NECO 2002 Physics · Question 16
What is the escape velocity of a satellite launched from the earth’s surface? (Take g as 10ms and the radius of the earth as 6.4 × 10m)
Options
- A.7.80 × 10k
- B.11.2 × 10k
- C.3.56 × 107
- D.6.00 × 107
- E.1.20 × 108
Answer: B. 11.2 × 10k
Explanation written with AI and checked against the answer key
Analyze the question: We need the escape velocity from Earth's surface, given g = 10 m/s² and R = 6.4 × 10⁶ m. Concept: Escape velocity is the minimum speed needed to leave Earth's gravitational field without further propulsion. Setting kinetic energy equal to the gravitational potential energy at the surface: ½mv² = GMm/R Since GM = gR², this becomes: v = √(2gR) Solution: v = √(2 × 10 × 6.4 × 10⁶) v = √(1.28 × 10⁸) v = √(12.8 × 10⁷) ≈ 1.13 × 10⁴ m/s This is about 11.3 km/s, or 11.3 × 10³ m/s. The accepted real value is 11.2 km/s. Matching the options: The options are garbled in the text. "10k" most likely stands for 10³, so option B reads 11.2 × 10³ m/s. This is the closest match to our result of about 11.3 × 10³ m/s. Option A (7.8 × 10³ m/s) is the orbital velocity near Earth's surface, which is √(gR) ≈ 8 × 10³ m/s. Options C, D and E are far too large, since they are of the order of 10⁷ to 10⁸ m/s, which would exceed or approach the speed of light. Tip: Remember two key values: orbital velocity = √(gR) ≈ 8 km/s, and escape velocity = √(2gR) ≈ 11.2 km/s. Escape velocity is always √2 times the orbital velocity. Also, if an option looks close to or beyond the speed of light (3 × 10⁸ m/s), you can rule it out immediately.AI can make mistakes. Report anything that looks wrong.
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