At a certain depth "d " below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height $$\mathrm{3 R}$$ above earth surface. Where $$\mathrm{R}$$ is Radius of earth (Take $$\mathrm{R}=6400 \mathrm{~km}$$ ). The depth $$\mathrm{d}$$ is equal to

If the gravitational field in the space is given as $$\left(-\frac{K}{r^{2}}\right)$$. Taking the reference point to be at $$\mathrm{r}=2 \mathrm{~cm}$$ with gravitational potential $$\mathrm{V}=10 \mathrm{~J} / \mathrm{kg}$$. Find the gravitational potential at $$\mathrm{r}=3 \mathrm{~cm}$$ in SI unit (Given, that $$\mathrm{K}=6 \mathrm{~Jcm} / \mathrm{kg}$$)

The time period of a satellite of earth is 24 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value, then its new time period will become.