A satellite is orbiting very close to a planet. Its periodic time depends only on

Options

A. density of the planet
B. mass of the planet
C. radius of the planet
D. mass of the satellite

Answer

To solve the problem of determining what the periodic time of a satellite orbiting very close to a planet depends on, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Scenario**:
- A satellite is orbiting very close to a planet. This means that the radius of the satellite's orbit is approximately equal to the radius of the planet.

2. **Identify Relevant Forces**:
- The gravitational force acting on the satellite provides the necessary centripetal force for its circular motion. The gravitational force can be expressed as:
\[ F_g = \frac{G M_p M_s}{r^2} \]
where \( G \) is the gravitational constant, \( M_p \) is the mass of the planet, \( M_s \) is the mass of the satellite, and \( r \) is the radius of the planet (approximately equal to the radius of the satellite's orbit).

3. **Centripetal Force Requirement**:
- The centripetal force required for the satellite to maintain its circular orbit is given by:
\[ F_c = \frac{M_s v^2}{r} \]
- Setting the gravitational force equal to the centripetal force:
\[ \frac{G M_p M_s}{r^2} = \frac{M_s v^2}{r} \]

4. **Canceling Mass of Satellite**:
- We can cancel \( M_s \) from both sides (assuming \( M_s \neq 0 \)):
\[ \frac{G M_p}{r^2} = \frac{v^2}{r} \]

5. **Rearranging for Velocity**:
- Rearranging gives us:
\[ v^2 = \frac{G M_p}{r} \]

6. **Relate Velocity to Angular Velocity**:
- The relationship between linear velocity \( v \) and angular velocity \( \omega \) is:
\[ v = r \omega \]
- Substituting this into the equation gives:
\[ (r \omega)^2 = \frac{G M_p}{r} \]
- Simplifying leads to:
\[ r^2 \omega^2 = \frac{G M_p}{r} \]
\[ \omega^2 = \frac{G M_p}{r^3} \]

7. **Finding the Period**:
- The period \( T \) of the satellite's orbit is related to angular velocity by:
\[ T = \frac{2\pi}{\omega} \]
- Substituting for \( \omega \):
\[ T = 2\pi \sqrt{\frac{r^3}{G M_p}} \]

8. **Expressing Mass of the Planet**:
- The mass of the planet can be expressed in terms of its density \( \rho \) and volume:
\[ M_p = \rho \cdot \frac{4}{3} \pi r^3 \]
- Substituting this into the period equation gives:
\[ T = 2\pi \sqrt{\frac{r^3}{G \left(\rho \cdot \frac{4}{3} \pi r^3\right)}} \]
- This simplifies to:
\[ T = 2\pi \sqrt{\frac{3}{4\pi G \rho}} \]

9. **Conclusion**:
- The periodic time \( T \) depends only on the density \( \rho \) of the planet and the gravitational constant \( G \).

### Final Answer:
The periodic time of a satellite orbiting very close to a planet depends only on the density of the planet.