If a satellite orbiting the Earth is 9 times closer to the Earth than the Moon, what is the time period of rotation of the satellite? Given rotational time period of Moon $=27$ days and gravitational attraction between the satellite and the moon is neglected.

A small point of mass $m$ is placed at a distance $2 R$ from the centre ' $O$ ' of a big uniform solid sphere of mass M and radius R . The gravitational force on ' m ' due to M is $\mathrm{F}_1$. A spherical part of radius $\mathrm{R} / 3$ is removed from the big sphere as shown in the figure and the gravitational force on m due to remaining part of $M$ is found to be $F_2$. The value of ratio $F_1: F_2$ is

A satellite of $$10^3 \mathrm{~kg}$$ mass is revolving in circular orbit of radius $$2 R$$. If $$\frac{10^4 R}{6} \mathrm{~J}$$ energy is supplied to the satellite, it would revolve in a new circular orbit of radius

(use $$g=10 \mathrm{~m} / \mathrm{s}^2, R=$$ radius of earth)