Answer
$2.55\times 10^7m$
Work Step by Step
We know that
$K.E_i+U_i=K.E_f+U_f$
$\implies \frac{1}{2}mv_i^2-G\frac{M_e m}{R_E}=\frac{1}{2}mv_f^2-G\frac{M_E m}{r_f}$
$\implies \frac{2GM_E}{r_f}=\frac{2GM_E}{R_E}+v_f^2-v_i^2$
As $v_f=\frac{1}{2}v_e=\frac{1}{2}\sqrt{2GM_E/R_E}$
Now $\frac{2GM_E}{r_f}=\frac{2GM_E}{R_E}+(\frac{1}{4}\frac{2GM_E}{R_E})-(\frac{2GM_E}{R_E})$
$\frac{1}{r_f}=\frac{1}{4R_E}$
$\implies r_f=4R_E$
We plug in the known values to obtain:
$\frac{1}{r_f}=4(6.37\times 10^6m)$
$\frac{1}{r_f}=2.55\times 10^7m$
