For a satellite orbiting very close to the surface of the Earth,what is the relationship between its escape velocity $(v_{e})$ and its orbital velocity $(v_{0})$?

The mass and radius of the Earth and Moon are $M_1, R_1$ and $M_2, R_2$ respectively. Their centres are at a distance $d$ apart. The minimum speed with which a body of mass $m$ should be projected from a distance $\frac{2d}{3}$ from the centre of $M_1$ so as to escape to infinity is:

$A$ body is projected vertically from the Earth's surface with a velocity equal to half the escape velocity. The maximum height reached by the body is ($R =$ radius of the Earth).

An object is thrown vertically upwards from the surface of the earth with a velocity $x$ times the escape velocity on the earth $(x < 1)$. The maximum height to which it rises from the center of the earth is (radius of earth is $R$):

What is the escape energy of an object of mass $m$ at a height of $4R_e$ from the surface of the Earth? ($R_e$ = radius of the Earth)

$A$ body is projected vertically upwards from the Earth's surface. If the velocity of projection is $\left(\frac{1}{3}\right)$ of the escape velocity,then the height up to which the body rises is $(R = \text{radius of Earth})$