Two particles of equal mass $m$ go around a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is

Two satellites of masses $m$ and $3\,m$ revolve around the earth in circular orbits of radii $r$ and $3r$ respectively. The ratio of orbital speeds of the satellites respectively is:

The distance of a geostationary satellite from the centre of the earth (Radius $R = 6400 \ km$) is nearest to (in $R$)

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

Two satellites $S_{1}$ and $S_{2}$ are revolving in circular orbits around a planet with radii $R_{1} = 3200 \, km$ and $R_{2} = 800 \, km$ respectively. The ratio of the speed of satellite $S_{1}$ to the speed of satellite $S_{2}$ in their respective orbits is $\frac{1}{x}$,where $x =$

$A$ satellite of mass $M$ is revolving in a circular orbit of radius $r$ around the Earth. The time period of revolution of the satellite is: