A comet moves in an elliptical orbit with an eccentricity of e = 0.20 around a star. The distance between the perihelion and the aphelion is `1.0 xx 10^(8)` km. If the speed of the comet at perihelion is 81 km/s, then speed of the comet at the aphelion is :

Options

A. 182 km/s
B. 36 km /s
C. 121.5 km/s
D. 54 km/s

Explanation

To solve the problem, we need to find the speed of the comet at aphelion given its speed at perihelion and the distance between the perihelion and aphelion.

### Step 1: Understand the terms
- **Perihelion (r₁)**: The point in the orbit where the comet is closest to the star.
- **Aphelion (r₂)**: The point in the orbit where the comet is farthest from the star.
- The distance between perihelion and aphelion is given as \(1.0 \times 10^8\) km.

### Step 2: Relate perihelion and aphelion distances
The distance between the perihelion and aphelion can be expressed as:
\[ d = r_2 - r_1 \]
where \(d\) is the distance between perihelion and aphelion.

### Step 3: Express perihelion and aphelion in terms of semi-major axis (a) and eccentricity (e)
For an elliptical orbit:
- The perihelion distance \(r_1 = a(1 - e)\)
- The aphelion distance \(r_2 = a(1 + e)\)

Thus, the distance between perihelion and aphelion can be expressed as:
\[ d = r_2 - r_1 = a(1 + e) - a(1 - e) = 2ae \]

### Step 4: Solve for semi-major axis (a)
Given that \(d = 1.0 \times 10^8\) km and \(e = 0.20\):
\[ 1.0 \times 10^8 = 2a(0.20) \]
\[ 1.0 \times 10^8 = 0.4a \]
\[ a = \frac{1.0 \times 10^8}{0.4} = 2.5 \times 10^8 \text{ km} \]

### Step 5: Calculate perihelion and aphelion distances
Now we can find \(r_1\) and \(r_2\):
\[ r_1 = a(1 - e) = 2.5 \times 10^8 (1 - 0.20) = 2.5 \times 10^8 \times 0.80 = 2.0 \times 10^8 \text{ km} \]
\[ r_2 = a(1 + e) = 2.5 \times 10^8 (1 + 0.20) = 2.5 \times 10^8 \times 1.20 = 3.0 \times 10^8 \text{ km} \]

### Step 6: Use conservation of angular momentum
The angular momentum \(L\) at perihelion and aphelion is conserved:
\[ L = m \cdot r_1 \cdot v_1 = m \cdot r_2 \cdot v_2 \]
where \(v_1\) is the speed at perihelion and \(v_2\) is the speed at aphelion.

Since mass \(m\) cancels out:
\[ r_1 \cdot v_1 = r_2 \cdot v_2 \]

### Step 7: Solve for speed at aphelion
Substituting the known values:
\[ (2.0 \times 10^8) \cdot (81) = (3.0 \times 10^8) \cdot v_2 \]
\[ 162 \times 10^8 = 3.0 \times 10^8 \cdot v_2 \]
\[ v_2 = \frac{162 \times 10^8}{3.0 \times 10^8} = 54 \text{ km/s} \]

### Final Answer
The speed of the comet at aphelion is **54 km/s**.