Suppose that thomass and radius of the Moon changes to 7.35 x 10^22 kg and 1.7 × 10º m respectively. The escape velocity of the Moon will then be(Note: The value of Gravitational constant is 6.63 x 10^-11 Nm²/kg².)
Correct answer: C. 2.4 x 10^3 m/s
- A. 1.0 x 10^3 m/s
- B. 1.1 x 10^4 m/s
- C. 2.4 x 10^3 m/s
- D. 8.2 x 10^6 m/s
Explanation
The escape velocity (ve) is calculated using the formula: ve = √(2GM/R), where G is the gravitational constant, M is the mass of the celestial body, and R is its radius. Plugging the values: M = 7.35 × 1022 kg, R = 1.7 × 106 m, and G = 6.63 × 10-11 Nm²/kg², we find ve = √(2 * 6.63 × 10-11 * 7.35 × 1022 / 1.7 × 106) ≈ 2.4 x 103 m/s. Therefore, the correct escape velocity is approximately 2.4 x 103 m/s, which corresponds to Option C. The other options are incorrect because they either underestimate or overestimate the escape velocity based on the given values.
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