Question

Two planets \(P_1\) and \(P_2\) with equal mass have radii \(R_1\) and \(R_2\), respectively, where \[ R_2=\frac{R_1}{2} \] The escape speeds of \(P_1\) and \(P_2\) are \(v_1\) and \(v_2\), respectively. Then the value of \[ \frac{v_2}{v_1} \] is:

• A. \(2\)
• B. \(\frac{1}{\sqrt{2}}\)
• C. \(1\)
• D. \(\sqrt{2}\)

Hint:

Escape velocity is proportional to \(1/\sqrt{R}\) when mass remains constant. A smaller planet radius results in a larger escape velocity. Always begin with the formula \(v_e=\sqrt{2GM/R}\). Use ratios to simplify calculations quickly.

Answer 1

The Correct Option is D

Concept:

• Escape velocity is the minimum speed required for a body to escape the gravitational field of a planet without further propulsion.

• The escape velocity from the surface of a planet is given by \[ v_e=\sqrt{\frac{2GM}{R}} \] where \(M\) is the mass of the planet and \(R\) is its radius.

• For planets having equal masses, escape velocity varies inversely as the square root of the radius.

Step 1: Write the escape velocity for planet \(P_1\)
For planet \(P_1\), \[ v_1=\sqrt{\frac{2GM}{R_1}} \] where \(M\) is the mass of the planet.

Step 2: Write the escape velocity for planet \(P_2\)
For planet \(P_2\), \[ v_2=\sqrt{\frac{2GM}{R_2}} \] Since both planets have equal masses, the value of \(M\) remains the same.

Step 3: Take the ratio of the two escape velocities
Dividing the two expressions, \[ \frac{v_2}{v_1} = \sqrt{ \frac{\frac{2GM}{R_2}} {\frac{2GM}{R_1}} } \] \[ \frac{v_2}{v_1} = \sqrt{\frac{R_1}{R_2}} \]

Step 4: Substitute the given relation between radii
Given, \[ R_2=\frac{R_1}{2} \] Substituting, \[ \frac{v_2}{v_1} = \sqrt{ \frac{R_1} {R_1/2} } \] \[ \frac{v_2}{v_1} = \sqrt{2} \]

Step 5: Write the final result
Therefore, \[ \boxed{\frac{v_2}{v_1}=\sqrt{2}} \] Hence the correct option is \[ \boxed{\text{Option (D)}} \]