$A$ planet of mass $m$ is moving in an elliptical orbit about the sun with time period $T$. If $A$ is the area of the orbit,then its angular momentum is:

If the radius of the Earth's orbit is made $\frac{1}{4}$ of its original value,the duration of a year will become:

The period of a planet around the sun is $8$ times that of earth. The ratio of the radius of the planet's orbit to the radius of the earth's orbit 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 is revolving around the sun in an elliptical path as shown. The orbital velocity of the planet will be minimum at:

If the distance of the earth from the Sun is $1.5 \times 10^8 \, km$. Then the distance of an imaginary planet from the Sun,if its period of revolution is $2.83$ years,is $............. \times 10^8 \, km$.