$A$ satellite moves around the Earth in a circular orbit of radius $R$ making one revolution per day. $A$ second satellite moving in a circular orbit moves around the Earth once in $8$ days. The radius of the orbit of the second satellite is

$A$ planet orbits in an elliptical path of eccentricity $e$ around a massive star considered fixed at one of the foci. The point in space,where it is closest to the star is denoted by $P$ and the point,where it is farthest is denoted by $A$. Let $v_P$ and $v_A$ be the respective speeds at $P$ and $A$,then

The period of revolution of planet $A$ around the sun is $8$ times that of $B$. The distance of $A$ from the sun is how many times greater than that of $B$ from the sun?

The kinetic energies of a planet in an elliptical orbit about the Sun,at positions $A, B$ and $C$ are $K_A, K_B$ and $K_C$,respectively. $AC$ is the major axis and $SB$ is perpendicular to $AC$ at the position of the Sun $S$ as shown in the figure. Then

The time period of a satellite of Earth is $5\, h$. If the separation between the centre of the Earth and the satellite is increased to $4$ times the previous value,the new time period will become ....... $h$.

Assertion $(A)$: Angular speed,linear speed,and kinetic energy change with time,but angular momentum remains constant for a planet orbiting the sun.
Reason $(R)$: Angular momentum is constant as no external torque acts on the planet.