The height vertically above the earth's surface at which the acceleration due to gravity becomes $$1 \%$$ of its value at the surface is

An uniform sphere of mass $$M$$ and radius $$R$$ exerts a force of $$F$$ on a small mass $$m$$ placed at a distance of 3R from the centre of the sphere. A spherical portion of diameter $$R$$ is cut from the sphere as shown in the fig. The force of attraction between the remaining part of the disc and the mass $$\mathrm{m}$$ is

The acceleration due to gravity at a height of $$7 \mathrm{~km}$$ above the earth is the same as at a depth d below the surface of the earth. Then d is

If the earth were to spin faster, acceleration due to gravity at the poles