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:

The distance of Venus from the sun is $0.72\, AU$. The orbital period of Venus is ............ days.

Kepler's third law states that the square of the period of revolution $(T)$ of a planet around the sun is proportional to the cube of the average distance $(r)$ between the sun and the planet,i.e.,$T^2 = Kr^3$,where $K$ is a constant. If the masses of the sun and the planet are $M$ and $m$ respectively,then according to Newton's law of gravitation,the force of attraction between them is $F = \frac{GMm}{r^2}$,where $G$ is the gravitational constant. The relation between $G$ and $K$ is:

If $r_p, v_p, L_p$ and $r_a, v_a, L_a$ are radii,velocities and angular momenta of a planet at perihelion and aphelion of its elliptical orbit around the Sun respectively,then

$A$ planet takes $200$ days to complete one revolution around the Sun. If the distance of the planet from the Sun is reduced to one-fourth of the original distance,how many days will it take to complete one revolution?

Draw the areal velocity versus time graph for Mars.