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 ..........

$A$ planet orbits in an elliptical path of eccentricity $e$ around a massive star considered fixed at one of the foci. The point in space,where it is closest to the star is denoted by $P$ and the point,where it is farthest is denoted by $A$. Let $v_P$ and $v_A$ be the respective speeds at $P$ and $A$,then

The time period of a satellite of Earth is $24 \text{ hours}$. If the separation between the Earth and the satellite is decreased to one-fourth of the previous value, then its new time period will become: (in $\text{ hours}$)

$A$ satellite $S_1$ of mass $m$ is moving in an orbit of radius $r$. Another satellite $S_2$ of mass $2m$ is moving in an orbit of radius $2r$. The ratio of the time period of satellite $S_2$ to that of $S_1$ is

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:

$A$ planet revolving in an elliptical orbit has:
$(A)$ a constant velocity of revolution.
$(B)$ the least velocity when it is nearest to the sun.
$(C)$ its areal velocity is directly proportional to its velocity.
$(D)$ areal velocity is inversely proportional to its velocity.
$(E)$ a trajectory such that the areal velocity is constant.
Choose the correct answer from the options given below: