The figure shows the elliptical path $abcd$ of a planet around the sun $S$ such that the area of triangle $csa$ is $\frac{1}{4}$ of the area of the ellipse. With $db$ as the major axis and $ca$ as a chord perpendicular to the major axis,if $t_1$ is the time taken for the planet to travel along the path $abc$ and $t_2$ is the time taken for the path $cda$,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?

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:

The distance of Neptune and Saturn from the Sun is nearly $10^{13} \ m$ and $10^{12} \ m$ respectively. Assuming that they move in circular orbits,their periodic times will be in the ratio:

$A$ planet revolves around the Sun in an elliptical orbit as shown in the figure. Which of the following statements is correct?

The figure shows the motion of a planet around the sun in an elliptical orbit with the sun at the focus. The shaded areas are equal. If $t_1$ and $t_2$ represent the time taken for the planet to move from $A$ to $B$ and $C$ to $D$ respectively,then: