The weight of the atmosphere above 1 m^2 of Earth's surface is about 100,000 N. Density, of course, becomes less with altitude. But suppose the density of air were a constant 1.2 kg/m/s^3. Calculate where the top of the atmosphere would be.
To calculate where the top of the atmosphere would be if the density of air were a constant 1.2 kg/m^3, we can use the concept of atmospheric pressure.
In this case, we know that the weight of the atmosphere above 1 m^2 of Earth's surface is about 100,000 N. This weight is essentially the atmospheric pressure exerted on the Earth's surface.
We can use the formula for pressure, which is given by P = ρgh, where P is pressure, ρ is density, g is the acceleration due to gravity, and h is the height.
Given that the atmospheric pressure is 100,000 N, we can rearrange the formula to solve for the height (h) of the atmosphere:
h = P / (ρg)
Substituting the given values into the formula:
h = 100,000 N / (1.2 kg/m^3 * 9.8 m/s^2)
Simplifying the calculation:
h ≈ 8,680 meters
Therefore, if the density of air were a constant 1.2 kg/m^3, the top of the atmosphere would be approximately 8,680 meters above the Earth's surface.