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The escape velocity of a body on the surface of the earth is 11.2 km/s. If the earth's mass increases to twice its present value and the radius of the earth becomes half, the escape velocity would become:

1. 44.8 km/s
2. 22.4 km/s
3. 11.2 km/s (remains unchanged)
4. 5.6 km/s

Correct answer: 44.8 km/s

Solution

Escape velocity is given by \( v_e = \sqrt{\frac{2GM}{R}} \). If the mass \( M \) doubles and the radius \( R \) halves, \( v_e \) becomes \( \sqrt{\frac{2G(2M)}{R/2}} = 2v_e \). Thus, the new escape velocity is \( 2 \times 11.2 = 44.8 \, \text{km/s} \).

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