Answer
by using Equation $13-28$ we can get the escape speed: $$v=\sqrt{\frac{2 G m}{r}} \approx 0.02 \mathrm{m} / \mathrm{s}$$
Work Step by Step
by using Equation $13-28$ we can get the escape speed: $$v=\sqrt{\frac{2 G m}{r}} \approx 0.02 \mathrm{m} / \mathrm{s}$$
