A planet has a mass $M$ and a radius $R$ - AQA - A-Level Physics - Question 12 - 2022 - Paper 2

To derive the period of rotation of the planet, we can start by considering the forces acting on the loose material at the equator. For the material to remain in contact with the surface, the gravitational force must equal the centripetal force required to keep it moving in a circular path. The gravitational force is given by:

F_g = rac{GMm}{R^2}

where:

• $G$ is the gravitational constant,
• $M$ is the mass of the planet,
• $m$ is the mass of the loose material,
• $R$ is the radius of the planet.

The centripetal force required for a mass moving in a circle is:

F_c = rac{mv^2}{R}

where:

• $v$ is the tangential velocity of the loose material.

Setting the gravitational force equal to the centripetal force:

rac{GMm}{R^2} = rac{mv^2}{R}

We can cancel $m$ from both sides and rearranging gives:

v^2 = rac{GM}{R}

The tangential speed $v$ can also be expressed in terms of the period $T$ of rotation:

v = rac{2 heta}{T}

where $heta$ is the angular displacement in radians (with $heta = 2 ext{π}$ for one complete rotation). Thus:

v = rac{2 ext{π}R}{T}

Substituting this back into our earlier equation:

rac{2 ext{π}R}{T}^2 = rac{GM}{R}

Rearranging this leads to:

T^2 = rac{4 ext{π}^2R^3}{GM}

Taking the square root of both sides, we find:

T = 2 ext{π} imes rac{R^{3/2}}{ ext{√}{GM}}

Thus, the correct expression for the period of rotation of the planet is:

T = 2 ext{π} rac{R^{3/2}}{ ext{√}{GM}}

Among the given options, this is aligned with option C.