Why does a satellite not have an escape velocity?

Given the radius of Earth $R$ and the length of a day $T$,the height of a geostationary satellite is: [$G$ = Gravitational Constant,$M$ = Mass of Earth]

If the orbital speed of a body revolving in a circular path near the surface of the earth is $8 \ km s^{-1}$,then the orbital speed of a body revolving around the earth in a circular orbit at a height of $19,200 \ km$ from the surface of the earth is (Radius of the earth $R = 6400 \ km$). (in $km s^{-1}$)

Two satellites $A$ and $B$ of masses $200 \, kg$ and $400 \, kg$ are revolving around the Earth at heights of $600 \, km$ and $1600 \, km$ respectively. If $T_{A}$ and $T_{B}$ are the time periods of $A$ and $B$ respectively,then find the value of $T_{B} - T_{A}$.
[Given: Radius of Earth $R = 6400 \, km$,Mass of Earth $M = 6 \times 10^{24} \, kg$,$G = 6.67 \times 10^{-11} \, Nm^{2}/kg^{2}$]

Assuming that the earth is revolving around the sun in a circular orbit of radius $R$,the angular momentum is directly proportional to $R^{n}$. The value of $n$ is

$A$ test particle is moving in a circular orbit in the gravitational field produced by a mass density $\rho(r) = \frac{K}{r^2}$. Identify the correct relation between the radius $R$ of the particle's orbit and its period $T$.