Two spherical planets P and Q have the same uniform density `rho,` masses `M_p and M_Q` and surface areas A and 4A respectively. A spherical planet R also has uniform density `rho` and its mass is `(M_P + M_Q).` The escape velocities from the plantes P,Q and R are `V_P V_Q and V_R` respectively. Then

Options

A. `V_Q gt V_R gt V_P`
B. `V_R gt V_Q gt V_P`
C. `V_P/V_P =3`
D. `V_P/ V_ Q =(1)/(2)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the Masses of Planets P and Q
Given the surface area of planet P is \( A \) and the surface area of planet Q is \( 4A \), we can express their masses in terms of their densities and volumes.

1. **For Planet P:**
- Surface Area \( A = 4\pi R_P^2 \)
- Therefore, \( R_P = \sqrt{\frac{A}{4\pi}} \)
- Volume \( V_P = \frac{4}{3}\pi R_P^3 = \frac{4}{3}\pi \left(\sqrt{\frac{A}{4\pi}}\right)^3 = \frac{A^{3/2}}{6\sqrt{\pi}} \)
- Mass \( M_P = \rho V_P = \rho \cdot \frac{A^{3/2}}{6\sqrt{\pi}} \)

2. **For Planet Q:**
- Surface Area \( 4A = 4\pi R_Q^2 \)
- Therefore, \( R_Q = \sqrt{A/\pi} \)
- Volume \( V_Q = \frac{4}{3}\pi R_Q^3 = \frac{4}{3}\pi \left(\sqrt{\frac{A}{\pi}}\right)^3 = \frac{4A^{3/2}}{3\sqrt{\pi^3}} \)
- Mass \( M_Q = \rho V_Q = \rho \cdot \frac{4A^{3/2}}{3\sqrt{\pi^3}} \)

### Step 2: Calculate the Mass of Planet R
The mass of planet R is given as \( M_R = M_P + M_Q \).

- Substitute the expressions for \( M_P \) and \( M_Q \):
\[ M_R = \rho \cdot \frac{A^{3/2}}{6\sqrt{\pi}} + \rho \cdot \frac{4A^{3/2}}{3\sqrt{\pi^3}} = \rho \cdot \left(\frac{A^{3/2}}{6\sqrt{\pi}} + \frac{4A^{3/2}}{3\sqrt{\pi^3}}\right) \]

### Step 3: Calculate the Escape Velocities
The escape velocity \( V \) from a planet is given by the formula:
\[ V = \sqrt{\frac{2GM}{R}} \]
Where \( G \) is the gravitational constant.

1. **For Planet P:**
\[ V_P = \sqrt{\frac{2G M_P}{R_P}} \]

2. **For Planet Q:**
\[ V_Q = \sqrt{\frac{2G M_Q}{R_Q}} \]

3. **For Planet R:**
\[ V_R = \sqrt{\frac{2G M_R}{R_R}} \]

### Step 4: Relate the Escape Velocities
Using the expressions for masses and radii, we can express \( V_Q \) and \( V_R \) in terms of \( V_P \).

- Since \( M_Q = 8M_P \) and \( R_Q = 2R_P \):
\[ V_Q = \sqrt{\frac{2G (8M_P)}{2R_P}} = 2\sqrt{\frac{2G M_P}{R_P}} = 2V_P \]

- For planet R, we find its radius \( R_R \) and mass \( M_R \):
\[ R_R = 2R_P \quad \text{and} \quad M_R = 9M_P \]
\[ V_R = \sqrt{\frac{2G (9M_P)}{(2R_P)}} = \frac{3}{\sqrt{2}} \sqrt{\frac{2G M_P}{R_P}} = \frac{3}{\sqrt{2}} V_P \]

### Step 5: Compare the Escape Velocities
From the calculations:
- \( V_Q = 2V_P \)
- \( V_R = \frac{3}{\sqrt{2}} V_P \)

### Conclusion
Thus, the escape velocities can be ordered as:
\[ V_Q > V_R > V_P \]