Solution:
Using conservation of energy: \(v_\infty = \sqrt{v^2 - v_e^2}\). Given \(v = 22.4\text{ km/s} = 2v_e\), we find \(v_\infty = \sqrt{(2v_e)^2 - v_e^2} = v_e\sqrt{3} = 11.2\sqrt{3}\text{ km/s}\).
The escape velocity of a body on the earth surface is 11.2 km/s. If the same body is projected upward with velocity 22.4 km/s, the velocity of this body at infinite distance from the centre of the earth will be
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