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Consider a spacecraft in an elliptical orbit around the earth. At the low point, or perigee, of its orbit, it is 400 km above the earth’s surface; at the high point, or apogee, it is 4000 km above the earth’s surface. What is the period of the spacecraft’s orbit?
  • a)
    7510 s
  • b)
    7310 s
  • c)
    6910 s
  • d)
    7910 s
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
Consider a spacecraft in an elliptical orbit around the earth. At the ...
Given parameters:
- Perigee altitude: 400 km
- Apogee altitude: 4000 km

We need to find the period of the spacecraft's orbit.

1. Finding the semi-major axis of the orbit:
- The semi-major axis is the average distance of the spacecraft from the center of the earth.
- It can be calculated using the formula: semi-major axis = (perigee altitude + apogee altitude) / 2
- Substituting the values, we get: semi-major axis = (400 + 4000) / 2 = 2200 km

2. Finding the eccentricity of the orbit:
- The eccentricity of the orbit measures how elongated the orbit is.
- It can be calculated using the formula: eccentricity = (apogee altitude - perigee altitude) / (apogee altitude + perigee altitude)
- Substituting the values, we get: eccentricity = (4000 - 400) / (4000 + 400) = 0.9

3. Finding the period of the orbit:
- The period of the orbit is the time taken by the spacecraft to complete one orbit around the earth.
- It can be calculated using the formula: period = 2π * √(semi-major axis³ / GM), where GM is the product of earth's mass and gravitational constant.
- Substituting the values, we get: period = 2π * √((2200 km)³ / (3.986 × 10⁵ km³/s²)) = 7910 seconds.

Therefore, the correct answer is option 'D' (7910 seconds).
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