Gravitation — Easy JEE physics MCQ
The ratio of escape velocity to orbital velocity for a satellite orbiting close to Earth's surface is:
- A. $1$
- B. $\sqrt{2}$
- C. $2$
- D. $\frac{1}{\sqrt{2}}$
Solution
Orbital velocity for a satellite close to Earth's surface:
$$v_o = \sqrt{\frac{GM}{R}} = \sqrt{gR}$$
Escape velocity from Earth's surface:
$$v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$$
Ratio:
$$\frac{v_e}{v_o} = \frac{\sqrt{2gR}}{\sqrt{gR}} = \sqrt{2}$$