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

Two satellites $A$ and $B$ revolve around the same planet in coplanar circular orbits. Their periods of revolution are $1\, h$ and $8\, h$,respectively. The radius of the orbit of $A$ is $10^{4}\, km$. The speed of $B$ relative to $A$ when they are closest is (in $km/h$):

$A$ satellite $S$ is moving in an elliptical orbit around the Earth. The mass of the satellite is very small compared to the mass of the Earth.

$A$ planet of mass $M$ has two natural satellites with masses $m_1$ and $m_2$. The radii of their circular orbits are $R_1$ and $R_2$ respectively. Ignore the gravitational force between the satellites. Define $v_1, L_1, K_1$ and $T_1$ to be,respectively,the orbital speed,angular momentum,kinetic energy,and time period of revolution of satellite $1$; and $v_2, L_2, K_2$ and $T_2$ to be the corresponding quantities of satellite $2$. Given $m_1/m_2 = 2$ and $R_1/R_2 = 1/4$,match the ratios in List-$I$ to the numbers in List-$II$.

List-$I$	List-$II$
$P. \frac{v_1}{v_2}$	$1. \frac{1}{8}$
$Q. \frac{L_1}{L_2}$	$2. 1$
$R. \frac{K_1}{K_2}$	$3. 2$
$S. \frac{T_1}{T_2}$	$4. 8$

If the mass of a satellite is doubled and the time period remains constant,the ratio of the orbital radii in the two cases will be:

Even though the gravitational force of the Earth acts on the Moon towards its center,why does the Moon not fall onto the Earth?