The fastest possible of rotation of a planet is that for which the gravitational force on material at the equtor barely provides the centripetal force needed for the rotation. Show then that the corresponding shortest period of rotation is given by `T=sqrt((3pi)/(Grho))` where `rho` is the density of the planet, assumed to be homogeneous. Evalute the rotation period assuming a density of `3.0 gm//cm^(2)`, typical of many planets, satellites, and asteroids. No such object is found to be spinning with a period shorter than found by this analysis.

Explanation

To solve the problem, we need to derive the expression for the shortest period of rotation \( T \) of a planet, given that the gravitational force at the equator barely provides the centripetal force needed for the rotation.

### Step-by-Step Solution:

1. **Understanding Forces at the Equator**:
At the equator of a rotating planet, the gravitational force \( g \) must equal the centripetal force required for circular motion. The effective gravitational force \( g' \) at the equator is given by:
\[ g' = g - r \omega^2 \]
where \( r \) is the radius of the planet and \( \omega \) is the angular velocity.

2. **Condition for Fastest Rotation**:
For the fastest rotation, the effective gravitational force \( g' \) becomes zero:
\[ g - r \omega^2 = 0 \]
Rearranging gives:
\[ g = r \omega^2 \]

3. **Finding Angular Velocity**:
From the equation \( g = r \omega^2 \), we can solve for \( \omega \):
\[ \omega = \sqrt{\frac{g}{r}} \]

4. **Relating Angular Velocity to Period**:
The angular velocity \( \omega \) is related to the period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]
Substituting this into the equation for \( \omega \):
\[ \frac{2\pi}{T} = \sqrt{\frac{g}{r}} \]
Rearranging gives:
\[ T = \frac{2\pi}{\sqrt{\frac{g}{r}}} = 2\pi \sqrt{\frac{r}{g}} \]

5. **Substituting for \( g \)**:
The gravitational acceleration \( g \) at the surface of a planet can be expressed in terms of its mass \( M \) and radius \( r \):
\[ g = \frac{GM}{r^2} \]
where \( G \) is the gravitational constant. The mass \( M \) can be expressed in terms of density \( \rho \):
\[ M = \rho V = \rho \left(\frac{4}{3} \pi r^3\right) \]
Substituting this into the equation for \( g \):
\[ g = \frac{G \left(\rho \frac{4}{3} \pi r^3\right)}{r^2} = \frac{4\pi G \rho r}{3} \]

6. **Final Expression for the Period**:
Substituting \( g \) back into the expression for \( T \):
\[ T = 2\pi \sqrt{\frac{r}{\frac{4\pi G \rho r}{3}}} = 2\pi \sqrt{\frac{3r}{4\pi G \rho}} \]
Simplifying gives:
\[ T = \sqrt{\frac{3\pi}{G \rho}} \]

### Evaluation of the Rotation Period:

1. **Substituting Density**:
Given \( \rho = 3 \, \text{g/cm}^3 = 3000 \, \text{kg/m}^3 \) (since \( 1 \, \text{g/cm}^3 = 1000 \, \text{kg/m}^3 \)), and using \( G = 6.674 \times 10^{-11} \, \text{m}^3/\text{kg s}^2 \):
\[ T = \sqrt{\frac{3\pi}{G \cdot 3000}} = \sqrt{\frac{3 \times 3.14}{6.674 \times 10^{-11} \times 3000}} \]

2. **Calculating**:
\[ T \approx \sqrt{\frac{9.42}{2.0022 \times 10^{-7}}} \approx \sqrt{4.699 \times 10^7} \approx 6850 \, \text{s} \approx 1.9 \, \text{hours} \]

### Final Answer:
The shortest period of rotation \( T \) is approximately \( 1.9 \) hours.