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

Step 1: State the data.
Two planets have equal mass $M$ with radii $R_1$ and $R_2 = R_1/2$. We want $v_2/v_1$ for their escape speeds.
Step 2: Escape speed formula.
The escape speed from a planet's surface is $v_e = \sqrt{\dfrac{2GM}{R}}$.
Step 3: Write each escape speed.
\[ v_1 = \sqrt{\frac{2GM}{R_1}}, \qquad v_2 = \sqrt{\frac{2GM}{R_2}} \]
Step 4: Take the ratio.
Equal masses mean $2GM$ cancels, leaving only the radii.
\[ \frac{v_2}{v_1} = \sqrt{\frac{R_1}{R_2}} \]
Step 5: Insert the radius relation.
With $R_2 = R_1/2$, the ratio $R_1/R_2 = 2$.
\[ \frac{v_2}{v_1} = \sqrt{2} \]
Step 6: Conclusion.
The smaller planet has the larger escape speed, by a factor $\sqrt{2}$.
\[ \boxed{\dfrac{v_2}{v_1} = \sqrt{2}} \]