If the escape velocity of a body from the surface of the earth is $11.2 \,km \,s^{-1}$, then the orbital velocity of a satellite in an orbit which is at a height equal to the radius of the earth is

$A$ satellite is orbiting just above the surface of the earth with period $T$. If $d$ is the density of the earth and $G$ is the universal constant of gravitation,the quantity $\frac{3 \pi}{G d}$ represents :

The radius of the orbit of a geostationary satellite is (mean radius of earth is $R$,angular velocity about own axis is $\omega$,and acceleration due to gravity on earth's surface is $g$).

Two satellites $A$ and $B$ go round a planet in circular orbits having radii $4R$ and $R$,respectively. If the speed of satellite $A$ is $3v$,then the speed of satellite $B$ is:

An artificial satellite is revolving around a planet of radius $R$ in a circular orbit of radius $a$. If the time period of revolution of the satellite $T \propto a^{3/2} g^x R^y$,then the values of $x$ and $y$ are respectively. [Note: $g$ is the acceleration due to gravity at the surface of the planet.]

The period of a satellite,in a circular orbit near an equatorial plane,will not depend on