Chapter- 2(Lecture-2) Satellite Communication Engineering.

Discussion

A satellite is in a 322 km high circular orbit. Determine:

a. The orbital angular velocity in radians per second;

b. The orbital period in minutes; and

c. The orbital velocity in meters per second.

Note: assume the average radius of the earth is 6,378.137 km and Kepler’s constant has

the value 3.986004418x10^5 km^3/s^2.

Solution

It is actually easier to answer the three parts of this question backwards, beginning with

the orbital velocity, then calculating the period, and hence the orbital angular velocity.

First we will find the total radius of the orbit r = 322 + 6,378.137 km = 6700.137 km

(c) From eqn. (2.5), the orbital velocity v = (m/r)^1/2 = (3.986004418x10^5 / 6700.137)^1/2 =7.713066 km/s = 7,713.066 m/s.

(b) From eqn. (2.6), T = (2pir^3/2)/(m^1/2) = (2pi6,700.137^3/2)/( 3.986004418x10^5)^1/2 =

(3,445,921.604)/(631.3481146) = 5,458.037372 seconds = 90.9672895 minutes = 90.97

minutes.

(a) The orbital period from above is 5,458.037372 seconds. One revolution of the earth

covers 360^o or 2pi radians. Hence 2pi radians are covered in 5,458.037372 seconds,

giving the orbital angular velocity as 2pi/5,458.037372 radians/s = 0.0011512 radians/s.

An alternative calculation procedure would calculate the distance traveled in one orbit

(2pir = 2pi6700.137 = 42,098.20236 km). This distance is equivalent to 2pi radians and so 1 km is equivalent to 2pi/42,098.20236 radians = 0.0001493 radians. From above, the

orbital velocity was 7.713066 km/s = 7.713066x0.0001493 radians/s = 0.0011512

radians/s.

Summary: After completion of this lecture you will be able to calculate Orbital Velocity in meter per second, Orbital Period in minutes and Orbital Angular Velocity in radian per second Mathematically.