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 centripetal force required to keep an object of mass $m$ moving in a circle of radius $R$ with angular velocity $heta$ is:

F_c = m rac{v^2}{R}

where $v = R heta$ (the linear velocity).

For the material to remain on the surface, the gravitational force must equal the centripetal force:

rac{GMm}{R^2} = m rac{(R heta)^2}{R}

Cancelling $m$ from both sides and simplifying, we get:

rac{GM}{R^2} = R heta^2

Rearranging the equation gives:

heta^2 = rac{GM}{R^3}

Taking the square root:

heta = rac{ ext{sqrt}(GM)}{R^{3/2}}

The period $T$ is related to angular velocity by:

T = rac{2 heta}{2 ext{π}}

Substituting our expression for $heta$:

T = rac{2 ext{π}}{ heta} = 2 ext{π} rac{R^{3/2}}{ ext{sqrt}(GM)}

Thus, the final expression for the period of rotation $T$ becomes:

T = 2 ext{π} rac{R^{3/2}}{ ext{sqrt}(GM)}

This matches the answer option C in the context of the choices.