If a planet revolves around the sun with an angular velocity $(\omega)$ and distance $(r)$,then the areal velocity of the planet is proportional to which of the following?

If $A$ is the areal velocity of a planet of mass $M$,its angular momentum is

The mean radius of the earth's orbit around the sun is $1.5 \times 10^{11} \ m$. The mean radius of the orbit of Mercury around the sun is $6 \times 10^{10} \ m$. Mercury will revolve around the sun in:

The time period of a satellite of the Earth is $5 \text{ hours}$. If the separation between the Earth and the satellite is increased to four times the previous value, the new time period of the satellite will be: (in $\text{ hours}$)

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:

$A$ planet revolves around the Sun in an elliptical orbit,where the semi-major axis $a$ is double the semi-minor axis $b$ $(a = 2b)$. The Sun is at the focus. Given that the planet takes $24$ hours to travel through the path $bed$ as shown in the figure,find the time taken by the planet to travel along the path $dab$.