A satellite revolves around the earth in an elliptical orbit. Its speed is

Options

A. same at all points on the orbit.
B. greatest when it farthest from the earth
C. greatest when it is closest to the earth
D. greatest neither when it is closest nor when it is farthest from the earth, but at some other point.

Explanation

To determine the speed of a satellite revolving around the Earth in an elliptical orbit, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Nature of the Orbit**:
- A satellite in an elliptical orbit experiences varying speeds at different points in its orbit. This is due to the gravitational force acting as a central force towards the Earth, which is located at one of the foci of the ellipse.

**Hint**: Remember that the gravitational force is not constant throughout the orbit; it varies with distance from the Earth.

2. **Applying Kepler's Laws**:
- According to Kepler's laws, particularly the law of areas, a line segment joining a planet and a satellite sweeps out equal areas during equal intervals of time. This implies that the satellite moves faster when it is closer to the Earth (at periapsis) and slower when it is farther away (at apoapsis).

**Hint**: Kepler’s laws are crucial for understanding the motion of satellites in elliptical orbits.

3. **Angular Momentum Conservation**:
- The angular momentum (L) of the satellite is conserved. It can be expressed as:
\[ L = mvr \]
where \(m\) is the mass of the satellite, \(v\) is its speed, and \(r\) is the distance from the center of the Earth.

**Hint**: Since angular momentum is conserved, any change in distance \(r\) will affect the speed \(v\).

4. **Relationship Between Speed and Distance**:
- From the conservation of angular momentum, we can deduce that:
\[ mvr = \text{constant} \]
This implies that:
\[ v \propto \frac{1}{r} \]
Therefore, as the distance \(r\) decreases (i.e., when the satellite is closer to the Earth), the speed \(v\) increases.

**Hint**: The inverse relationship between speed and distance is key to solving this problem.

5. **Evaluating the Options**:
- Based on the relationship \(v \propto \frac{1}{r}\):
- The speed is greatest when the satellite is closest to the Earth (at periapsis).
- The speed is least when the satellite is farthest from the Earth (at apoapsis).
- Therefore, the correct statement is that the satellite's speed is greatest when it is closest to the Earth.

**Hint**: Analyze each option carefully, using the relationship derived from angular momentum.

### Conclusion:
The correct answer is that the satellite's speed is greatest when it is closest to the Earth.