Question

Escape velocity on earth is \(V_e=\sqrt{2gR}\). A planet has half the radius of earth and density equal to half that of earth. If escape speed from that planet is \(\dfrac{V_0}{N}\), find \(N\).

• A. \(4\)
• B. \(6\)
• C. \(2\)
• D. \(8\)

Hint:

Escape velocity depends on planet density and radius as: \[ V_e \propto R\sqrt{\rho} \] This relation is very useful for comparing planets.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Escape velocity on a planet's surface can be expressed in terms of its radius and mass.
Since the problem involves comparing planets with the same density, we should express the mass in terms of density and radius.
Step 2: Key Formula or Approach:
The formula for escape velocity is:
\[ V_e = \sqrt{\frac{2GM}{R}} \]
Since mass $M = \text{Volume} \times \text{Density} = \frac{4}{3}\pi R^3 \rho$, substitute $M$:
\[ V_e = \sqrt{\frac{2G}{R} \left(\frac{4}{3}\pi R^3 \rho\right)} = \sqrt{\frac{8\pi G \rho}{3} R^2} \]
This yields the proportionality: $V_e \propto R \sqrt{\rho}$.
Step 3: Detailed Explanation:
Given that the new planet has the same density as Earth ($\rho' = \rho$), the escape velocity is directly proportional to its radius:
\[ V \propto R \]
Therefore, if the radius of the planet is half that of Earth ($R' = \frac{R}{2}$), its escape velocity $V'$ will also be half that of Earth:
\[ V' = \frac{V_e}{2} \]
The problem states the escape velocity is $\frac{V_0}{N}$ (treating $V_0$ as synonymous with $V_e$ based on context). Thus, comparing both sides, we get:
\[ N = 2 \]
Step 4: Final Answer:
The value of N is 2.