Question

A satellite is revolving around the Earth in a closed orbit. The height of the satellite above Earth's surface at perigee and apogee are 2500 km and 4500 km, respectively. Consider the radius of the Earth to be 6500 km. The eccentricity of the satellite's orbit is ________ (Round off to 1 decimal place).

Answer 1

Correct Answer: 0.1

The satellite's orbit has perigee and apogee distances of 9000 km and 11000 km from the center of the Earth, respectively, when adding Earth's radius (6500 km) to their heights above the surface. The semi-major axis (a) can be computed as the average of these distances:
\[a=\frac{9000\text{ km}+11000\text{ km}}{2}=10000\text{ km}\]
The eccentricity (e) of the orbit is defined as:
\[e=\frac{\text{apogee distance}-\text{perigee distance}}{\text{apogee distance}+\text{perigee distance}}\]
Substituting the distance values:
\[e=\frac{11000-9000}{11000+9000}=\frac{2000}{20000}=0.1\]