Question: The fastest possible rate of rotation of a planet is that for which the gravitational force on material at the equator just barely provides the centripetal force needed for the rotation. (Why?) (a) Show that the corresponding shortest period of rotation is$\mathit{T}\mathbf{=}\sqrt{\frac{\mathbf{3}\mathbf{\pi }}{\mathbf{G}\mathbf{\rho }}}$where $\mathbf{\rho }$is the uniform density (mass per unit volume) of the spherical planet. (b) Calculate the rotation period assuming a density of $\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$, typical of many planets, satellites, and asteroids. No astronomical object has ever been found to be spinning with a period shorter than that determined by this analysis.

The uniform density of the spherical planet is
$$\begin{gathered} \rho =3.0\frac{g}{c{m}^3} \\ =3.0\times {10}^3\frac{kg}{m^3} \end{gathered}$$

Every particle in the universe is attracted to every other particle with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance, according to Newton's Law of Universal Gravitation.

We can use the concept of Newton’s law of gravitation and centripetal force. The gravitational force on the planet provides the centripetal force on the planet. We can use the expression of density and orbital speed.

Formula:

$$\begin{gathered} F=\frac{GMm}{r^2} \\ {F}_c=\frac{m{v}^2}{r} \\ v=\frac{2\pi r}{T} \end{gathered}$$

Where,

F is the gravitational force

F_c is thecentripetal force

G is the gravitational constant ($6.67\times {10}^{-11} {\text{m}}^{\text{3}}\text{/kg} ·{\text{s}}^{\text{2}}$)

M is the mass of earth

V is the speed of object

m is the mass of object

T is the time period of object

According to Newton’s law of gravitation, the gravitational force acting on the spherical planet is

$F=\frac{GMm}{r^2}$

The gravitational force on the planet provides the centripetal force on the planet. Hence, the expression for centripetal force is

$F_c=\frac{m{v}^2}{r}$

The gravitational force can be balanced by the centripetal force. Therefore,

$\frac{GMm}{r^2}=\frac{m{v}^2}{r}$ …(i)

According to the expression of orbital velocity of star,

$$\begin{gathered} v=\frac{2\pi r}{T} \\ {v}^2=\frac{4{\pi }^2{r}^2}{T^2} \end{gathered}$$

Equation (i) becomes

$$\begin{gathered} \frac{GMm}{r^2}=\frac{m}{r}\frac{4{\pi }^2{r}^2}{T^2} \\ \frac{GM}{r^2}=\frac{4{\pi }^{2}r}{T^2} \end{gathered}$$

…(ii)

The expression of the density of the spherical planet is

$$\begin{gathered} \rho =\frac{M}{V} \\ \rho =\frac{M}{\frac{\text{4}}{\text{3}}\pi r^3} \\ M=\frac{4}{3}\pi r^3\rho \end{gathered}$$

Equation (ii) becomes

.$$\begin{gathered} \frac{G\left(\frac{4}{3}\pi r^3\rho \right)}{r^2}=\frac{4{\pi }^{2}r}{T^2} \\ T=\sqrt{\frac{3\pi }{G\rho }} \end{gathered}$$

We can calculate the rotational period of the spherical planet as

$$\begin{gathered} T=\sqrt{\frac{3\times 3.142}{\left(6.67\times {10}^{-11}\frac{N{m}^2}{k{g}^2}\right)\times \left(3.0\times {10}^3\frac{kg}{m^3}\right)}} \\ =6.86\times {10}^{3}s \end{gathered}$$

The rotation period of the planet is$6.86\times {10}^{3}s$