if a satellite orbits as close to the earth's surface as possible.

Options

A. its speed is maximum
B. time period of its revolution is minimum
C. the total energy of the earth plus satellite system is minimum
D. the total energy of the earth plus satellite system is maximum.

Explanation

To solve the question regarding a satellite orbiting as close to the Earth's surface as possible, we will analyze the relationships between the satellite's speed, time period of revolution, and total energy.

### Step-by-Step Solution:

1. **Understanding the Orbit**:
- When a satellite orbits close to the Earth's surface, it is at a distance \( r \) from the center of the Earth, where \( r \) is approximately equal to the radius of the Earth \( R \).

2. **Gravitational Force**:
- The gravitational force acting on the satellite can be expressed using Newton's law of gravitation:
\[ F = \frac{GMm}{r^2} \]
- Here, \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the satellite, and \( r \) is the distance from the center of the Earth.

3. **Centripetal Force**:
- For the satellite to maintain its circular orbit, the gravitational force must equal the centripetal force required to keep the satellite in orbit:
\[ F = \frac{mv^2}{r} \]
- Setting these two forces equal gives:
\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \]

4. **Solving for Orbital Speed \( v \)**:
- By canceling \( m \) and rearranging, we find:
\[ v^2 = \frac{GM}{r} \]
- Therefore, the orbital speed \( v \) is given by:
\[ v = \sqrt{\frac{GM}{r}} \]

5. **Time Period of Revolution \( T \)**:
- The time period \( T \) of the satellite's orbit can be calculated using the formula:
\[ T = \frac{2\pi r}{v} \]
- Substituting the expression for \( v \):
\[ T = \frac{2\pi r}{\sqrt{\frac{GM}{r}}} = 2\pi \sqrt{\frac{r^3}{GM}} \]

6. **Total Energy of the Satellite**:
- The total mechanical energy \( E \) of the satellite in orbit is the sum of its kinetic energy \( K \) and potential energy \( U \):
\[ K = \frac{1}{2}mv^2 = \frac{1}{2}m\left(\frac{GM}{r}\right) = \frac{GMm}{2r} \]
\[ U = -\frac{GMm}{r} \]
- Thus, the total energy \( E \) is:
\[ E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r} \]

7. **Analyzing the Relationships**:
- As \( r \) approaches \( R \) (the radius of the Earth):
- The speed \( v \) is maximum because it is inversely proportional to \( r \).
- The time period \( T \) is minimum because it is directly proportional to \( r^{3/2} \).
- The total energy \( E \) is minimum (most negative) because it is inversely proportional to \( r \).

### Conclusion:
- Therefore, when a satellite orbits as close to the Earth's surface as possible:
- Its speed is maximum.
- The time period of revolution is minimum.
- The total energy of the Earth plus satellite is minimum (most negative).