Question

The mass of a planet is six times that of the earth. The radius of the planet is twice that of the earth. If the escape velocity from the earth is '$V_e$', then the escape velocity from the planet is

• A. $\sqrt{3}V_e$
• B. $\sqrt{2}V_e$
• C. $V_e$
• D. $\sqrt{5}V_e$

Hint:

For scaling problems in gravitation, simply substitute the multipliers directly into the core radical structure! Since mass increases by 6 and radius increases by 2, plug them into $\sqrt{M/R}$ to get $\sqrt{6/2} = \sqrt{3}$ instantly.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem asks for the escape velocity of a planet given how its mass and radius scale relative to the Earth.

Step 2: Key Formula or Approach:
The escape velocity ($V_{\text{esc}}$) from the surface of a spherical celestial body of mass $M$ and radius $R$ is given by the gravitational formula: $$V_{\text{esc}} = \sqrt{\frac{2GM}{R}}$$ This reveals a proportional scaling relationship: $$V_{\text{esc}} \propto \sqrt{\frac{M}{R}}$$

Step 3: Detailed Explanation:
Let's write down the given parameter scaling rules for the planet ($M_p$ and $R_p$) relative to Earth ($M_e$ and $R_e$): $$M_p = 6M_e$$ $$R_p = 2R_e$$ Set up the ratio of escape velocity of the planet ($V_p$) to that of the Earth ($V_e$): $$\frac{V_p}{V_e} = \sqrt{\frac{M_p}{M_e} \times \frac{R_e}{R_p}}$$ Substituting the scaling ratios into the equation: $$\frac{V_p}{V_e} = \sqrt{\frac{6M_e}{M_e} \times \frac{R_e}{2R_e}} = \sqrt{\frac{6}{2}} = \sqrt{3}$$ Isolating the planet's escape velocity $V_p$: $$V_p = \sqrt{3}V_e$$

Step 4: Final Answer:
The escape velocity from the planet is $\sqrt{3}V_e$, which corresponds to option (A).