Practice question
Question
A projectile is launched at 6km/s from Earth’s surface. What is its maximum distance from the center? (Escape speed = 11.2km/s,RE\=6.4×106m)
Explanation
12vi2−ve22 = −ve22REr. 18−62.72 = −62.72REr. rRE = 62.7244.72≈1.40. r = 1.40×6.4×106≈8.96×106m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 9.0 × 10⁶ m. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
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