Source: High school physics (Chinese)
Problem Sets:
Problem
As previously discussed, for an artificial body launched from Earth's surface to become a satellite in orbit, it must reach a minimum speed known as the first cosmic velocity, given by $v_1 = \sqrt{GM/R} = \sqrt{Rg} = 7.9 \times 10^3 \text{ m/s}$.
To launch an artificial body from Earth's surface such that it escapes the Earth's gravitational pull, it must reach a minimum speed known as the second cosmic velocity.
To escape Earth's gravity, the satellite's total mechanical energy must be zero. We apply the principle of conservation of mechanical energy. Let $m$ be the mass of the satellite, $M$ be the mass of the Earth, and $R$ be its radius. The second cosmic velocity is the escape velocity, $v_2$.
The initial mechanical energy at the Earth's surface is the sum of its kinetic and gravitational potential energy:
$$E_i = \frac{1}{2}mv_2^2 - \frac{GMm}{R}$$For the satellite to just escape to infinity, its final velocity and potential energy will be zero. Thus, its final mechanical energy is:
$$E_f = 0$$By conservation of energy, $E_i = E_f$:
$$\frac{1}{2}mv_2^2 - \frac{GMm}{R} = 0$$ $$\frac{1}{2}mv_2^2 = \frac{GMm}{R}$$ $$v_2 = \sqrt{\frac{2GM}{R}}$$The problem gives the first cosmic velocity as $v_1 = \sqrt{GM/R}$. We can express the second cosmic velocity in terms of the first:
$$v_2 = \sqrt{2} \sqrt{\frac{GM}{R}} = \sqrt{2}v_1$$Substituting the given value for $v_1$:
$$v_2 = \sqrt{2} \times (7.9 \times 10^3 \text{ m/s})$$