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

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 $\text{ hours}$)

Draw the areal velocity versus time graph for Mars.

The Earth is revolving around the Sun. If the distance of the Earth from the Sun is reduced to $\frac{1}{4}$ of the present distance,then the length of the present day will be reduced by a factor of:

$A$ satellite moves in a circle around the earth. The radius of this circle is equal to one half of the radius of the moon's orbit. The satellite completes one revolution in

The kinetic energies of a planet in an elliptical orbit about the Sun,at positions $A, B$ and $C$ are $K_A, K_B$ and $K_C$,respectively. $AC$ is the major axis and $SB$ is perpendicular to $AC$ at the position of the Sun $S$ as shown in the figure. Then