Final answer:
The radial position of the satellite at time 1 = 10 h after perigee passage is 42,356 km. The speed of the satellite at that time is 2.303 km/s. The radial component of the velocity is -1.271 km/s.
Step-by-step explanation:
To determine the radial position of the earth-orbiting satellite at time 1 = 10 h after perigee passage, we can calculate the angle swept by the satellite in that time period. Using the equation θ = (2π/P)*t, where θ is the angle, P is the period, and t is the time, we can find the angle to be 10.47 radians. To find the radial position, we can use the formula r = a(1 - e²)/(1 + e*cos(θ)), where r is the radial position, a is the semi-major axis, e is the eccentricity, and θ is the angle. Given that the perigee radius is 10,000 km and the semi-major axis is the average of the perigee and apogee radii, we can calculate the radial position to be 42,356 km.
To calculate the speed of the satellite at time 1 = 10 h after perigee passage, we can use the formula v = sqrt(mu*(2/r - 1/a)), where v is the speed, mu is the standard gravitational parameter of Earth, r is the radial position, and a is the semi-major axis. The calculated speed is 2.303 km/s.
To find the radial component of the velocity at time 1 = 10 h after perigee passage, we can use the formula vr = v*cos(θ), where vr is the radial component of velocity, v is the speed, and θ is the angle. The calculated radial component of velocity is -1.271 km/s.