Question

Two spherical planets \( P \) and \( Q \) have the same uniform density \( \rho \), and masses \( M_P \) and \( M_Q \), and surface areas \( A \) and \( 4A \) respectively. Another spherical planet \( R \) also has the same uniform density \( \rho \), and its mass is \( M_P + M_Q \). The escape velocities from these planets is

• A. \( V_R > V_Q > V_P \)
• B. \( V_R < V_Q < V_P \)
• C. \( V_R = V_Q > V_P \)
• D. \( V_R > V_Q = V_P \)

Hint:

For planets of same density, escape velocity is directly proportional to radius.

Answer 1

The Correct Option is A

Step 1: Use surface area of sphere.
\[ A = 4\pi R^2 \]

Step 2: Compare radii of planets \( P \) and \( Q \).
Since surface areas are \( A \) and \( 4A \), radius of \( Q \) is twice radius of \( P \):
\[ R_Q = 2R_P \]

Step 3: Use mass-density relation.
For same density:
\[ M \propto R^3 \]

Step 4: Compare masses.
\[ M_Q = 8M_P \]

Step 5: Find mass of planet \( R \).
\[ M_R = M_P + M_Q \]
\[ M_R = M_P + 8M_P = 9M_P \]

Step 6: Relate escape velocity with radius.
Escape velocity is:
\[ V = \sqrt{\frac{2GM}{R}} \]
For same density, \( M \propto R^3 \), so:
\[ V \propto R \]

Step 7: Compare escape velocities.
\[ R_Q = 2R_P \]
\[ R_R = \sqrt[3]{9}R_P \]
Since \( \sqrt[3]{9} > 2 \), therefore:
\[ V_R > V_Q > V_P \]