A planet revolves around the sun in an elliptical orbit. If `v_(p)` and `v_(a)` are the velocities of the planet at the perigee and apogee respectively, then the eccentricity of the elliptical orbit is given by :

Options

A. `(v_(p))/(v_(a))`
B. `(v_(a)-v_(p))/(v_(a)+v_(p))`
C. `(v_(p)+v_(a))/(v_(p)-v_(a))`
D. `(v_(p)-v_(a))/(v_(p)+v_(a))`

Explanation

To find the eccentricity of the elliptical orbit of a planet revolving around the Sun, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Orbit**:
- The planet moves in an elliptical orbit around the Sun, with the closest point being the perigee (P) and the farthest point being the apogee (A).
- Let \( v_p \) be the velocity at perigee and \( v_a \) be the velocity at apogee.

2. **Identifying Distances**:
- Let \( R_P \) be the distance from the Sun to the planet at perigee and \( R_A \) be the distance from the Sun to the planet at apogee.
- In terms of semi-major axis \( a \) and the distance from the center to the focus \( c \) (where the Sun is located), we have:
- \( R_P = a - c \)
- \( R_A = a + c \)

3. **Conservation of Angular Momentum**:
- The angular momentum \( L \) of the planet is conserved, so we can write:
\[ m R_P v_P = m R_A v_A \]
- Here, \( m \) is the mass of the planet, which cancels out:
\[ R_P v_P = R_A v_A \]

4. **Expressing Velocities**:
- Rearranging the above equation gives:
\[ \frac{v_P}{v_A} = \frac{R_A}{R_P} \]

5. **Substituting Distances**:
- Substitute the expressions for \( R_A \) and \( R_P \):
\[ \frac{v_P}{v_A} = \frac{a + c}{a - c} \]

6. **Relating Eccentricity**:
- The eccentricity \( e \) of the orbit is defined as:
\[ e = \frac{c}{a} \]
- We can express \( c \) in terms of \( a \) and \( e \):
\[ c = ae \]

7. **Substituting for \( c \)**:
- Substitute \( c = ae \) into the velocity ratio:
\[ \frac{v_P}{v_A} = \frac{a + ae}{a - ae} \]
- Simplifying gives:
\[ \frac{v_P}{v_A} = \frac{a(1 + e)}{a(1 - e)} = \frac{1 + e}{1 - e} \]

8. **Cross Multiplying**:
- Cross-multiplying gives:
\[ v_P - e v_P = v_A + e v_A \]
- Rearranging leads to:
\[ v_P - v_A = e(v_P + v_A) \]

9. **Solving for Eccentricity**:
- Finally, we can isolate \( e \):
\[ e = \frac{v_P - v_A}{v_P + v_A} \]

### Final Result:
The eccentricity \( e \) of the elliptical orbit is given by:
\[ e = \frac{v_P - v_A}{v_P + v_A} \]