Choose the factor on which the orbital velocity does not depend when the satellite is orbiting close to the earth’s surface

The correct option is "The mass of the satellite".

To understand why the orbital velocity does not depend on the mass of the satellite, let us derive the expression for orbital velocity from first principles.

When a satellite of mass $m$ is orbiting close to the Earth's surface in a circular orbit of radius $r$ (which is approximately equal to the radius of the Earth, $R$), the gravitational force of attraction between the Earth (of mass $M$) and the satellite provides the necessary centripetal force required to keep the satellite in its circular orbit.

The gravitational force ($F_g$) acting on the satellite is given by Newton's Law of Universal Gravitation:

$F_g=\frac{G·M·m}{r^2}$

where $G$ is the universal gravitational constant.

The centripetal force ($F_c$) required to keep the satellite of mass $m$ moving with orbital velocity $v$ is given by:

$F_c=\frac{m·v^2}{r}$

For a stable orbit, the gravitational force must balance the centripetal force:

$F_c=F_g$

$\frac{m·v^2}{r}=\frac{G·M·m}{r^2}$

Notice that the mass of the satellite ($m$) appears on both sides of the equation. We can divide both sides by $m$ to eliminate it:

$\frac{v^2}{r}=\frac{G·M}{r^2}$

Solving for the orbital velocity $v$ by multiplying both sides by $r$ and taking the square root, we get:

$v^2=\frac{G·M}{r}$

$v=\sqrt{\frac{G·M}{r}}$

If the satellite is orbiting very close to the Earth's surface, the orbital radius $r$ is approximately equal to the radius of the Earth ($R$). Thus, the formula becomes:

$v\approx \sqrt{\frac{G·M}{R}}$

From this final expression, it is clear that the orbital velocity $v$ depends on:
1. The universal gravitational constant ($G$)
2. The mass of the Earth ($M$)
3. The radius of the Earth ($R$) or the orbital radius ($r$)

Since the mass of the satellite ($m$) cancels out during the derivation, the orbital velocity does not depend on the mass of the satellite.