Question

An Earth satellite (μ=398,600 km 3/sec2) is in an elliptical orbit with the perigee at 600 km altitude and an eccentricity of e = 0.866. Determine the radius at apogee, radius at perigee, and the eccentricity of the new satellite trajectory, if a rocket propulsion system is fired in the direction of flight giving an incremental velocity of 200 m/sec when (a) fired at apogee, (b) fired at perigee, and (c) fired at perigee, but in the opposite direction, reducing the velocity.

Answer 1

The radius at apogee is 8870 km, the radius at perigee is 6870 km, and the eccentricity of the new satellite trajectory is approximately -0.127.

Step-by-step explanation:

The radius at apogee can be found by adding the altitude of apogee (2500 km) to the radius of the Earth (6370 km). Therefore, the radius at apogee is 8870 km.

The radius at perigee can be found by adding the altitude of perigee (500 km) to the radius of the Earth (6370 km). Therefore, the radius at perigee is 6870 km.

The eccentricity of the new satellite trajectory can be found using the formula:

e = (Rp - Ra) / (Rp + Ra)

Where e is the eccentricity, Rp is the radius at perigee, and Ra is the radius at apogee.

Substituting the values we have:

e = (6870 km - 8870 km) / (6870 km + 8870 km)

e = -2000 km / 15740 km

e ≈ -0.127

Therefore, the eccentricity of the new satellite trajectory is approximately -0.127.