The period of a satellite in a circular orbit of radius $R$ is $T$. The period of another satellite in a circular orbit of radius $4R$ is:

The figure shows the orbit of a planet $P$ around the sun $S.$ $AB$ and $CD$ are the minor and major axes of the ellipse,respectively.
If $t_1$ is the time taken by the planet to travel along the path $ACB$ and $t_2$ is the time taken to travel along the path $BDA,$ then:

$A$ geostationary satellite is orbiting the earth at a height of $4R$ above the surface of the earth, where $R$ is the radius of the earth. Another satellite is orbiting the earth at a height $1.5R$ from the surface of the earth with periodic time $T$ in hours. The value of $T$ in hours is:

The orbit of a planet around a star is:

$A$ planet is revolving around the sun in an elliptical path as shown. The orbital velocity of the planet will be minimum at:

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: