Satellite Communications and Orbits in Engineering
University of Diyala
College of Engineering
Department of Communications
Engineering
Satellite Communications
By:
Dr. Majidah Hameed
Majeed
Lecture # 2
1.5 Spe
c
ial
Types
of
Inclined Orbit
s
.
–
A
sa
t
ellite
orbiting
in
a
plane
that
coincides
with
the
equatorial
plane
of
the
e
a
rth
is
in
an EQ
U
ATORIAL
O
R
B
I
T.
A
satellite
orbiting
in
an inclined
orbit with
an ang
l
e
of
inclination
of
90
degre
e
s
or
near
90
d
e
gre
e
s
is
in
a
POLAR
OR
B
IT.
As
shown
in
fig 1.8
fig
1.8
1.6
The
Elli
ptic
Orbit
The
only
orbit type we
will
discuss
here
is
the
elliptic
orb
i
t
as
s
hown in
fig
(
1.
9)
.
All
sa
t
ellites
that
orbit
around
the
E
a
r
t
h
are
in
a
elliptic
orbit.
A
parabolic
and
hyperbolic
orbit
are
no
n
-periodic,
and
hence
represent
esca
p
e
orbits,
that
is,
the
satellite
in
these
orbits
leaves
t
he
Earth.
The
para
b
olic
orbit
is
the
m
ini
m
u
m
energy
esc
a
p
e
orbit.
fig
1.9
In the
figure
we
can
identify
the following
ite
m
s:
Occupied
focus
loc
a
tion
of the
Earth
or
c
e
ntral
attracting body
a
=
s
e
m
i
-
ma
jor
axis
e
=
orbit
e
c
centricity
(
0<e<1
for
elliptic orbits,
e
=
0
is
circular
orbit)
Note
that
r is
measured
fro
m
the
center
of
the
E
a
rt
h
,
hence
we
c
an
note
that:
where
h =
height
abo
v
e
Earth’s
surfa
c
e,
R
e
=
earth radi
u
s
Often
t
i
me
s
the
pe
r
igee
and
apogee
d
i
stances
are
given
in
te
r
m
s
of
the
height
above
the
E
a
rth’s
surface.
This
is
not
corre
c
t,
and
the
radi
u
s
of
the
E
a
rth
m
ust
b
e
added
to
the
height
to get
the
perigee
and
apogee
radii.
The
polar
equation
for
the
elliptic
or
bit,
with
th
e
origin
at
one
focus
is
given
by:
…………….(1)
...
…
……………….(2)
………………..(3)
Example
The
inte
r
national
spa
c
e
stat
i
on
is
in
a
“372
x
381
km
orbit”,
what
is
the
ecc
e
ntricity
of
th
e
orbit?
Solution:
The
two
n
u
m
bers
re
f
er
to
the
peri
g
ee
he
i
ght
(372
k
m
)
and
the
apogee
height
(381
k
m
).
However,
these
are
the
heights
above
the
E
a
rth’s
surfa
c
e
and
m
ust
be
converted
to perigee
and
apogee
radii
by
a
dding
the E
a
rth’s
radius.
Hence
the
internati
o
n
a
l
space
station
is
in a
near
c
irc
u
lar orbit.
In
or
d
e
r
to
fin
d
the
a
pog
e
e
a
n
d
p
e
rig
e
e
h
e
i
gh
t
.
T
h
e
l
e
ng
th
of
th
e
r
a
d
iu
s
v
ec
tors
at
a
pog
ee
a
n
d
per
i
g
ee
c
an
be
obt
a
in
e
d
from
th
e
ge
o
m
e
t
ry
of
th
e
el
l
ip
s
e:
from equation
(2)
,
W
h
ere
r
a
:
ra
d
iu
s
v
e
c
t
or at
a
po
gee
p
e
rig
ee
r
p
:
rad
i
us
v
e
c
t
or
at
e
:
E
cc
e
ntr
i
c
it
y
T
h
e
n
w
e
can
calc
ul
a
t
e
h
a
a
n
d
h
p
a
s
fo
l
l
o
w
:
1.7
C
i
rcular
Orb
i
ts
We
sta
t
ed
previously
that
a
circular
orbit
is
a
special
type
of
elliptical
orbit.
It
should
realize
a
circular
orbit
is
one
in
which
the
ma
jor
and
m
inor
ax
i
s
distances
are
equal
or
approxi
ma
tely
equal.
Mean
height
above
e
a
rth,
i
nstead
of
peri
g
ee
and
apoge
e
,
is
used
in
descri
b
ing
a
circular
orbit.
A
sa
t
ellite
in
a
circu
l
ar
orbit
at
a
height
of
approxi
m
ately
36000
Km
above
the
earth
is
in
a
synchronous
orbit.
At
this
altitude
the
peri
o
d
of
rotation
of
the
satellite
is
24
hours,
the
sa
m
e
as
t
he
rotation
p
e
rio
d
of
the
e
a
rth.
In
other
wor
ds,
the
orbit
of
the
satellite
is
in
sync
w
i
th
the
rotational
m
otion
of
the
e
a
r
th.
Although
inclined
and
polar
synchro
n
ous
orbits
are
possible,
the
te
r
m
synchronous
usually
refers
to
a
synchronous
equatorial
orbit.
In
this
type
of
orbit,
satellites
appear
to
hover
m
o
t
ionlessly
in
the
sky.
Fi
g
ure
1-8
shows
how
one
of
t
hese
satellites
can
p
r
ovide
coverage
to
al
m
ost
half
the
surface
of
the
e
a
r
t
h.
Fi
g
ure
1
-8
- Illu
m
ination
from
a
synchronous
satellite.
Three
of
t
hese sa
t
ellites
c
a
n
p
r
ovide
coverage
over
m
ost of
the
earth
(except for
the
extre
m
e
n
orth
and
south
polar
regi
ons).
A
polar
projection
of
the
global
covera
g
e
of
a
thre
e
-
satellite
syst
e
m
is
shown in
figure
1-9.
Fi
g
ure
1-
9
. -
Worldwide
sy
n
chronous
satellite
system
viewed
from
a
bove
the
North
Pol
e
.
1.8
Ke
pl
e
r's
La
w
:
Johann
Kepler
deter
m
in
ed
three
laws
chara
c
terizing
orbital
moti
o
n,
These
laws
c
a
n
be
proven
ma
thematically
using
Ne
w
ton's
law
of
gravitation.
Kepler's
laws
are
paraphrased
below
along
with
the
corresponding
p
h
ysical
i
m
plications.
These
laws
apply
dir
e
ctly
to
satellite
orbit
a
l
m
otion, thus
the
laws are
f
r
om the
point
of
view
of
an
Eart
h
-orbiting
satellite.
Ke
pl
e
r's
First
L
a
w:
Satellite
orbits
are
e
lliptical
Paths with
the
E
a
rth
at
one
focus of
the
ell
i
pse
Ke
pl
e
r's
Second
L
a
w
:
A
line
be
t
ween
the
center
of
the
Earth
and
the
sate
l
lite
sweeps
out
equal
areas
in
equal
inte
r
vals
of
time.
Ke
pl
e
r's
Third
L
a
w
:
The
square
of
the
orbital
period
is
proportional
to
the
cube
of
the
the
orbit's
semi
major
axis.
9.
Elev
atio
n
angle –
T
he
angle
f
rom
th
e
hor
i
z
o
n
tal
to
the
point
on
the
center
of
the
main
beam
of
the
antenna
w
h
en
the
antenna
is
poin
t
ed
dire
c
tly
at the
satellite
Cove
r
age
angle
-
the
measure
of
the
portion
of
the
e
arth's
surface
visible
to
the
satellite
•
Reaso
n
s
affecting
minimum
elevation
angle
of
earth
st
a
tion’s
antenna
o
(>0
)
–
Buildin
gs,
tre
e
s,
and
other
terrestrial objects
bl
o
ck
the
line
of
sight
–
Atmosphe
r
ic
attenuati
o
n
is
g
r
eater
at
lo
w
elevation
angles
–
Electrical
noise
generated
b
y
the earth's
heat
near
its
surface
adversely
affects
recep
t
ion
Fi
g
ure
1-
1
0.
Elevati
o
n
angle
Or
bit
He
i
ght
h
Cove
r
age
Angle β
Elevation
Angle
θ
𝑹
𝑹+ 𝒉
=
𝐜𝐨𝐬(𝜷+ 𝜽)
𝐜𝐨𝐬(𝜽)
Ex
:
A
sa
t
ellite
at
a
distance
of
10000
k
m
from
a
point
on
the
earths
surfa
c
e
,
t
h
e
Cove
r
age
Angle
β
=
4
draw
and
fin
d
the
Elevation
Angle
θ.
R=
6378
*
��
𝟑
km
Fc
=
(
m
v
2
)/r
1.10 The
Forces
Acting
on Satellite
When
a
satellite
is
launched,
it
is
pla
c
ed
i
n
orbit
around
the
Earth.
The
Earth's
gravity
holds
the
satellite
in
a
c
e
rtain
path
as
it
goes
around
the
E
a
r
th,
and
that
path
is
called
an
"or
bit."
There
are
t
wo
ma
in
forces
a
c
t
i
ng
on
satellite
:
a
c
e
ntrifugal
force
due
t
o
the
kinetic
energy
of
the
satellite,
wh
i
ch
atte
m
pts
to
fling
the
satellite
into
a
higher
orbit.
A
n
d
a
c
e
ntripetal
force
due
to
gravitational
attraction
of
the
planet
about
which
satelli
t
e
is
orbiting,
which
atte
m
pts
to
pull
the
sa
t
ellite
down
forward
the
planet.
If
these
two
forces
a
re
equal,
the
satellite
will
re
m
ain
i
n
stable
orbit.
The
fo
r
m
ula
f
or
centrifugal
for
c
e
is:
42,300k
m
.
Since the radi
u
s
of
t
he
Earth
is
6
3
78 km the
heig
ht
of the
geostationary
The
fo
r
m
ula
for
the
gravitational
force
be
t
ween
two
bodies
of
m
ass
M a
n
d
m
is
Fg=
(GM
m
)
/
r
2
Where
M
is
the
mass
of the
Eart
h,
m
is
t
he
ma
ss of
th
e
satell
i
te
G
(Newton's
Gravitational
Constant,
r is
the
orbital
radius
v
the
speed
of
satellite.
For
the
satellite
stable
in or
bit ,
Fc=
Fg,
then
(m
v
2
/r)
=
(GMm)/r
2
v
2
/r
=
(GM
)
/
r
2
No
w
,
v =
2πr
/
T
then
(2πr/
T
)^2
/r
= GM
/r^2
=>
(4
π
2
r)/
T
2
=
(
GM)/r
2
=>
r
3
=
(
GMT
2
)/4
π
2
We
know
that
T
is o
ne
day,
since
this is
the
period
of
the
E
a
r
t
h.
This
is
8.64
x
1
0
4
seconds. We
also know
that
M
is th
e
mass
of the
Eart
h,
which
is
6
x
1
0
24
kg.
L
a
stly,
we
know
that
G
(N
ewton's
Gravitational
Constant)
is
11
m
3
/kg.
s
2
So we
c
a
n
work
out
r.
r
3
=
7.57
x 1
0
22
Therefore,
r
=
4.23 x
1
0
7
=
42,3
0
0
k
m
.
6.67
x 1
0
-
So the
orbital
radius required
for a
geostationary,
or geosynchronous
orbi
t is
orbit
above
the
Eart
h
'
s
surface
is
~36
0
00
k
m
.
Fig
1.11
Gravitati
o
nal
force
i
s
inversely
pr
o
portional
to
the
square
of
th
e
distance
between
th
e
c
e
nters
of
gravity of
the
sat
e
llite
and
the
p
lanet
the
satellite is
orbiting,
in
this
case
the
e
a
rth.
The
gravitational
for
c
e
inward
(
F
I
N
,
the centripetal
force =
Fc)
is
directed
toward the
c
e
nter
of
g
ravity
of
the
earth.
The
kinetic
energy of
th
e
satel
l
ite
(
F
OU
T
,
the
centrifugal
fo
rc
e
=
F
g) is
dir
e
cted
opposite
to the
gravitational
force.
Kinet
i
c
energy
is
p
r
oportional
to
th
e
square
of
th
e
velocity
of the
satell
i
te.
When
t
hese
inwa
r
d
and outward
forces
a
re
b
alanc
e
d,
the
satellite
moves
a
round
the
e
a
rth
in
a
“free
fall” trajec
tory:
the
satellite’s
or
bit.
Fig
1.12 The
forc
e
s
actin
g
on
satellite
1.
1
1
Or
b
i
tal
V
e
lo
c
i
ty
T
h
e
v
e
lo
c
it
y
r
e
q
ui
re
d
t
o
m
a
i
n
t
a
i
n
a
s
a
t
e
llit
e
a
t
th
e
o
rbit
r
a
d
iu
s
r
ca
n
b
e
ca
l
c
ul
a
t
e
d
as
fo
llo
w
:
-
D
e
p
e
nd
i
n
g
o
n
th
e
l
a
w
s
of
m
oti
o
n
f
i
rst
d
e
v
e
lo
pe
d
b
y
K
e
pl
er
a
n
d
N
e
w
t
o
n
.
T
h
e
c
o
m
p
e
t
in
g
f
o
rces a
c
t
o
n
th
e
s
a
t
e
ll
i
t
e
;
gr
a
v
it
y
t
e
n
d
s
to
pull
th
e
s
a
t
e
ll
i
t
e
in
to
w
a
rd
s
th
e
ea
rth,
w
h
i
l
e
i
t
s
orb
i
t
a
l
v
e
l
o
c
it
y
t
e
n
d
s to
pu
l
l
th
e
s
a
t
e
llit
e
a
w
a
y
fro
m
t
h
e
ear
t
h,
a
s
in
1.13
T
h
e
g
r
a
v
i
t
a
t
i
o
n
a
l
f
o
r
c
e
,
F
in,
a
n
d
t
h
e
a
ng
u
l
a
r
vel
o
c
it
y
fo
r
c
e
,
F
out,
can
b
e
r
e
pr
e
s
en
t
e
d
a
s:
For
the
satellite
stable
in or
bit ,
F
c(
Fin
)
=
Fg
(Fo),
then
(m
v
2
/
=(GMm)/
r
2
v
2
/r
=
(GM
)
/
r
2
Where
v
: satellite
velocity
for
circular
or
bit
in k
m
/sec
F
i
g
ure
1.1
3
F
or
c
e
s
a
c
ti
n
g
o
n
a
s
a
t
e
l
l
i
te
N
ot
e
th
a
t
for
th
e
d
is
cus
s
i
o
n
a
bov
e
a
l
l
o
th
e
r
forces
a
c
tin
g
o
n
th
e
s
a
t
e
llit
e,
s
u
c
h
a
s
t
he
gr
a
v
i
t
y
forc
e
s
from
th
e
m
oon,
sun,
a
n
d
o
th
er
b
odi
e
s,
i
s
n
e
g
l
ec
t
e
d.
Example
H
u
m
an
-
ma
de
satell
i
tes
typic
a
l
ly
orbit
at
heights
of
400
m
iles
from
the
surfa
c
e
of
the
E
a
rth
(about
640
kilo
m
eter
s,
or
6.4
×
1
0
5
me
ters).
What’s
the
spe
e
d
of
such
a
satellite?
All
you have
to
do
is
p
ut
in the
nu
m
bers:
This
c
onverts to
about 16,800
m
i
l
es
per hour.
1.
1
2
Launching
Satellites
into
Or
bi
t
.
Placing
a
satellite
into
geosynchronous orbit
requires
an enor
m
ous
a
m
ount
of
energy.
T
h
e
launch
process
can
be
divid
e
d
into
two
phases:
the
launch
phase
and
the
orbit
inje
c
tion
phase.
The
Launch
Phase
During the launch
phase, the
launch
vehicle
places
t
he
satellite
into
the
transfer
orbit--an
eliptic
a
l
orbit
that
has
at its farth
est
point
from
earth
(apoge
e
)
the
geosynchronous
e
levation
of
22,238
m
iles
a
nd
a
t its
nearest
point
(perigee)
an
elevation of
usually
not
less
than
100
m
i
l
es
a
s
sho
w
n
below
in
Fi
g
ure
1
-14.
The
Orbit
Inje
c
tio
n
Phase
The
energy
required
to
m
ove
the
sa
t
ellite
from
the
elliptical
transfer
or
b
i
t
into
the
geosynchronous
orbi
t is
s
u
ppli
e
d
by
the
satellite’s
a
pogee
kick
m
otor (AKM).
This
is
known
as
the
or
bit
injection
p
hase.
Figure 1
-
1
4:
The
Elliptical
Transfer
Orbit
A
geostation
a
ry
orbit
is
an
orbit
in
which a
sp
a
cecraft
c
a
n
appear
to
hover
over
a
fix
ed
point on E
a
rth.
That
is
particularly
useful
for
com
m
unication or
observation
satellites.
This
is
a
little
different
than
a
geosynchronous orbi
t
,
w
hich
is
one
in
which
a
s
p
ace
c
r
aft
will
pass
over
the
same
point,
once
p
er
day.
So,
if
we
want
o
ur spacecraft
t
o
appe
a
r
t
o
hover
ov
e
r
a
fixed
point
on
E
a
rth,
two
criteria
m
ust
be
met. The
first is
that
the
orbit
m
ust
be
equatorial
(inclination
of
0
degre
e
s)
and
the
second that
the
a
ngular velocity
must
ma
tch th
a
t
of the
surface of
the
E
a
rth.
That
turns
o
ut
to be
a
round
35,900 km (2
2
,300
m
iles
)
.
That
gives
us
two
c
h
allenges
to solve.
T
h
e
first is
that
we
need to
get
to
that
high
altitude
and
the
second
is
that
w
e
likely need
to
c
h
ange
our
inclination
.
Fi
g
ure
1-
15
Dependi
n
g
on
the
mission
and
launch vehicle,
one
of
two
things
happe
n
s
first.
The
rocket
often
takes
t
he
space
c
r
aft
to
a
parking
orbit
at
an
altitude
of
18
0
-200
km
,
this
c
a
lled
the
parking
orbit
,
but
it
c
a
n
som
e
t
i
m
es
skip
the
parking
o
r
bit
and
go
straight
to
what
is
called
a
g
eost
a
tionary
transfer
orbit.
Parking
orbi
t
s
are
often
used
to
test
vehicle
syste
m
s
b
efore
com
m
i
ttin
g
to
f
urther
action.
Fi
g
ure
1-
16
The
Geostationary
Trans
f
er
Orbit (GTO)
is
a
hig
h
ly
e
lliptical
orbit
with
a
peri
g
ee
of
180-200
km
above
the
Eart
h
'
s
surface
and
an
ap
o
gee
of
ar
o
und
35,9
0
0
k
m
.
T
h
e
spa
c
ecraft
has
an
en
g
ine
that
is
c
a
lled
the
Apogee
K
i
ck
Motor
(AKM).
T
hat
m
otor
is
fired
at
apogee
to circularize the
or
bit.
But
it
usually
is
not
all
done
at
onc
e
.
Usually the
AKM
is
used
three
t
i
me
s.
Figure 1
-
17
Each
of these
bu
r
ns has
two
o
b
jectives:
1)
increase
the
peri
g
ee
of the
or
b
i
t
and
2)
decrease
the
inclination of
th
e
orbit.
1.
1
3
Side
re
al
T
i
m
e
Vs
Sol
a
r
T
i
m
e
S
i
der
e
al
T
i
m
e
:
i
s
a
t
i
m
e
s
c
a
l
e
th
a
t
i
s
b
a
s
e
d
o
n
th
e
E
a
rth
'
s
r
a
t
e
of
rot
a
t
io
n
m
ea
su
red
re
l
a
tiv
e
to
th
e
fix
e
d
s
t
a
r
s
.
