Answer
$v=\sqrt{2GM(\frac{3}{R}-\frac{1}{r_1})}$
Work Step by Step
We use conservation of energy to find:
$\frac{1}{2}mv^2-\frac{1}{2}mv_0^2=GMm(\frac{1}{R}-\frac{1}{r_1})$
$\frac{1}{2}v^2-\frac{1}{2}v_0^2=GM(\frac{1}{R}-\frac{1}{r_1})$
$v=\sqrt{2GM(\frac{1}{R}-\frac{1}{r_1})+v_0^2}$
We now need to substitute a value for $v_0^2$. We know that escape speed equals:
$\sqrt{\frac{2GM}{R}}$
Since the actual speed is $\sqrt{2}$ greater, we find: $v_0^2 =\frac{4GM}{R}$.
Plugging this into the above equation gives:
$v=\sqrt{2GM(\frac{3}{R}-\frac{1}{r_1})}$
