According to Kepler, the period of revolution of a planet $(T)$ and its mean distance from the sun $(r)$ are related by the equation:

If a body describes a circular motion under an inverse square field,the time taken to complete one revolution $T$ is related to the radius of the circular orbit $r$ as:

An earth satellite $X$ is revolving around the earth in an orbit whose radius is one-fourth of the radius of the orbit of a communication satellite. The time period of revolution of $X$ is ..........

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:

The time period of a satellite in a circular orbit of radius $R$ is $T$. The period of another satellite in a circular orbit of radius $9R$ is............ $T$.

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: