A satellite is in an elliptical orbit around the earth with aphelion of `6R_E` and perihelion of `3R_E`, where `R_E` is the radius of the earth. The eccentricity of the orbit is

Options

A. `1/2`
B. `1/3`
C. `1/4`
D. `1/6`

Explanation

To find the eccentricity of the elliptical orbit of the satellite, we can use the definitions of aphelion and perihelion distances along with the formula for eccentricity.

### Step-by-step Solution:

1. **Identify the given values**:
- Aphelion distance \( D_A = 6R_E \)
- Perihelion distance \( D_P = 3R_E \)

2. **Recall the definitions**:
- The aphelion distance \( D_A \) is the farthest point from the Earth in the orbit.
- The perihelion distance \( D_P \) is the closest point to the Earth in the orbit.

3. **Use the formulas for aphelion and perihelion**:
- The formulas for the aphelion and perihelion in terms of the semi-major axis \( A \) and eccentricity \( e \) are:
\[ D_A = A(1 + e) \]
\[ D_P = A(1 - e) \]

4. **Set up the equations**:
- From the aphelion distance:
\[ 6R_E = A(1 + e) \quad \text{(1)} \]
- From the perihelion distance:
\[ 3R_E = A(1 - e) \quad \text{(2)} \]

5. **Divide the two equations**:
- Divide equation (1) by equation (2):
\[ \frac{D_A}{D_P} = \frac{A(1 + e)}{A(1 - e)} \implies \frac{6R_E}{3R_E} = \frac{1 + e}{1 - e} \]
- Simplifying gives:
\[ 2 = \frac{1 + e}{1 - e} \]

6. **Cross-multiply and solve for \( e \)**:
- Cross-multiplying gives:
\[ 2(1 - e) = 1 + e \]
- Expanding and rearranging:
\[ 2 - 2e = 1 + e \implies 2 - 1 = 2e + e \implies 1 = 3e \]
- Thus, we find:
\[ e = \frac{1}{3} \]

### Conclusion:
The eccentricity of the orbit is \( e = \frac{1}{3} \).