Difficulty: Medium
Correct Answer: 8.04 km/s at a height of about 285 km (circular orbit)
Explanation:
Introduction / Context:
Satellite motion around Earth depends on achieving the correct orbital speed for a given altitude. The “first cosmic velocity” near Earth’s surface is about 7.9–8.0 km/s, rising slightly with altitude for circular orbits. Distinguishing circular, escape, and geostationary conditions avoids common misconceptions in space mechanics.
Given Data / Assumptions:
Concept / Approach:
For a circular orbit, v = sqrt(μ/R), where R is orbital radius. At LEO heights (≈ 200–300 km), v is around 7.8–7.9 km/s; quoting 8.04 km/s at ~285 km is a reasonable rounded engineering value indicating a circular orbit condition. Escape speed (≈ 11.2 km/s) is not circular; geostationary speed is ~3.07 km/s, not 11.11 km/s.
Step-by-Step Solution:
Verification / Alternative check:
Back-of-envelope using v = sqrt(μ/R) with μ ≈ 3.986e14 m^3/s^2 and R ≈ (Earth radius 6,371 km + 285 km) gives v ≈ 7.75–7.8 km/s; small rounding differences are acceptable in MCQs.
Why Other Options Are Wrong:
They mix up escape speed, circular speed, and geostationary values or assert an incorrect constant sea-level circular orbit speed.
Common Pitfalls:
Assuming one fixed “orbital speed” for all altitudes; confusing escape speed with circular speed; ignoring atmospheric drag near the surface.
Final Answer:
8.04 km/s at a height of about 285 km (circular orbit)
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