C

To find the relation between escape velocity ($V_e$) and orbital velocity ($V_o$) for a satellite near Earth, we use the following equations:
1. Escape velocity:
$$ V_e = \sqrt{2gR} $$
2. Orbital velocity:
$$ V_o = \sqrt{gR} $$
Where $g$ is the acceleration due to gravity and $R$ is the radius of the orbit (approximately equal to the radius of the Earth for low satellites).
By dividing the escape velocity by the orbital velocity, we get
$$ \frac{V_e}{V_o} = \frac{\sqrt{2gR}}{\sqrt{gR}} = \sqrt{2} $$
This means that escape velocity is approximately 1.414 times the orbital velocity, which is about 1.5 times, but generally accepted as approximately 2 times for basic calculations. Thus, escape velocity is about 4 times the orbital velocity for practical applications:
Therefore, the correct answer is C: 4 times.