Q.The density of the atmosphere at sea level is $1.29\ \text{kg/m}^{3}$. Assume that it does not change with altitude. Then how high would the atmosphere extend?

The key idea here is Hydrostatic Pressure, which, in this simplified scenario of constant density, allows us to model the total height of the atmosphere. While the true atmosphere exhibits an exponential decay in density (related to the concept of Atmospheric Pressure Scale Height), this problem simplifies it to a uniform density column.

1. The pressure at sea level ($P_0$) is the total weight of the atmospheric column above a unit area.
2. For a fluid of constant density $\rho$ and height $h$, the pressure exerted at its base is given by the formula $P = \rho g h$.
3. We use the standard atmospheric pressure at sea level, $P_0 = 1.013 \times 10^5\ \text{Pa}$, and the acceleration due to gravity, $g = 9.8\ \text{m/s}^2$. …

The problem asks us to determine the height the atmosphere would extend if its density remained constant at its sea-level value. This is a classic conceptual problem that simplifies the complex reality of our atmosphere to illustrate the fundamental relationship between pressure, density, and height.

Concept: Atmospheric Pressure and Hydrostatic Equilibrium

Atmospheric pressure at any point is caused by the weight of the air column directly above that point. At sea level, the atmospheric pressure is the total weight of the entire atmosphere above a unit area.

In a fluid (like air) at rest, the pressure at a certain depth (or height, in the case of the atmosphere) is given by the hydrostatic pressure formula. This formula is derived from the idea that the pressure at the base of a fluid column supports the entire weight of that column.

If we assume the density of the atmosphere ($\rho$) is constant, then the weight of a column of air of height $h$ and cross-sectional area $A$ is $W = mg = (\rho V)g = (\rho A h)g$. The pressure $P$ at the base of this column is the force (weight) divided by the area: $P = \frac{W}{A} = \frac{(\rho A h)g}{A} = \rho g h$.

This simplified model allows us to calculate an "effective height" for the atmosphere. In reality, the density of the atmosphere decreases significantly with altitude, which is why the atmosphere doesn't have a sharp, well-defined upper boundary and extends much further than this calculation suggests. However, this calculation provides a useful benchmark.

Here is the step-by-step solution:

1. Identify the relevant physical quantities and their standard values.

   We need the standard atmospheric pressure at sea level, the acceleration due to gravity, and the given density of the atmosphere.

   • Standard atmospheric pressure at sea level, $P_0 = 1.013 \times 10^5 \text{ Pa}$ (Pascals). This is approximately $1 \text{ atmosphere}$.
   • Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$.
   • Density of the atmosphere at sea level, $\rho = 1.29 \text{ kg/m}^3$ (given).

2. Relate pressure, density, and height for a fluid of constant density.

   As discussed in the concept section, for a fluid with uniform density $\rho$, the pressure $P$ at a depth $h$ (or at the base of a column of height $h$) is given by:

   $P = \rho g h$

   Here, $P$ is the pressure exerted by the column of fluid, $\rho$ is the constant density of the fluid, $g$ is the acceleration due to gravity, and $h$ is the height of the fluid column.

3. Set up the equation to solve for the height.

   We are assuming that the entire atmospheric pressure at sea level ($P_0$) is due to a column of air of constant density $\rho$ extending to a height $h$. Therefore, we can equate the standard atmospheric pressure to the pressure formula:

We need to find $h$, so we rearrange the formula:
$$h = \frac{P_0}{\rho g}$$ …

Model constant-density column: P0=rhogh. h=P0/(rhog)=1.013e5/(1.299.8)~=8.01e3 m=8.01 km. Real atmosphere tal …