A seconds pendulum is placed in a space laboratory orbiting round the earth at a height '$$3 \mathrm{R}$$' from the earth's surface. The time period of the pendulum will be ( $$R=$$ radius of earth)

For a body of mass '$$m$$', the acceleration due to gravity at a distance '$$R$$' from the surface of the earth is $$\left(\frac{g}{4}\right)$$. Its value at a distance $$\left(\frac{R}{2}\right)$$ from the surface of the earth is ( $$R=$$ radius of the earth, $$g=$$ acceleration due to gravity)

The ratio of energy required to raise a satellite of mass '$$m$$' to height '$$h$$' above the earth's surface to that required to put it into the orbit at same height is [ $$\mathrm{R}=$$ radius of earth]

A pendulum is oscillating with frequency '$$n$$' on the surface of the earth. It is taken to a depth $$\frac{R}{2}$$ below the surface of earth. New frequency of oscillation at depth $$\frac{R}{2}$$ is

[ $$R$$ is the radius of earth]