The orbit of a certain a satellite has a semi-major axis of `1.5 xx 10^7` m and an eccentricity of 0.20. Its perigee (minimum distance) and apogee (maximum distance), respectively, are

Options

A. `6.0 xx 10^6 m, 9.0 xx 10^6 m`
B. `3.0 xx 10^6 m, 1.2 xx 10^7 m`
C. `1.2 xx 10^7 m, 1.8 xx 10^7 m`
D. `1.0 xx 10^7 m, 1.2 xx 10^7 m`

Answer

To find the perigee and apogee distances of a satellite given its semi-major axis and eccentricity, we can use the following formulas:

1. **Perigee (minimum distance)**:
\[ r_p = a(1 - e) \]
2. **Apogee (maximum distance)**:
\[ r_a = a(1 + e) \]

Where:
- \( r_p \) is the perigee distance,
- \( r_a \) is the apogee distance,
- \( a \) is the semi-major axis,
- \( e \) is the eccentricity.

### Given:
- Semi-major axis, \( a = 1.5 \times 10^7 \) m
- Eccentricity, \( e = 0.20 \)

### Step 1: Calculate the Perigee Distance
Using the formula for perigee:
\[ r_p = a(1 - e) \]
Substituting the values:
\[ r_p = 1.5 \times 10^7 \times (1 - 0.20) \]
\[ r_p = 1.5 \times 10^7 \times 0.80 \]
\[ r_p = 1.5 \times 0.80 \times 10^7 \]
\[ r_p = 1.2 \times 10^7 \text{ m} \]

### Step 2: Calculate the Apogee Distance
Using the formula for apogee:
\[ r_a = a(1 + e) \]
Substituting the values:
\[ r_a = 1.5 \times 10^7 \times (1 + 0.20) \]
\[ r_a = 1.5 \times 10^7 \times 1.20 \]
\[ r_a = 1.5 \times 1.20 \times 10^7 \]
\[ r_a = 1.8 \times 10^7 \text{ m} \]

### Final Answer:
The perigee (minimum distance) is \( 1.2 \times 10^7 \) m and the apogee (maximum distance) is \( 1.8 \times 10^7 \) m.

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