The weight of the atmosphere above 1 m2{\rm m}^{2} of Earth's surface is about 100,000 N{\rm N}. Density, of course, becomes less with altitude. But suppose the density of air were a constant 1.2 kg/m3{\rm kg}/{\rm m}^{3}.

Calculate where the top of the atmosphere would be.

How do I find atmosphere ?

Weight per area is pressure, it is given as 100kPa/m^2 above.

but this weight would equal density*volume*9.8N/kg, or
density*1m^2*height*9.8

set that equal to 100kN/m^2, and solve for height.

24.5

8.3x10^3m

To find the position of the top of the atmosphere, we can use the concept of pressure. The weight of the atmosphere above a unit area is the same as atmospheric pressure.

First, let's find the pressure at the surface of the Earth:
Pressure = Force / Area
Given that the weight of the atmosphere above 1 m² of Earth's surface is about 100,000 N, and the standard gravity is approximately 9.8 m/s², we can calculate the pressure at the surface:
Pressure = 100,000 N / 1 m² = 100,000 N/m² = 100 kPa

Now we need to determine the density at the surface, which is given as 1.2 kg/m³.

To find the altitude at which the density of air drops to a constant value of 1.2 kg/m³, we can use the hydrostatic equation, which relates pressure, density, and altitude in a fluid:

P = ρgh

Where:
P is the pressure,
ρ is the density,
g is the acceleration due to gravity (approximately 9.8 m/s²),
h is the height or altitude.

Since we want to find the altitude at which the density is 1.2 kg/m³, we can rearrange the equation to solve for h:

h = P / (ρg)

Substituting the values we already have:
h = 100,000 N/m² / (1.2 kg/m³ * 9.8 m/s²) ≈ 8,680 m

Hence, the top of the atmosphere would be approximately at an altitude of 8,680 meters (or 8.68 kilometers).