Answer to Question #167795 in Physics for Tithi

Question

Show that the escape velocity from the surface of earth is √2 times the velocity of projection of an artificial satellite orbiting close around the earth.

Answer 1

The escape velocity from the surface of earth is

$$v_e=\sqrt{\frac{2GM}{R}}$$

For orbiting around the earth:

$$v=\sqrt{\frac{GM}{R+h}}$$

For a satellite very close to Earth

$$v=\sqrt{\frac{GM}{R+0}}=\sqrt{\frac{GM}{R}}$$

Thus

$$\frac{v_e}{v}=\sqrt{2}$$