Two planets revolve around the sun with frequencies ${N_1}$ and ${N_2}$ revolutions per year. If their average orbital radii are ${R_1}$ and ${R_2}$ respectively,then ${R_1}/{R_2}$ is equal to:

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 time period of a satellite of the Earth is $5$ hours. If the separation between the Earth and the satellite is increased to four times the previous value,the new time period will become ......... $hours$.

The figure shows the motion of a planet around the sun in an elliptical orbit with the sun at the focus. The shaded areas $A$ and $B$ are also shown in the figure,which can be assumed to be equal. If ${t_1}$ and ${t_2}$ represent the time taken for the planet to move from $a$ to $b$ and $d$ to $c$ respectively,then:

$A$ satellite is launched into a circular orbit of radius $R$ around Earth,while a second satellite is launched into a circular orbit of radius $1.02 R$. The percentage difference in the time periods of the two satellites is:

The average distance of the Earth from the Sun is $L_{1}$. If one year of the Earth is $D$ days, then one year of another planet whose average distance from the Sun is $L_{2}$ will be: