BENY NURMANDA (1101120215)
TT-36-G2
CHAPTER 2 – SATELLITE COMMUNICATIONS (DENNIS RODDY)
2.9. The exact conversion factor between feet and meters is 1 ft = 0.3048 m. A satellite travels in an
unperturbed circular orbit of semi major axis a = 27,000 km. Determine its tangential speed in (a)
km/s, (b) ft/s, and (c) mi/h.
Answer :
a. 𝑣 = √
𝐺𝑀
𝑅
6.673𝑥10−11 𝑥5.97𝑥1024
27000𝑥103
= √
b.
3841.1𝑥103
0.3048
= 12602034.121 𝑓𝑡/𝑠
c.
3841.1𝑥103
0.000621
= 2386.748 𝑚𝑖/𝑠
= 3841.1 𝑘𝑚/𝑠
2.10. Explain what is meant by apogee height and perigee height. The Cosmos 1675 satellite has an
apogee height of 39,342 km and a perigee height of 613 km. Determine the semi major axis and the
eccentricity of its orbit. Assume a mean earth radius of 6371 km.
Answer :
Apogee : The point farthest from earth
Perigee : The point of closest approach to earth
COSMOS 1675
Apogee = 39342 km
Perigee = 613 km
Semi major axis =
ℎ𝑎 = 𝑟𝑎 − 𝑅
39342 − 6371 = 𝑟𝑎
𝑟𝑎 = 6410.342 𝑘𝑚
ℎ𝑝 = 𝑟𝑝 − 𝑅
𝑟𝑝 = 613 + 6371
𝑟𝑝 = 6984 𝑘𝑚
𝑟𝑎 = 𝑎(1 + 𝑒)
𝑟𝑝 = 𝑎(1 − 𝑒)
𝑎=
TUGAS-2 SISTEM KOMUNIKASI SATELIT
𝑟𝑝 + 𝑟𝑎
= 6697.171 𝑘𝑚
2
BENY NURMANDA (1101120215)
TT-36-G2
𝑟𝑎
𝑒 = ( ) − 1 = −0.043
𝑎
2.11. The Aussat 1 satellite in geostationary orbit has an apogee height of 35,795 km and a perigee
height of 35,779 km. Assuming a value of 6378 km for the earth’s equatorial radius, determine the
semimajor axis and the eccentricity of the satellite’s orbit.
ℎ𝑎 = 35795 𝑘𝑚
ℎ𝑝 = 35779 𝑘𝑚
𝑅 = 6378 𝑘𝑚
ℎ𝑎 = 𝑟𝑎 − 𝑅
𝑟𝑎 = 35795 + 6378
𝑟𝑎 = 6413.795 𝑘𝑚
ℎ𝑝 = 𝑟𝑝 − 𝑅
𝑟𝑝 = 35779 + 6378
𝑟𝑎 = 6413.779 𝑘𝑚
𝑟𝑝 + 𝑟𝑎
= 6413.787 𝑘𝑚
2
𝑟𝑎
𝑒 = − 1 = 1.25𝑥10−6
𝑎
𝑎=
2.17. A satellite in polar orbit has a perigee height of 600 km and an apogee height of 1200 km.
Calculate (a) the mean motion, (b) the rate of regression of the nodes, and (c) the rate of rotation of
the line of apsides. The mean radius of the earth may be assumed equal to 6371 km.
a. Mean motion =
𝑅𝑎 = 1200 + 6371 = 7571 𝑘𝑚
𝑅𝑝 = 600 + 6371 = 6971 𝑘𝑚
𝑆𝑀𝑎 =
𝑅𝑎 +𝑅𝑝
2
=
7571+6971
2
= 7271 𝑘𝑚 - Semi-major Axis
𝐺𝑀 6.673𝑥10−11 𝑥5.97𝑥1024
√
𝑚𝑒𝑎𝑛 𝑚𝑜𝑡𝑖𝑜𝑛 = 𝑛 = 𝑀𝑚 = √
= 7402.026
𝑆𝑀𝑎
7271𝑥103
b. The rate of regression of the nodes = 0 because of polar orbit
c. The rate of rotation of the line of apsides = -3.158 deg/day
2.23. The Molnya 3-(25) satellite has the following parameters specified: perigee height 462 km;
apogee height 40,850 km; period 736 min; inclination 62.8°. Using an average value of 6371 km for
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BENY NURMANDA (1101120215)
TT-36-G2
the earth’s radius, calculate (a) the semimajor axis and (b) the eccentricity. (c) Calculate the nominal
mean motion n0. (d) Calculate the mean motion. (e) Using the calculated value for a, calculate the
anomalistic period and compare with the specified value. Calculate (f ) the rate of regression of the
nodes, and (g) the rate of rotation of the line of apsides.
Answer :
a. Semi-major Axis
ℎ𝑎 = 𝑟𝑎 − 𝑅
40850 − 6371 = 𝑟𝑎
𝑟𝑎 = 34479 𝑘𝑚
ℎ𝑝 = 𝑟𝑝 − 𝑅
𝑟𝑝 = 613 + 6371
𝑟𝑝 = 6984 𝑘𝑚
𝑟𝑎 = 𝑎(1 + 𝑒)
𝑟𝑝 = 𝑎(1 − 𝑒)
𝑎=
𝑟𝑝 + 𝑟𝑎
= 20731.5 𝑘𝑚
2
b. Eccentricity
𝑒=
𝑟𝑎
− 1 = 0.6631
𝑎
c. Nominal mean motion
𝜇
3.986005𝑥1014
𝑛0 = √ 3 = √
= 2.115𝑥10−4
𝑎
(20731.5𝑥103 )3
d. Mean motion
𝐺𝑀 6.673𝑥10−11 𝑥5.97𝑥1024
√
𝑚𝑒𝑎𝑛 𝑚𝑜𝑡𝑖𝑜𝑛 = 𝑛 = 𝑀𝑚 = √
= 4383.614
𝑆𝑀𝑎
20731.5𝑥103
e. Anomalistic Period
𝑃𝐴 =
2𝜋
2𝑥3.14
=
= 1.432𝑥10−3 𝑠
𝑛
4383.614
2.30. A “no name” satellite has the following parameters specified: perigee height 197 km; apogee
height 340 km; period 88.2 min; inclination 64.6°. Using an average value of 6371 km for the earth’s
radius, calculate (a) the semimajor axis and (b) the eccentricity. (c) Calculate the nominal mean
motion n0. (d) Calculate the mean motion. (e) Using the calculated value for a, calculate the
TUGAS-2 SISTEM KOMUNIKASI SATELIT
BENY NURMANDA (1101120215)
TT-36-G2
anomalistic period and compare with the specified value. Calculate (f ) the rate of regression of the
nodes, and (g) the rate of rotation of the line of apsides.
a. Semi-major Axis
ℎ𝑎 = 𝑟𝑎 − 𝑅
340 + 6371 = 𝑟𝑎
𝑟𝑎 = 6711 𝑘𝑚
ℎ𝑝 = 𝑟𝑝 − 𝑅
𝑟𝑝 = 197 + 6371
𝑟𝑝 = 6568 𝑘𝑚
𝑟𝑎 = 𝑎(1 + 𝑒)
𝑟𝑝 = 𝑎(1 − 𝑒)
𝑎=
𝑟𝑝 + 𝑟𝑎
= 6639.5 𝑘𝑚
2
b. Eccentricity
𝑒=
𝑟𝑎
− 1 = 0.0107
𝑎
c. Nominal mean motion
𝜇
3.986005𝑥1014
𝑛0 = √ 3 = √
= 570418.0413
𝑎
(0.0107𝑥103 )3
d. Mean motion
𝐺𝑀 6.673𝑥10−11 𝑥5.97𝑥1024
√
𝑚𝑒𝑎𝑛 𝑚𝑜𝑡𝑖𝑜𝑛 = 𝑛 = 𝑀𝑚 = √
= 774.604
𝑆𝑀𝑎
6639.5𝑥103
e. Anomalistic Period
𝑃𝐴 =
2𝜋
2𝑥3.14
=
= 8.107𝑥10−3 𝑠
𝑛
774.604
2.36. A satellite is in an exactly polar orbit with apogee height 7000 km and perigee height 600 km.
Assuming a spherical earth of uniform mass and radius 6371 km, calculate (a) the semimajor axis, (b)
the eccentricity, and (c) the orbital period. (d) At a certain time the satellite is observed ascending
directly overhead from an earth station on latitude 49°N. Give that the argument of perigee is 295°
Calculate the true anomaly at the time of observation.
a. Semi-major Axis
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BENY NURMANDA (1101120215)
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ℎ𝑎 = 𝑟𝑎 − 𝑅
7000 + 6371 = 𝑟𝑎
𝑟𝑎 = 13371 𝑘𝑚
ℎ𝑝 = 𝑟𝑝 − 𝑅
𝑟𝑝 = 600 + 6371
𝑟𝑝 = 6971 𝑘𝑚
𝑟𝑎 = 𝑎(1 + 𝑒)
𝑟𝑝 = 𝑎(1 − 𝑒)
𝑎=
𝑟𝑝 + 𝑟𝑎
= 10171 𝑘𝑚
2
b. Eccentricity
𝑒=
𝑟𝑎
− 1 = 0.314
𝑎
c. Nominal mean motion
𝜇
3.986005𝑥1014
𝑛0 = √ 3 = √
= 6.154𝑥10−4
𝑎
(10171𝑥103 )3
d. Mean motion
𝐺𝑀 6.673𝑥10−11 𝑥5.97𝑥1024
√
𝑚𝑒𝑎𝑛 𝑚𝑜𝑡𝑖𝑜𝑛 = 𝑛 = 𝑀𝑚 = √
= 6258.43
𝑆𝑀𝑎
10171𝑥103
e. Anomalistic Period
𝑃𝐴 =
2𝜋
2𝑥3.14
=
= 1.0034𝑥10−3 𝑠
𝑛
6258.43
TUGAS-2 SISTEM KOMUNIKASI SATELIT