What would be the height of the atmosphere if the air density (a) were constant and (b) decreased linearly to zero with height? Assume a sea-level density of 1.21 kg/m³.

To determine the height of the atmosphere under the given conditions, we need to consider two scenarios: when air density remains constant and when it decreases linearly to zero with altitude. Let's break this down step-by-step for clarity.

Scenario A: Constant Air Density

If we assume that air density remains constant throughout the atmosphere, we can use the concept of pressure and the hydrostatic equation. The hydrostatic equation states that the change in pressure with height is equal to the negative product of density, gravitational acceleration, and the change in height:

Mathematically, this can be expressed as:

P = P0 - ρgH

Where:

• P = pressure at height H
• P0 = sea-level pressure (approximately 101325 Pa)
• ρ = air density (1.21 kg/m³)
• g = acceleration due to gravity (approximately 9.81 m/s²)

As we move upward, pressure decreases until it reaches zero at the top of the atmosphere. Setting P to zero, we can solve for H:

0 = P0 - ρgH

H = P0 / (ρg)

Plugging in the values:

H = 101325 Pa / (1.21 kg/m³ * 9.81 m/s²) ≈ 8445 meters

This indicates that if air density were constant at sea level values, the atmosphere would extend approximately 8445 meters high.

Scenario B: Linearly Decreasing Air Density

Now, let’s consider the case where air density decreases linearly with height. This means that as we go higher, the density drops proportionally until it reaches zero at some height H. We can express air density as:

ρ(H) = 1.21 kg/m³ * (1 - (H/Hmax))

Where Hmax is the height of the atmosphere we want to find. Using the hydrostatic balance again, we can derive the pressure equation. The pressure at height H can be expressed as:

P(H) = P0 - ∫(ρ(H)g dH)

Substituting our expression for ρ(H) and integrating, we find that:

P(H) = 101325 Pa - ∫(1.21(1 - (H/Hmax))g dH

Solving this integral from 0 to Hmax, we can find that:

P(Hmax) = P0 - (1.21g/2)Hmax

Setting P(Hmax) to zero gives us:

0 = 101325 Pa - (1.21*9.81/2)Hmax

Now we can solve for Hmax:

Hmax = 101325 Pa / (1.21 * 9.81 / 2) ≈ 16790 meters

So, if the air density decreases linearly to zero, the atmosphere could extend approximately 16790 meters high.

Summary

In summary, if air density were constant, the height of the atmosphere would be around 8445 meters. However, if the air density decreases linearly to zero, the atmosphere could reach about 16790 meters. These calculations illustrate how different assumptions about air density can significantly impact the estimated height of the atmosphere.