How does the rotation of the Earth affect the acceleration due to gravity on its surface?

The depth $d$ below the surface of the earth where the value of acceleration due to gravity becomes $\left(\frac{1}{n}\right)$ times the value at the surface of the earth is ($R$ = radius of the earth).

Two satellites $P$ and $Q$ are moving in different circular orbits around the Earth (radius $R$). The heights of $P$ and $Q$ from the Earth's surface are $h_p$ and $h_Q$,respectively,where $h_p = R / 3$. The accelerations of $P$ and $Q$ due to Earth's gravity are $g_p$ and $g_Q$,respectively. If $g_p / g_Q = 36 / 25$,what is the value of $h_Q$?

Assuming the density of the Earth is constant, which graph correctly represents the variation of acceleration due to gravity $(g)$ with the distance $(r)$ from the center of the Earth (radius of the Earth $= R$)?

Assertion $(A)$: $A$ particle of mass $m$ dropped into a hole made along the diameter of the Earth from one end to the other possesses simple harmonic motion.
Reason $(R)$: Gravitational force between any two particles is inversely proportional to the square of the distance between them.

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,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$).