At what altitude will the acceleration due to gravity be $25\%$ of that at the earth's surface (given radius of earth is $R$)?

The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n-1}$ times the value at the earth's surface is ($R=$ radius of the earth).

Earth is assumed to be a sphere of radius $R$. If $g_{\phi}$ is the value of effective acceleration due to gravity at latitude $30^{\circ}$ and $g$ is the value at the equator,then the value of $|g - g_{\phi}|$ is ($\omega$ is the angular velocity of rotation of the Earth,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$).

$A$ body weighs $250\,N$ on the surface of the earth. How much will it weigh halfway down to the center of the earth (in $,N$)?

If the radius of the earth were to shrink by $1\%$ with its mass remaining the same,the acceleration due to gravity on the earth's surface would

At which location,the equator or the poles of the Earth,is the value of acceleration due to gravity $g$ higher? Why?