The variation of acceleration due to gravity $g$ with distance $d$ from the centre of the earth is best represented by ($R =$ Earth's radius)

The mass of the moon is $\frac{1}{81}$ of the earth,but the gravitational pull (acceleration due to gravity) is $\frac{1}{6}$ of the earth. This is due to the fact that:

Consider Earth to be a sphere of radius $R_e$ rotating about its own axis with angular speed $\omega$. If $g_{E}$ and $g_{P}$ are the accelerations due to gravity at the equator and the poles respectively,then $(g_{P}-g_{E})$ is given by $\left[\cos (0^{\circ})=\sin (\frac{\pi}{2})=1, \sin (0^{\circ})=\cos (\frac{\pi}{2})=0\right]$

The dependence of acceleration due to gravity $g$ on the distance $r$ from the centre of the earth,assumed to be a sphere of radius $R$ of uniform density,is as shown in the figure below. The correct figure is:

There is a planet which is $8$ times more massive and $27$ times denser than the Earth. If $g^{\prime}$ and $g$ are the accelerations due to gravity on the surfaces of the planet and the Earth respectively,then:

When a body is taken from the pole to the equator,its weight: