A satellite having mass much smaller than the mass of earth is revolving in an elliptical orbit around the earth. In this case

Options

A. The acceleration of satellite is always directed towards the centre of earth.
B. The total mechanical energy of satellite remains constant.
C. The magnitude and direction of angular momentum remain conserved.
D. The linear momentum of satellite is maximum when it is farther from earth.

Answer

To solve the problem regarding a satellite revolving in an elliptical orbit around the Earth, we will analyze each statement provided in the question step by step.

### Step 1: Analyze the Direction of Acceleration
The first statement to evaluate is whether the acceleration of the satellite is always directed towards the center of the Earth.

- **Explanation**: The gravitational force acting on the satellite is always directed towards the center of the Earth. According to Newton's second law, the acceleration of an object is in the direction of the net force acting on it. Since the only significant force acting on the satellite is gravity, the acceleration must also be directed towards the center of the Earth.

**Conclusion**: The first option is correct.

### Step 2: Evaluate Total Mechanical Energy
Next, we need to check if the total mechanical energy of the satellite remains constant.

- **Explanation**: The total mechanical energy (E) of a satellite in orbit can be expressed as the sum of its kinetic energy (K) and potential energy (U). For a satellite in an elliptical orbit, the total mechanical energy is given by:
\[ E = K + U = -\frac{GMm}{2r} \]
where \( G \) 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 to the satellite. Since \( G \), \( M \), and \( m \) are constants and the average distance \( r \) remains constant over one complete orbit, the total mechanical energy remains constant.

**Conclusion**: The second option is also correct.

### Step 3: Check Angular Momentum Conservation
Now, we will determine if the magnitude and direction of angular momentum remain conserved.

- **Explanation**: The angular momentum \( L \) of the satellite is given by:
\[ L = mvr \]
In an elliptical orbit, while both the radius \( r \) and the velocity \( v \) of the satellite change, the product \( mvr \) remains constant due to the conservation of angular momentum. This is in accordance with Kepler's laws of planetary motion, which state that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

**Conclusion**: The third option is correct.

### Step 4: Analyze Linear Momentum
Finally, we need to assess the statement regarding the linear momentum of the satellite being maximum when it is farther from the Earth.

- **Explanation**: Linear momentum \( p \) is given by:
\[ p = mv \]
In an elliptical orbit, when the satellite is farther from the Earth (at apogee), its speed \( v \) is lower compared to when it is closer (at perigee). Thus, even though the distance \( r \) is maximum, the velocity \( v \) is minimum, leading to a minimum linear momentum at apogee. Therefore, the statement that linear momentum is maximum when the satellite is farther from Earth is incorrect.

**Conclusion**: The fourth option is incorrect.

### Final Answer
The correct options are A, B, and C.