S
id
ere
a
l
day
=
2
3
h
5
6
m
04
.0
9
0
5
3
s
S
ol
a
r
T
i
m
e
:
i
s
a
rec
k
o
ni
n
g
of
th
e
pa
ss
a
ge
of
ti
m
e
b
a
s
e
d
on
th
e
S
u
n
'
s
p
osi
ti
o
n
in
th
e
s
k
y
.
S
ol
ar
day
=
2
4
h
S
a
t
e
l
l
i
te
o
rb
i
ts
c
o
o
r
d
i
n
a
tes
ar
e
s
p
e
c
i
f
i
ed
i
n
si
d
e
r
e
al
t
i
m
e
r
a
th
e
r
t
h
a
n
i
n
s
ol
a
r
t
i
m
e
.
S
o
l
a
r
t
i
m
e
,
w
hi
c
h
f
or
m
s
t
he
b
a
s
is
of
a
l
l
g
l
o
b
a
l
t
i
m
e
s
t
an
d
ard
s
.
The
e
a
rth
rotates
once per
sid
ere
a
l
d
ay
of
The
e
a
rth
rotates
once per
sid
ere
a
l
d
ay
of
23 h
56
m
i
n
4.09s.
Use
23 h
56
m
in 4.09s.
1.14
Footprin
t
:
The
area
on
E
a
r
th
that
the
satellite
c
a
n
'
see'
(
or
rea
c
h
w
ith
its
antennas)
is
c
a
lled
the
satellite
'
fo
otprint
'
.
A
satellite
'
s
f
ootprint
refers
to
the
area over
which
the
satellite
operat
e
s: th
e
intersection
of
a
sat
e
llite
anten
n
a
trans
m
ission
pattern
and the
surface
of
the
E
a
r
th
Fi
g
ure
1-
18
Exploring the intricacies of satellite communications and orbits in engineering, focusing on special types of inclined orbits, circular orbits, elliptic orbits, and the calculation of eccentricity. Examples such as the International Space Station's orbit are used to illustrate concepts. The content delves into key formulas, diagram interpretations, and practical applications within the field of communications engineering.
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University of Diyala College of Engineering Department of Communications Engineering Satellite Communications By: Dr. Majidah Hameed Majeed
1.5 Special Types of Inclined Orbits. A satellite orbiting in a plane that coincides with the equatorial plane of the earth is in an EQUATORIAL ORBIT. A satellite orbiting in an inclined orbit with an angle of inclination of 90 degrees or near 90 degrees is in a POLAR ORBIT. As shown in fig 1.8 fig 1.8 1.6 The Elliptic Orbit The only orbit type we will discuss here is the elliptic orbit as shown in fig ( 1.9) . All satellites that orbit around the Earth are in a elliptic orbit. A parabolic and hyperbolic orbit are non-periodic, and hence represent escape orbits, that is, the satellite in these orbits leaves the Earth. The parabolic orbit is the minimum energy escape orbit.
fig 1.9 In the figure we can identify the followingitems: Occupied focus location of the Earth or central attracting body a = semi-major axis r = radius from center of Earth to satellite rp = Earth perigee distance ra = Earth apogee distance e = orbit eccentricity (0<e<1 for elliptic orbits, e = 0 is circular orbit) Note that r is measured from the center of the Earth, hence we can note that: where h = height above Earth s surface, Re = earth radius Often times the perigee and apogee distances are given in terms of the height above the Earth s surface. This is not correct, and the radius of the Earth must be added to the height to get the perigee and apogee radii. The polar equation for the elliptic orbit, with the origin at one focus is given by:
.(1) ... .(2) ..(3) Example The international space station is in a 372 x 381 km orbit , what is the eccentricity of the orbit? Solution: The two numbers refer to the perigee height (372 km) and the apogee height (381 km). However, these are the heights above the Earth s surface and must be converted to perigee and apogee radii by adding the Earth s radius. Hence the international space station is in a near circular orbit. In order to find the apogee and perigee height. The length of the radius vectors at apogee and perigee can be obtained from the geometry of the ellipse:
from equation (2) , Where ra: radius vector at apogee rp: radius vector at perigee e : Eccentricity Then we can calculate haand hpas follow: 1.7 Circular Orbits We stated previously that a circular orbit is a special type of elliptical orbit. It should realize a circular orbit is one in which the major and minor axis distances are equal or approximately equal. Mean height above earth, instead of perigee and apogee, is used in describing a circular orbit. A satellite in a circular orbit at a height of approximately 36000 Km above the earth is in a synchronous orbit. At this altitude the period of rotation of the satellite is 24 hours, the same as the rotation period of the earth. In other words, the orbit of the satellite is in sync with the rotational motion of the earth. Although inclined and polar synchronous orbits are possible, the term synchronous usually refers to a synchronous equatorial orbit. In this type of orbit, satellites appear to hover motionlessly in the sky. Figure 1-8 shows how one of these satellites can provide coverage to almost half the surface of the earth. Figure 1-8 - Illumination from a synchronous satellite.
