A particle of mass 'm' is projected with a velocity v = kV_e (k

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Question

A particle of mass 'm' is projected with a velocity v = kV_e (k < 1) from the surface of the Earth, where V_e is the escape velocity. The maximum height above the surface reached by the particle is

Accepted Answer

Correct Answer: A — Rk²/(1 - k²)Step-by-step Solution:Using conservation of mechanical energy between the surface and the maximum height h:(1/2)mv² - GMm/R = 0 - GMm/(R + h).Since v = kV_e and V_e² = 2GM/R, we have v² = k² × 2GM/R.Substituting: (1/2)m × k² × (2GM/R) - GMm/R = -GMm/(R + h).(k²GM/R) - (GM/R) = -GM/(R + h).Dividing by GM: (k² - 1)/R = -1/(R + h).(1 - k²)/R = 1/(R + h) → R + h = R/(1 - k²) → h = R/(1 - k²) - R = Rk²/(1 - k²).Why other options are incorrect: Options B, C, and D all have incorrect algebraic forms that do not reduce from the correct energy conservation equation.Key Concept: Conservation of energy with gravitational potential energy: U = -GMm/r (not mgh for large heights). At maximum height, KE = 0. Escape velocity V_e = √(2GM/R).

A particle of mass 'm' is projected with a velocity v = kV_e (k < 1) from the surface of the Earth, where V_e is the escape velocity. The maximum height above the surface reached by the particle is

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A particle of mass 'm' is projected with a velocity v = kV_e (k < 1) from the surface of the Earth, where V_e is the escape velocity. The maximum height above the surface reached by the particle is

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