The value of acceleration due to gravity $g$ is maximum at

What should be the angular speed of the Earth about its axis so that the acceleration due to gravity at the equator becomes zero?

$A$ simple pendulum is taken from the equator to the pole. Its period

If a planet has a mass and radius both half that of the Earth,the acceleration due to gravity at its surface would be ......... $m/s^2$ ($g$ on Earth $= 9.8\, m/s^2$)

Acceleration due to gravity at the surface of a planet is equal to that at the surface of the Earth,and its density is $1.5$ times that of the Earth. If the radius of the Earth is $R$,the radius of the planet is:

Choose the correct alternative:
$(a)$ Acceleration due to gravity increases/decreases with increasing altitude.
$(b)$ Acceleration due to gravity increases/decreases with increasing depth (assume the Earth to be a sphere of uniform density).
$(c)$ Acceleration due to gravity is independent of mass of the Earth/mass of the body.
$(d)$ The formula $-G M m(1 / r_{2}-1 / r_{1})$ is more/less accurate than the formula $m g(r_{2}-r_{1})$ for the difference of potential energy between two points $r_{2}$ and $r_{1}$ distance away from the centre of the Earth.