Question

What is the minimum required velocity in km/s of the satellite to escape from the gravitational attraction of the earth? Assumed a circular orbit with an altitude of 4,220 km.

Answer 1

The escape velocity at that height would be

$$v_e=\sqrt{\frac{2GM}{R_e+r}}=\sqrt{\frac{2(6.673·10^{-11})(6·10^{24})}{6375·10^{3}+4220·10^3}}=\\\space\\=8694\text{ m/s}.$$