1. Net425:Satellite Communications
Networks and Communication Department
Tutorial 1

2. Instructor information
Name: Maram AlMuhareb
Office #:
Website :
Networks and Communication Department

3. Grades grade Due week Evaluation
3% الاربعاء التسليم الساعه 12 الليل دروب بوكس
Week 7
Assignment
5% الاثنين الساعه 12 الظهر المكتب
Week 11
Mini research
project
5% المحاضره الجايه
Week 13
LAB FINAL quiz 2%
Attendence , participation
15%
TOTAL
Networks and Communication Department

4. Revision - Newton’s Law - Satellite Motion - Orbital Period 6-Dec-18
Networks and Communication Department

5. Figure 2.1 (p. 18) Forces acting on a satellite in a stable orbit around the earth (from Fig. 3.4 of reference 1). Gravitational force is inversely proportional to the square of the distance between the centers of gravity of the satellite and the planet the satellite is orbiting, in this case the earth. The gravitational force inward (FIN, the centripetal force) is directed toward the center of gravity of the earth. The kinetic energy of the satellite (FOUT, the centrifugal force) is directed diametrically opposite to the gravitational force. Kinetic energy is proportional to the square of the velocity of the satellite. When these inward and outward forces are balanced, the satellite moves around the earth in a “free fall” trajectory: the satellite’s orbit. For a description of the units, please see the text.

6. Centripetal force acting on the satellite,
Fin = m x ( µ / r2 )
= m x (GME / r2 )
Centrifugal force acting on satellite,
Fout = m x a
= m x ( v2 / r )
Where a = centrifugal acceleration and
a = v2 / r

7. If the forces on satellite are balanced:
Fin = Fout
m x (µ / r2) = m x (v2 / r)
Therefore velocity of the satellite in circular orbit, v:
v = (µ / r)1/2 or (GME / r)1/2

8. Orbital Period Orbital Period, T T = (2Лr)/v = (2Лr) x (µ / r)-1/2

9. Problem
6-Dec-18
Networks and Communication Department

10. Exercises
A satellite is in a circular orbit around the earth. The altitude of the satellite’s orbit above the surface of the earth is 1400 km.
Givens: Re= km, kepler’s const. = *10^5 km^3/S^2.
What are the centripetal and Centrifugal accelerators acting on the satellite in its orbit?
What is the velocity of the satellite in this orbit?
What is the orbital period of the satellite in this orbit? (hours, minutes, and seconds)
6-Dec-18
Networks and Communication Department

11. Answers
A-r=re+h= = km 𝑎 𝑖𝑛 = 𝜇 𝑟 2 = ∗ 10 5 ( ) 2 = km/ 𝑆 2 𝑎 𝑖𝑛 = 𝑎 𝑜𝑢𝑡 Note: so sine the satellite was in stable orbit the centrifugal acceleration must be equal to centripetal acceleration (we need only calculate one of them) B- 𝑣= 𝜇 𝑟 = ∗ = km /s
6-Dec-18
Networks and Communication Department

12. Answers
C- T= 2Л(r)3/2/ µ1/2
= 2𝜋 ( ) ( ∗ 10 5 ) = = S
= 1hours 53 min Sec
6-Dec-18
Networks and Communication Department

13. How to Convert from seconds to h , m , s ?
Take the time in seconds S divide it by 3600 : / 3600 =
The real part will be the hours which is 1 hour
Multiply the rational part by 60 to get the minutes * 60 =
The real part will be the minutes which is 53 min
Multiply the rational part by 60 to get the seconds * 60 =46.9 seconds