expanding or contracting, and while there is some up and down motion, those motions balance out.

This means that each parcel of the atmosphere is buoyed up by pressure from below and pulled down by gravity and pressure from above. To be in an equilibrium state, this means that $$\frac{dP}{dh} = -\rho g \tag{1}$$ where $$P$$ is pressure, $$h$$ is altitude above sealevel, $$\rho$$ is density, and $$g$$ is the downward acceleration due to gravity. Another way to look at hydrostatic equilibrium is that the pressure at any given layer of the atmosphere is just enough to support the weight of all of the atmosphere above that layer. Note that both $$\rho$$ and $$g$$ are strictly positive. This means that pressure must necessarily decrease with increasing altitude.

Assuming the ideal gas law, $$PV=nRT \tag{2}$$ where $$P$$ once again is pressure, $$V$$ is volume, $$n$$ is the number of particles, $$R$$ is the universal gas constant, and $$T$$ is absolute temperature, another way to express this law is $$P = \rho R^\ast\, T \tag{3}$$ where $$\rho$$ is density, $$R^\ast$$ is the specific gas constant, and $$T$$ is temperature. Combining equations (1) and (3) results in $$\frac{dP}{dh} = -P \frac{g}{R^\ast\, T} \tag{4}$$

Assuming that $$g$$, $$R^\ast$$, and $$T$$ are constant, equation (4) immediately results in an exponential decline in pressure with altitude. We know that temperature is not constant; mountain climbers have known that for a long, long time. Adding a decreasing temperature to equation (4) results in a super exponential decline in pressure with altitude.

The lowest three layers of the atmosphere exhibit a temperature that either decreases with altitude (the thermosphere and the mesosphere) or remains more or less constant with altitude (the stratosphere). The pressure (and hence density) at the top of the mesosphere is very low. The top of the mesosphere was one of the definitions of the Kármán line, where the air becomes so thin that an airplane couldn't fly. (The Kármán line is now arbitrarily set at 100 km.)

So to whom should credit be given? It wasn't just one person. Newton certainly played a role, but so did Bernoulli, Kármán, and many others.

• Was thinking about writing something similar for a while. There is also a math.stackexchange post with the same point of view. Mar 9 at 21:50
• The question is how to determine that freedom from drag (enabling an unpowered orbit) is possible at the height of a low earth orbit. Newton identified, in passages cited and linked above, that the atmosphere at 200 miles up is vanishingly thin, effectively a vacuum, and that in vacuo, including above the atmosphere, resistance by the medium (drag) disappears and orbital motion persists. Apart from absence of reference (acknowledged) to temperature effects, what part of the answer to the question had Bernoulli, Karman and 'many others' to supply? Descriptions and citations please! Mar 10 at 2:53

Sea level air pressure is 101.3 kPa (14.7 psi). This means there are 1.03 kg of air in 1 cm² column of air stretching from sea level to space (this is 14.7 pounds of air in a 1” square column). One kilogram of air occupies around 816 liter at sea level at 15°C (one pound of air occupies 12.4 cubic foot) at sea level pressure… a 1” square column 1785 feet high.

If we go up to 1785 foot altitude, there are now only 13.7 pounds of air above us and local pressure is 13.7psi. The air is 13.7/14.7 as dense as sea level, so a pound of air occupies 14.7/13.7 times as much or a column 1915 feet high.

If we go up to 1785+1915=3700ft, the air pressure will be 14.7-2=12.7psi.

And so on until you have determined pressure at the altitude of interest.

This method does not correct for effects of temperature, but is simple enough it could be performed pre-calculus (17th century)

Note: This calculator https://www.mide.com/air-pressure-at-altitude-calculator gives pressure of 12.84psi at 3700 ft altitude as a reality check.

• @TheMatrixEquation-balance the boundary between the Earth's uppermost atmosphere and the thermosphere isn't relevant to this question or this answer. Unpowered orbits are viable below the magnetosphere for extended periods. Mar 7 at 0:42
• This answer gives a possible way of figuring out that space is a vacuum, but the question asks how it was done. Note the "history" tag. Is this how it was done? If so, by who? Mar 7 at 1:34
• This answer is wrong; one cannot take sea level air density as constant, because it falls exponentially. It's also illegible in most of the world due to its choice of idiosyncratic units. I tried to correct it, but gave up because the linear approximation is wrong to begin with. Mar 7 at 7:31
• @gerrit: The answer is a stepwise approximation: approximate the density as constant for a small mass of air, figure out where that mass would end, compute a new density at the top of that column, and repeat. Smaller steps give better approximations. In the limit, you get that exponential falloff, but the answer assumes no knowledge of concepts like limits. Mar 7 at 10:11
• "This method does not correct for effects of temperature, but is simple enough it could be performed pre-calculus (17th century)" This is a linear extrapolation technique that would lead to the wrong answer at 100km altitude, and wouldn't even reproduce the correct variation of the atmospheric scale height with altitude in the isothermal approximation, as you need the weakening of gravity for that. The calculator you link is inconsistent with this pre-calculus technique you describe. The units you use are obscure. Hence a clear downvote. Mar 7 at 17:16

Depends how far back you want to go.

People would've climbed mountains a very long time ago and would've noticed less air. It would be reasonable to conclude that if you went even higher up there would be less and less and at some point none.