At what height over the earth's pole,the free fall acceleration decreases by one percent? (Assume the radius of earth to be $6400 \, km$)

If the Earth suddenly stops rotating,then the weight of an object of mass $m$ at the equator will $[\omega$ is the angular speed of the Earth and $R$ is its radius]

If the Earth has no rotational motion,the weight of a person on the equator is $W$. Determine the speed with which the Earth would have to rotate about its axis so that the person at the equator will weigh $\frac{3}{4} W$. The radius of the Earth is $6400 \ km$ and $g = 10 \ m/s^2$.

$A$ narrow but tall cabin is falling freely near the earth's surface. Inside the cabin,two small stones $A$ and $B$ are released from rest (relative to the cabin). Initially $A$ is much above the centre of mass and $B$ is much below the centre of mass of the cabin. $A$ close observation of the motion of $A$ and $B$ will reveal that

If the mass of the Earth decreases by $25 \%$ and its radius increases by $50 \%$,then the acceleration due to gravity at its surface decreases by nearly ......... $\%$

The acceleration due to gravity at a height $(1/20)^{th}$ of the radius of the Earth above the Earth's surface is $9 m s^{-2}$. Its value at an equal depth below the surface of the Earth is: (in $m s^{-2}$)