Final answer:
The eccentricity of a satellite's orbit with a perigee of 300 km and apogee of 3000 km is approximately 0.8182. When the satellite crosses the y-axis, its height above Earth's surface is approximately 948.54 km.
Step-by-step explanation:
The question is asking for the eccentricity of an Earth satellite's orbit given its perigee and apogee and then to determine the satellite's height when it crosses the y-axis. The perigee is the closest point a satellite gets to Earth during its orbit, while the apogee is the farthest point.
The semi-major axis (a) of an ellipse is the average of the perigee and apogee distances from the center of the Earth. Thus, a = \((300 km + 3000 km) / 2\) = 1650 km. The eccentricity (e) of an orbit is calculated as \(e = (apogee - perigee) / (2a)\), so substituting the values gives \(e = (3000 km - 300 km) / (2 \times 1650 km) = 2700 km / 3300 km = 0.8182\).
When the satellite crosses the y-axis, it is at its semi-minor axis (b). The semi-minor axis can be found from the relationship \(b = a \times \sqrt{1 - e^2}\). So, \(b = 1650 km \times \sqrt{1 - 0.8182^2} = 1650 km \times \sqrt{1 - 0.6696} = 1650 km \times \sqrt{0.3304} \approx 1650 km \times 0.5748 = 948.54 km\).
Since the satellite's height when it crosses the y-axis corresponds to the semi-minor axis, the satellite's height above Earth's surface is approximately 948.54 km.