Question

The escape velocity from a planet is ${\text{v}}_e$. A particle starts from rest at a large distance from the planet, reaches the planet only under gravitational attraction, and passes through the smooth tunnel through its center, its speed at the center of the planet will be :

• A. $\sqrt{1.5}{\text{v}}_e$
• B. $\frac{{\text{v}}_e}{\sqrt{2}}$
• C. ${\text{v}}_e$
• D. Zero

Answer 1

The correct option is A $\sqrt{1.5}{\text{v}}_e$

From mechanical energy conservation,
$\frac{1}{2}m{v}^2=\frac{3GMm}{2R}$

$\Rightarrow v=\sqrt{\frac{3GM}{R}}$
$\Rightarrow v=\sqrt{\frac{3}{2}\frac{2GM}{R}}$
$\Rightarrow v=\sqrt{1.5}{\text{v}}_e$