$A$ body is dropped from a satellite in a parking orbit. Which of the following describes its behaviour correctly?

The time period of a geostationary satellite is (in $\text{ h}$)

$A$ 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 heavenly bodies $S_1$ and $S_2$,not far from each other,are seen to revolve in orbits:

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

If the gravitational force between two objects were proportional to $\frac{1}{R}$ (and not as $\frac{1}{R^2}$) where $R$ is the separation between them,then a particle in a circular orbit under such a force would have its orbital speed $v$ proportional to