If the radius of the Earth becomes $n$ times its present value,without any change in its mass,then the duration of the day becomes:

Three point masses,each of mass $m$,are kept at the corners of an equilateral triangle of side $L$. The system rotates about the center of the triangle. The period of rotation $T$ is directly proportional to:

Two particles,each of mass $m$,are moving in a circle of radius $R$ under the influence of their mutual gravitational attraction. What is their speed?

The planet Mars has two moons,Phobos and Deimos.
$(i)$ Phobos has a period of $7$ hours,$39$ minutes and an orbital radius of $9.4 \times 10^{3} \; km$. Calculate the mass of Mars.
$(ii)$ Assume that Earth and Mars move in circular orbits around the Sun,with the Martian orbit being $1.52$ times the orbital radius of the Earth. What is the length of the Martian year in days?

Mean solar day is the time interval between two successive noons when the sun passes through the zenith point (meridian). $A$ sidereal day is the time interval between two successive transits of a distant star through the zenith point (meridian). By drawing an appropriate diagram showing the earth's spin and orbital motion,show that the mean solar day is $4\,\text{min}$ longer than the sidereal day. In other words,distant stars would rise $4\,\text{min}$ early every successive day.

Two point masses of mass $4m$ and $m$ respectively,separated by a distance $d$,are revolving under their mutual force of attraction. The ratio of their kinetic energies is: