Practice question
Question
A projectile is launched at 9km/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. 40.5−62.72 = −62.72REr. rRE = 62.7222.22≈2.82. r = 2.82×6.4×106≈1.80×107m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.8 × 10⁷ m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
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