The planets with radii `R_(1)` and `R_(2)` have densities `p_(1),p_(2)` respectively. Their atmospheric pressues are `p_(1)` and `p_(2)` respectively.Therefore, the ratio of masses of their atmospheres, neglecting variation of g within the limits of atmoshpere is

Options

A. `rho_(1)R_(2)p_(1)//rho_(2)R_(1)rho_(2)`
B. `p_(1)R_(2)rho_(2)//pP_(2)R_(1)rho_(1)`
C. `p_(1)R_(1)rho_(1)//p_(2)R_(2)rho_(2)`
D. `p_(1)R_(1)rho_(2)//p_(2)R_(2)rho_(1)`

Answer

To solve the problem, we need to find the ratio of the masses of the atmospheres of two planets based on their given parameters. Let's break down the solution step by step.

### Step 1: Understanding the relationship between gravity, density, and radius
The acceleration due to gravity \( g \) on the surface of a planet can be expressed as:
\[ g = \frac{G \cdot M}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is its radius.

The mass \( M \) of a planet can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ M = \rho \cdot V \]
For a spherical planet, the volume \( V \) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, the mass becomes:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]

### Step 2: Expressing gravity in terms of density and radius
Substituting the expression for mass into the formula for gravity, we get:
\[ g = \frac{G \cdot \left(\rho \cdot \frac{4}{3} \pi R^3\right)}{R^2} = \frac{4}{3} \pi G \rho R \]
This shows that gravity is proportional to the product of density and radius:
\[ g \propto \rho R \]

### Step 3: Finding the ratio of gravitational accelerations
For two planets with densities \( \rho_1, \rho_2 \) and radii \( R_1, R_2 \), the ratio of their gravitational accelerations \( g_1 \) and \( g_2 \) can be expressed as:
\[ \frac{g_1}{g_2} = \frac{\rho_1 R_1}{\rho_2 R_2} \]
This will be our Equation (1).

### Step 4: Relating atmospheric pressure to mass
The atmospheric pressure \( P \) on a planet can be expressed as:
\[ P = \frac{W}{A} \]
where \( W \) is the weight of the atmosphere and \( A \) is the surface area of the planet. The weight \( W \) can be expressed as:
\[ W = m \cdot g \]
Thus, we can write:
\[ P = \frac{m \cdot g}{A} \]
Rearranging gives:
\[ m = \frac{P \cdot A}{g} \]

### Step 5: Finding the ratio of masses of the atmospheres
For the two planets, we can express the masses of their atmospheres \( m_1 \) and \( m_2 \) as:
\[ m_1 = \frac{P_1 \cdot A_1}{g_1}, \quad m_2 = \frac{P_2 \cdot A_2}{g_2} \]
The surface area \( A \) of a sphere is given by \( A = 4 \pi R^2 \). Therefore:
\[ A_1 = 4 \pi R_1^2, \quad A_2 = 4 \pi R_2^2 \]
Substituting these into the mass equations:
\[ m_1 = \frac{P_1 \cdot (4 \pi R_1^2)}{g_1}, \quad m_2 = \frac{P_2 \cdot (4 \pi R_2^2)}{g_2} \]
Now, taking the ratio:
\[ \frac{m_1}{m_2} = \frac{P_1 \cdot (4 \pi R_1^2) / g_1}{P_2 \cdot (4 \pi R_2^2) / g_2} \]
The \( 4 \pi \) cancels out:
\[ \frac{m_1}{m_2} = \frac{P_1 R_1^2 g_2}{P_2 R_2^2 g_1} \]

### Step 6: Substituting \( g_1 \) and \( g_2 \)
Using Equation (1) to substitute \( g_1 \) and \( g_2 \):
\[ \frac{g_2}{g_1} = \frac{\rho_2 R_2}{\rho_1 R_1} \]
Thus, we can write:
\[ g_2 = g_1 \cdot \frac{\rho_2 R_2}{\rho_1 R_1} \]
Substituting this into the mass ratio equation:
\[ \frac{m_1}{m_2} = \frac{P_1 R_1^2 \cdot g_1 \cdot \frac{\rho_2 R_2}{\rho_1 R_1}}{P_2 R_2^2 \cdot g_1} \]
The \( g_1 \) cancels out:
\[ \frac{m_1}{m_2} = \frac{P_1 R_1^2 \rho_2 R_2}{P_2 R_2^2 \rho_1} \]
This simplifies to:
\[ \frac{m_1}{m_2} = \frac{P_1 R_1 \rho_2}{P_2 R_2 \rho_1} \]

### Final Answer
Thus, the ratio of the masses of their atmospheres is:
\[ \frac{m_1}{m_2} = \frac{P_1 R_1 \rho_2}{P_2 R_2 \rho_1} \]