Question

A planet has mass \( 81 M_E \) and radius \( 9 R_E \), where \( M_E \) and \( R_E \) are the mass and radius of Earth respectively. The escape velocity from the planet is:

• A. Equal to Earth's escape velocity
• B. 3 times Earth's escape velocity
• C. \( \frac{1}{3} \) times Earth's escape velocity
• D. 9 times Earth's escape velocity

Hint:

Escape velocity scales as \( v_e \propto \sqrt{\frac{M}{R}} \). If mass is multiplied by \( k_1 \) and radius by \( k_2 \), the escape velocity changes by a factor of \( \sqrt{\frac{k_1}{k_2}} \). Here, \( \sqrt{\frac{81}{9}} = \sqrt{9} = 3 \).

Answer 1

The Correct Option is B

Step 1: Recall what escape speed depends on.
Escape speed is $v_e=\sqrt{\dfrac{2GM}{R}}$, so only the ratio $\dfrac{M}{R}$ matters once $G$ is fixed.
Step 2: Use a ratio instead of full numbers.
Compare the planet to Earth by taking $\dfrac{v_P}{v_E}=\sqrt{\dfrac{M_P/R_P}{M_E/R_E}}$, which kills the constant $2G$.
Step 3: Plug in the given multiples.
Here $M_P = 81M_E$ and $R_P = 9R_E$, so \[ \frac{M_P/R_P}{M_E/R_E} = \frac{81/9}{1} = 9 \]
Step 4: Take the square root of the ratio.
\[ \frac{v_P}{v_E} = \sqrt{9} = 3 \]
Step 5: Interpret the surface-gravity view.
Since $g\propto \dfrac{M}{R^2}$ gives $\dfrac{81}{81}=1$ (same surface gravity), but escape speed also carries an extra factor of $\sqrt{R}$, the radius being 9 times larger boosts $v_e$ by $\sqrt{9}=3$. Both pictures agree.
Step 6: State the result.
The planet's escape speed is 3 times that of Earth. \[ \boxed{v_P = 3\,v_E} \]