Three of these satellites can provide coverage over most of the earth (except for the extreme north and south polar regions). A polar projection of the global coverage of a three-satellite system is shown in figure 1-9. Figure 1-9. - Worldwide synchronous satellite system viewed from above the North Pole. 1.8 Kepler's Law: Johann Kepler determined three laws characterizing orbital motion, These laws can be proven mathematically using Newton's law of gravitation. Kepler's laws are paraphrased below along with the corresponding physical implications. These laws apply directly to satellite orbital motion, thus the laws are from the point of view of an Earth-orbiting satellite. Kepler's First Law: Satellite orbits are elliptical Paths with the Earth at one focus of the ellipse Kepler's Second Law: A line between the center of the Earth and the satellite sweeps out equal areas in equal intervals of time.
Kepler's Third Law: The square of the orbital period is proportional to the cube of the the orbit's semi major axis. 9. Elevation angle The angle from the horizontal to the point on the center of the main beam of the antenna when the antenna is pointed directly at the satellite Coverage angle - the measure of the portion of the earth's surface visible to the satellite Reasons affecting minimum elevation angle of earth station s antenna o (>0 ) Buildings, trees, and other terrestrial objects block the line of sight Atmospheric attenuation is greater at low elevation angles Electrical noise generated by the earth's heat near its surface adversely affects reception
Figure 1-10. Elevation angle Orbit Height h CoverageAngle ElevationAngle ?+ ?=???(?+ ?) ? ???(?) Ex : A satellite at a distance of 10000 km from a point on the earths surface , the Coverage Angle = 4 draw and find the Elevation Angle . R= 6378 * ?km 1.10 The ForcesActing on Satellite When a satellite is launched, it is placed in orbit around the Earth. The Earth's gravity holds the satellite in a certain path as it goes around the Earth, and that path is called an "orbit." There are two main forces acting on satellite : a centrifugal force due to the kinetic energy of the satellite, which attempts to fling the satellite into a higher orbit. And a centripetal force due to gravitational attraction of the planet about which satellite is orbiting, which attempts to pull the satellite down forward the planet. If these two forces are equal, the satellite will remain in stable orbit. The formula for centrifugal force is: Fc = (mv2)/r
The formula for the gravitational force between two bodies of mass M and m Fg= (GMm)/r2 is Where M is the mass of the Earth, m is the mass of the satellite G (Newton's Gravitational Constant, r is the orbital radius v the speed of satellite. For the satellite stable in orbit , Fc= Fg, then (mv2/r) = (GMm)/r2 v2/r = (GM)/r2 Now, v = 2 r /T then (2 r/T)^2 /r = GM /r^2 => (4 2 r)/T2 = (GM)/r2 => r3 = (GMT2)/4 2 We know that T is one day, since this is the period of the Earth. This is 8.64 x 104 seconds. We also know that M is the mass of the Earth, which is 6 x 1024 kg. Lastly, we know that G (Newton's Gravitational Constant) is 6.67 x 10- 11 m3/kg.s2 So we can work out r. r3 = 7.57 x 1022 Therefore, r = 4.23 x 107 = 42,300 km. So the orbital radius required for a geostationary, or geosynchronous orbit is 42,300km. Since the radius of the Earth is 6378 km the height of the geostationary
orbit above the Earth's surface is ~36000 km. Fig 1.11 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 = Fc) is directed toward the center of gravity of the earth. The kinetic energy of the satellite (FOUT, the centrifugal force= Fg) is directed 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.
