# Mass of the earth's atmosphere given the density as a function of h

• swashbuckler77
In summary, the conversation discusses an exponential model for the density of the Earth's atmosphere and how to approximate the mass of the portion of the atmosphere from the surface to 100 meters above sea level. The equations used involve the surface area of a sphere and the density as a function of height. The solution can be approached through integration by parts, but it is important to consider the small variation in density over 100 meters, which can affect the accuracy of the calculations.
swashbuckler77

## Homework Statement

An exponential model for the density of the earth’s atmosphere says that if the temperature of the atmosphere were constant, then the density of the atmosphere as a function of height, h (in meters), above the surface of the Earth would be given by δ(h) = 1.28e(−0.000124h) kg/m3.

Write and evaluate a sum that approximates the mass of the portion of the atmosphere from h = 0 to h = 100 m (i.e., the ﬁrst 100 meters above sea level). Assume the radius of the Earth is 6400 km.

## Homework Equations

Since this is like a density distribution problem but in three dimensions, I surmised that now, instead of using mass=2∏∫r*δ(r) dr from a to b (the circumference of a circle and the formula provided in the textbook's example problem), I will use mass=4∏∫h2*δ(h) dh, which is the surface area of a sphere with density as a function δ(h) distributed over 6400000 to 6400100, because going from circumference of a circle to surface area of a sphere seems like a logical transition from 2D to 3D.

## The Attempt at a Solution

mass=4∏∫h2*δ(h) dh from 6400000 to 6400100

=4∏∫h2*1.28e(−0.000124h) dh from 6400000 to 6400100

=5.12∏∫h2*e(−0.000124h) dh from 6400000 to 6400100

I'm unsure if I am going in the right direction because this looks like a very ugly integral with a lot of zeroes that make me rather uncomfortable.

Last edited:
It is not a very hugly integral.
It is easy to do by integration by parts.
However, check that the variation of density over 100 meters is very small.
For a physics course, you could as well assume it is constant or it varies linearly (Taylor or interpolation).
Anyway, if you want to see the effect of the exponential dependence over only 100m, you need to care for all the digits.

I checked that:

- the error is 1% if you assume constant density
- the error is 0.005% by Taylor expansion
- the error is 0.001% by linear interpolation

## 1. What is the density of the earth's atmosphere?

The density of the earth's atmosphere varies depending on location and altitude. On average, it is around 1.2 kg/m^3 near the surface.

## 2. How does the density of the earth's atmosphere change with altitude?

The density of the earth's atmosphere decreases as altitude increases. This is due to the decrease in air pressure and the gravitational pull of the earth.

## 3. What is the relationship between density and altitude in the earth's atmosphere?

The relationship between density and altitude in the earth's atmosphere is inverse. As altitude increases, density decreases.

## 4. How is the mass of the earth's atmosphere calculated using the density as a function of altitude?

The mass of the earth's atmosphere can be calculated by integrating the density function with respect to altitude. This will give the total mass of the atmosphere above a certain altitude.

## 5. What are some factors that can affect the density of the earth's atmosphere?

Several factors can affect the density of the earth's atmosphere, including temperature, humidity, and air pressure. Changes in these factors can cause fluctuations in the density of the atmosphere at a given altitude.

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