Suppose a planet goes around the Sun with a linear speed twice as fast as that of the Earth. What will be its orbit size compared to that of the Earth? (Radius of Earth $= R$)

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 figure shows an elliptical orbit of a planet of mass $m$ about the sun $S.$ The shaded area $SCD$ is twice the shaded area $SAB.$ If $t_1$ is the time taken for the planet to move from $C$ to $D$ and $t_2$ is the time taken to move from $A$ to $B,$ then:

$A$ satellite is launched into a circular orbit of radius $R$ around the Earth. $A$ second satellite is launched into an orbit of radius $1.03 R$. The time period of revolution of the second satellite is larger than the first one approximately by: (in $\%$)

$A$ satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the centre of the Earth in the new orbit is $2$ times that of the earlier orbit. The time period in the second orbit is:

The distance between the Sun and the Earth is $R$. What will be the duration of the year if the distance between the Sun and the Earth becomes $3R$?