Consider a light planet revolving around a massive star in a circular orbit of radius '$$r$$' with time period '$$T$$'. If the gravitational force of attraction between the planet and the star is proportional to $$\mathrm{r}^{\frac{7}{2}}$$, then $$\mathrm{T}^2$$ is proportional to

The radius of the orbit of a geostationary satellite is (mean radius of earth is '$$R$$', angular velocity about own axis is '$$\omega$$' and acceleration due to gravity on earth's surface is '$$g$$')

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

The radius of earth is $$6400 \mathrm{~km}$$ and acceleration due to gravity $$\mathrm{g}=10 \mathrm{~ms}^{-2}$$. For the weight of body of mass $$5 \mathrm{~kg}$$ to be zero on equator, rotational velocity of the earth must be (in $$\mathrm{rad} / \mathrm{s}$$ )