Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Gravitation

Question:

A particle is launched from the surface of earth with speed v. For the particle to move as a satellite, which statement is correct?

Options:

$\frac{v_{e}}{2} < v < v_{e}$

$\frac{v_{e}}{\sqrt{2}} < v < v_{e}$

$v_{e} < v <\sqrt{2 v_{e}}$

$\frac{v_{e}}{\sqrt{2}} < v < \frac{v_{e}}{2}$

Correct Answer:

$\frac{v_{e}}{\sqrt{2}} < v < v_{e}$

Explanation:

$v_e=\sqrt{\frac{2 G m}{r}}, \quad v_0=\sqrt{\frac{G m}{r}}$, we have

$v_0<v<v_{e}$

$\Rightarrow \frac{v_{e}}{\sqrt{2}}<v<v_{e}$

$\left\{∵ v_0=\sqrt{\frac{GM}{r}}  ~\text{and}~  v_{e}=\sqrt{\frac{2 GM}{r}}\right\}$