The escape velocity from the Earth is 11.2 km/s. If a body is projected with a velocity twice the escape velocity, what will be its velocity in interstellar space?

C. 3 11.2 km/s — Using conservation of energy, (1/2)mv^2 - GMm/R = (1/2)mv_f^2. Substituting v = 2v_e and v_e^2 = 2GM/R, we find v_f = sqrt(3)v_e.

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$The escape velocity from the Earth is v_{e} = 11.2 km/s . If a body is projected with a velocity v = 2 v_{e}, what will be its velocity v_{f} in interstellar space?$

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Correct answer: C — $3 \times 11.2 km/s$

By energy conservation, \frac{1}{2} m v^{2} - \frac{GM m}{R} = \frac{1}{2} m v_{f}^{2} . With v = 2 v_{e} and v_{e}^{2} = \frac{2 GM}{R}, we get \frac{1}{2} m (4 v_{e}^{2}) - \frac{1}{2} m v_{e}^{2} = \frac{1}{2} m v_{f}^{2}, so v_{f}^{2} = 3 v_{e}^{2}, hence v_{f} = 3 v_{e} .