There is a second's pendulum on the surface of earth. It is taken to the surface of planet whose mass and radius are twice that of earth. The period of oscillation of second's pendulum on the planet will be

For a satellite orbiting around the earth in a circular orbit, the ratio of potential energy to kinetic energy at same height is

Periodic time of a satellite revolving above the earth's surface at a height equal to radius of the earth '$$R$$' is [ $$g=$$ acceleration due to gravity]

Consider a planet whose density is same as that of the earth but whose radius is three times the radius '$$R$$' of the earth. The acceleration due to gravity '$$\mathrm{g}_{\mathrm{n}}$$' on the surface of planet is $$\mathrm{g}_{\mathrm{n}}=\mathrm{x}$$. $$\mathrm{g}$$ where $$\mathrm{g}$$ is acceleration due to gravity on surface of earth. The value of '$$\mathrm{x}$$' is