A planet of mass $M$ and radius $R$ rotates so quickly that material at its equator only just remains on its surface - AQA - A-Level Physics - Question 10 - 2019 - Paper 2

The tangential velocity $v$ can be expressed in terms of the period $T$ as:

v = rac{2\pi R}{T}

Substituting the expression for $v$ back into the force equation gives:

rac{GMm}{R^2} = rac{m}{R} \left(\frac{2\pi R}{T}\right)^2

After simplifying, we can cancel $m$ and multiply both sides by $T^2$, leading to:

$GM = \frac{4\pi^2 R}{T^2}$

Rearranging this for $T$ gives:

$T^2 = \frac{4\pi^2 R^3}{GM}$

Thus,

$T = 2\pi \sqrt{\frac{R^3}{GM}}$

The derived formula for the period of rotation matches option C from the question, which is:

$2\pi \sqrt{\frac{R^3}{GM}}$