The time period of a satellite of earth is $24 \ hours$. 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 $....... \ hours$.

According to Kepler,planets move in

Two planets move around the sun. The periodic times and the mean radii of the orbits are $T_1, T_2$ and $r_1, r_2$ respectively. The ratio $T_1/T_2$ is equal to

The time period of a satellite of the Earth is $5$ hours. If the separation between the Earth and the satellite is increased to four times the previous value,the new time period will become ......... $hours$.

Two planets are at mean distances $d_1$ and $d_2$ from the sun,and their orbital frequencies are $n_1$ and $n_2$ respectively. Then:

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: