An artificial satellite is in an elliptical orbit around the earth with aphelion of `6R` and perihelion of `2R` where `R` is radius of the earth `=6400 km`. Calculate the eccentricity of the elliptical orbit.

Options

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

Explanation

To find the eccentricity of the elliptical orbit of the artificial satellite, we can follow these steps:

### Step 1: Define the terms
- Let the perihelion distance (closest point to Earth) be \( r_p = 2R \).
- Let the aphelion distance (farthest point from Earth) be \( r_a = 6R \).
- Here, \( R \) is the radius of the Earth, which is \( 6400 \, \text{km} \).

### Step 2: Use the formulas for perihelion and aphelion
In an elliptical orbit, the distances of perihelion and aphelion can be expressed in terms of the semi-major axis \( a \) and eccentricity \( e \):
- \( r_p = a(1 - e) \)
- \( r_a = a(1 + e) \)

### Step 3: Set up equations
From the definitions:
1. \( 2R = a(1 - e) \) (1)
2. \( 6R = a(1 + e) \) (2)

### Step 4: Solve for \( a \) and \( e \)
Add equations (1) and (2):
\[ r_a + r_p = a(1 + e) + a(1 - e) = 2a \]
Substituting the values:
\[ 6R + 2R = 2a \implies 8R = 2a \implies a = 4R \]

Now, subtract equation (1) from equation (2):
\[ r_a - r_p = a(1 + e) - a(1 - e) = 2ae \]
Substituting the values:
\[ 6R - 2R = 2ae \implies 4R = 2ae \implies ae = 2R \]

### Step 5: Substitute \( a \) into the equation for \( e \)
Now we have \( a = 4R \), substitute this into \( ae = 2R \):
\[ (4R)e = 2R \implies e = \frac{2R}{4R} = \frac{1}{2} \]

### Conclusion
The eccentricity \( e \) of the elliptical orbit is \( \frac{1}{2} \).