In the asteroid belt a pebble is in close orbit around a shperical rock having density nearly same as that of earth (close meaning that the pebble goes around the rock very near to the rock's surface) orbital period of the pebble around the rock is of the order of

Options

A. 1 day
B. 1 month
C. 1 hr
D. 1 yr

Explanation

To find the orbital period of a pebble in close orbit around a spherical rock with a density similar to that of Earth, we can follow these steps:

### Step 1: Understand the Forces
The pebble is in a close orbit around the spherical rock, which means that the gravitational force acting on the pebble provides the necessary centripetal force for its circular motion. Therefore, we can set the gravitational force equal to the centripetal force.

### Step 2: Write the Equations
The gravitational force \( F_g \) acting on the pebble is given by:
\[ F_g = \frac{GMm}{R^2} \]
where:
- \( G \) is the gravitational constant (\( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)),
- \( M \) is the mass of the rock,
- \( m \) is the mass of the pebble,
- \( R \) is the radius of the rock (and the orbit).

The centripetal force \( F_c \) required to keep the pebble in circular motion is given by:
\[ F_c = m \omega^2 R \]
where \( \omega \) is the angular velocity of the pebble.

### Step 3: Set the Forces Equal
Setting the gravitational force equal to the centripetal force gives us:
\[ \frac{GMm}{R^2} = m \omega^2 R \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{GM}{R^2} = \omega^2 R \]

### Step 4: Solve for Angular Velocity \( \omega \)
Rearranging the equation to solve for \( \omega^2 \):
\[ \omega^2 = \frac{GM}{R^3} \]

### Step 5: Express Mass \( M \)
The mass \( M \) of the rock can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ M = \rho V = \rho \left(\frac{4}{3} \pi R^3\right) \]
Substituting this into the equation for \( \omega^2 \):
\[ \omega^2 = \frac{G \left(\rho \frac{4}{3} \pi R^3\right)}{R^3} = \frac{4\pi G \rho}{3} \]

### Step 6: Find the Orbital Period \( T \)
The orbital period \( T \) is related to angular velocity \( \omega \) by:
\[ T = \frac{2\pi}{\omega} \]
Substituting for \( \omega \):
\[ T = 2\pi \sqrt{\frac{3}{4\pi G \rho}} \]

### Step 7: Substitute Values
Now we can substitute the known values:
- \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- \( \rho \) (density of Earth) \( \approx 5.51 \times 10^3 \, \text{kg/m}^3 \)

Calculating \( T \):
\[ T = 2\pi \sqrt{\frac{3}{4\pi (6.67 \times 10^{-11})(5.51 \times 10^3)}} \]

### Step 8: Final Calculation
Perform the calculation:
1. Calculate \( 4\pi G \rho \).
2. Take the square root of \( \frac{3}{4\pi G \rho} \).
3. Multiply by \( 2\pi \).

After performing the calculations, we find that the orbital period \( T \) is approximately \( 5062 \) seconds.

### Summary
The orbital period of the pebble around the spherical rock is of the order of \( 5062 \) seconds.

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