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(A) The escape velocity is given by $v_E = \sqrt{\frac{2GM}{R}}$.At the Earth's surface,the total mechanical energy is $E_1 = KE_1 + PE_1 = \frac{1}{2

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$\frac{1}{3}$

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(A) The escape velocity is given by $v_E = \sqrt{\frac{2GM}{R}}$.At the Earth's surface,the total mechanical energy is $E_1 = KE_1 + PE_1 = \frac{1}{2}mv^2 - \frac{GMm}{R}$.Substituting $v = \alpha v_E = \alpha \sqrt{\frac{2GM}{R}}$,we get $KE_1 = \frac{1}{2}m(\alpha^2 \cdot \frac{2GM}{R}) = \frac{GMm\alpha^2}{R}$.So,$E_1 = \frac{GMm\alpha^2}{R} - \frac{GMm}{R} = \frac{GMm}{R}(\alpha^2 - 1)$.At the maximum height $h$,the velocity is $0$,so $KE_2 = 0$ and $PE_2 = -\frac{GMm}{R+h}$.By the law of conservation of energy,$E_1 = E_2$:$\frac{GMm}{R}(\alpha^2 - 1) = -\frac{GMm}{R+h}$.$\alpha^2 - 1 = -\frac{R}{R+h} = -\frac{6400}{6400+800} = -\frac{6400}{7200} = -\frac{8}{9}$.$\alpha^2 = 1 - \frac{8}{9} = \frac{1}{9}$.Therefore,$\alpha = \frac{1}{3}$.

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