Question

The escape velocity for a planet is $v_e$. A particle is projected from its surface with a speed $v$. For this particle to move as a satellite around the planet,

• A. $\frac{v_e}{2}<v<v_e$
• B. $\frac{v_e}{\sqrt{2}}<v<v_e$
• C. $v_e<v<\sqrt{2}{v}_e$
• D. $\frac{v_e}{\sqrt{2}}<v<\frac{v_e}{2}$

Answer 1

The correct option is B $\frac{v_e}{\sqrt{2}}<v<v_e$

The minimum velocity for sattelite is at the nearest possible orbit, i.e., very near to the earth's surface.

So to launch sattelite to orbit very near to the earth's surface,

$mg=m{v}^2/R\Rightarrow v=\sqrt{gR}=\sqrt{GM/R}=v_e/\sqrt{2}$

The maximum velocity for sattelite to orbit around earth is the escape velocity, $v_e$.

So, $v_e/\sqrt{2}<v<v_e$