Practice question
Question
A projectile is launched at 5km/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. 12.5−62.72 = −62.72REr. rRE = 62.7250.22≈1.25. r = 1.25×6.4×106 = 8.0×106m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 8.0 × 10⁶ m. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.