Practice question
Question
A projectile is launched from Earth’s surface with speed 8km/s. How far from the center does it reach? (Escape speed = 11.2km/s,RE\=6.4×106m)
Explanation
Energy conservation: 12mvi2−GMEmRE = −GMEmr. ve2 = 2GMERE, so GMERE = ve22. vi22−ve22 = −ve22REr. (8)22−(11.2)22 = −(11.2)22REr. 32−62.72 = −62.72REr. rRE = 62.7230.72≈2.04. r = 2.04×6.4×106≈1.31×107m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.3 × 10⁷ m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
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