Fig 1.12 The forces acting on satellite 1.11 Orbital Velocity The velocity required to maintain a satellite at the orbit radius r can be calculated as follow:- Depending on the laws of motion first developed by Kepler and Newton. The competing forces act on the satellite; gravity tends to pull the satellite in towards the earth, while its orbital velocity tends to pull the satellite away from the earth, as in 1.13 The gravitational force, Fin, and the angular velocity force, Fout, can be represented as: For the satellite stable in orbit , Fc( Fin ) = Fg (Fo), then (mv2/ =(GMm)/r2 v2/r = (GM)/r2 Where v : satellite velocity for circular orbit in km/sec
Figure 1.13 Forces acting on a satellite Note that for the discussion above all other forces acting on the satellite, such as the gravity forces from the moon, sun, and other bodies, is neglected. Example Human-made satellites typically orbit at heights of 400 miles from the surface of the Earth (about 640 kilometers, or 6.4 105meters). What s the speed of such a satellite?All you have to do is put in the numbers: This converts to about 16,800 miles per hour. 1.12 Launching Satellites into Orbit. Placing a satellite into geosynchronous orbit requires an enormous amount of energy. The launch process can be divided into two phases: the launch phase and the orbit injection phase. The Launch Phase During the launch phase, the launch vehicle places the satellite into the transfer orbit--an eliptical orbit that has at its farthest point from earth (apogee) the
geosynchronous elevation of 22,238 miles and at its nearest point (perigee) an elevation of usually not less than 100 miles as shown below in Figure 1-14. The Orbit Injection Phase The energy required to move the satellite from the elliptical transfer orbit into the geosynchronous orbit is supplied by the satellite s apogee kick motor (AKM). This is known as the orbit injection phase. Figure 1-14: The EllipticalTransfer Orbit Ageostationary orbit is an orbit in which a spacecraft can appear to hover over a fixed point on Earth. That is particularly useful for communication or observation satellites. This is a little different than a geosynchronous orbit, which is one in which a spacecraft will pass over the same point, once per day. So, if we want our spacecraft to appear to hover over a fixed point on Earth, two criteria must be met. The first is that the orbit must be equatorial (inclination of 0 degrees) and the second that the angular velocity must match that of the surface of the Earth. That turns out to be around 35,900 km (22,300 miles). That gives us two challenges to solve. The first is that we need to get to that high
altitude and the second is that we likely need to change our inclination . Figure 1-15 Depending on the mission and launch vehicle, one of two things happens first. The rocket often takes the spacecraft to a parking orbit at an altitude of 180-200 km , this called the parking orbit , but it can sometimes skip the parking orbit and go straight to what is called a geostationary transfer orbit. Parking orbits are often used to test vehicle systems before committing to further action. Figure 1-16
The GeostationaryTransfer Orbit (GTO) is a highly elliptical orbit with a perigee of 180-200 km above the Earth's surface and an apogee of around 35,900 km. The spacecraft has an engine that is called theApogee Kick Motor (AKM). That motor is fired at apogee to circularize the orbit. But it usually is not all done at once. Usually theAKM is used three times. Figure 1-17 Each of these burns has two objectives: 1) increase the perigee of the orbit and 2) decrease the inclination of the orbit. 1.13 SiderealTime Vs SolarTime Sidereal Time: is a time scale that is based on the Earth's rate of rotation measured relative to the fixed stars. Sidereal day = 23h 56 m 04.09053 s Solar Time: is a reckoning of the passage of time based on the Sun's position in the sky. Solar day = 24 h Satellite orbits coordinates are specified in sidereal time rather than in solar time. Solartime, which forms the basis of all global time standards.
The earth rotates once per sidereal day of The earth rotates once per sidereal day of 23 h 56 min 4.09s. Use 23 h 56 min 4.09s. 1.14 Footprint: The area on Earth that the satellite can 'see' (or reach with its antennas) is called the satellite 'footprint'.Asatellite's footprint refers to the area over which the satellite operates: the intersection of a satellite antenna transmission pattern and the surface of the Earth Figure 1-18
