If all objects on the equator of the earth feel weightless,then the duration of the day will nearly become ....... $hr$.

Imagine Earth to be a solid sphere of mass $M$ and radius $R$. If the value of acceleration due to gravity at a depth $d$ below Earth's surface is the same as its value at a height $h$ above its surface and equal to $\frac{g}{4}$ (where $g$ is the value of acceleration due to gravity on the surface of Earth),the ratio of $\frac{h}{d}$ will be

Consider a planet in some solar system which has a mass double the mass of Earth and density equal to the average density of Earth. If the weight of an object on Earth is $W$,then the weight of the same object on that planet will be

$A$ tunnel is dug along a diameter of the Earth. If $M_e$ and $R_e$ are the mass and radius of the Earth respectively,then the force on a particle of mass $m$ placed in the tunnel at a distance $r$ from the center 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.

What should be the angular speed of the Earth so that a body lying on the equator may appear weightless? $(g = 10\,m/s^2, R = 6400\,km)$