Chapter 11:Satellites And Satellite Communication

Example 1,Page No:567

In [1]:
import math

# Variable Declaration
h       = 150;              # height of satellite from earth in km
G       = 6.67*10**-11;      # Gravitational constant
M       = 5.98*10**24;       # mass of the earth in kg
Re      = 6370;             # radius of earth in km

# Calculations
u       = G*M
V       = math.sqrt(u/float(((Re + h)*10**3))) # orbital velocity
V1      = V/float(1000);                       # orbital velocity in km/s

# Result
print'Orbital velocity = %3.3f km/s'%V1;
Orbital velocity = 7.821 km/s

Example 2,Page No:568

In [1]:
import math

# Variable Declaration
Ap_Pe_diff  = 30000;        # difference between apogee and perigee in Km
a           = 16000;        # semi major axis of orbit

# Calculations
e           =  Ap_Pe_diff/float((2*a));    # Eccentricity

# Result
print'Eccentricity = %3.2f'%e;
Eccentricity = 0.94

Example 3,Page No:568

In [2]:
import math

# Variable Declaration
a1      = 18000;            # semi major axis of the elliptical orbits of satellite 1
a2      = 24000;            # semi major axis of the elliptical orbits of satellite 2

# Calculations
# T    = 2*math.pi*math.sqrt(a**3/float(u));
#let K = T2/float(T1);
K       = (a2/float(a1))**(3/float(2));    # Ratio of orbital periods

# Result
print'The orbital period of satellite-2 is %3.2f times the orbital period of satellite-1'%K;
The orbital period of satellite-2 is 1.54 times the orbital period of satellite-1

Example 4,Page No:569

In [6]:
import math

# Variable Declaration
h       = 35800;            # height of satellite orbit from earth in km
G       = 6.67*10**-11;      # Gravitational constant
M       = 5.98*10**24;       # mass of the earth in kg
Re      = 6364;             # radius of earth in km
i       = 2;                # inclination angle

# Calculations
u       = G*M
r       = Re+h;
Vi      = (math.sqrt(u/float(r*10**3)))*math.tan((i*math.pi)/float(180));     # magnitude of velocity impulse
V       = Vi                                # magnitude of velocity impulse in m/s

# Result
print'Magnitude of velocity impulse = %d m/s'%V;
Magnitude of velocity impulse = 107 m/s

Example 5,Page No:571

In [7]:
import math

# Variable Declaration
h       = 13622;        # ht of circular orbit from earth's surface
Re      = 6378;         # Radius of earth in km

# Calculations
R       = Re+h;         # Radius of circular orbit
pimax   = 180 - (2*math.acos(Re/float(R)))*(180/float(math.pi));   # Maximum shadow angle
eclipmax_time = (pimax/float(360))*24;             # maximum daily eclipse duration

# Result
print'maximum shadow angle = %3.1f°\n'%pimax,'Maximum daily eclipse duration = %3.2fhours'%eclipmax_time;
maximum shadow angle = 37.2°
Maximum daily eclipse duration = 2.48hours

Example 6,Page No:572

In [13]:
import math
# Given data
h       = 35786;        # ht of geo.stationary orbit above earth surface
T       = 365;          # time in days
r       = 6378          # radius of earth in km

# ie(t) = 23.4*sin(2*%pi*t/T)
# for a circular orbit of 20000 km radius ,phi = 37.4° ,Therefore, the time from first day of eclipse to equinox is given by substituting ie(t) = 37.4/2 = 18.7°
phi     = 37.4
ie      = (phi/2)*(math.pi/180)
k       = 23.4*(math.pi/180)
t       = (365/(2*math.pi))*math.asin((ie/k)) 
# for geostationary orbit
phimax  = 180 - 2*(math.acos(r/(r+h)))*(180/math.pi)
t_geo   = (365/(2*math.pi))*math.asin((8.7*math.pi/180)/k)

# Output
print'Total time from first day of eclipse to last day of eclipse = %3.1f days'%t
print '\nTotal time from first day of eclipse to last day of eclipse for geostationary orbit = %3.2f days'%t_geo
Total time from first day of eclipse to last day of eclipse = 53.8 days

Total time from first day of eclipse to last day of eclipse for geostationary orbit = 22.13 days

Example 7,Page No:600

In [10]:
import math
# Given data
m       = 100;      # mass of satellite
V       = 8000;     # orbital velocity in m/s
Re      = 6370;     # radius of earth in Km
H       = 200;      # satellite height above earth surface

# Calculations
CF      = (m*V**2)/float(((Re+H)*10**3));       # centrifugal force

# output
print'Centrifugal Force = %d Newtons'%CF;
Centrifugal Force = 974 Newtons

Example 8,Page No:601

In [11]:
import math

# Variable Declaration
Apogee  = 30000;        # Apogee pt of satellite elliptical orbit
Perige  = 1000;         # perigee pt of satellite elliptical orbit

# Calculations
a       = (Apogee + Perige)/float(2);  # semi major axis

# Result
print'Semi-major axis = %d Km'%a;
Semi-major axis = 15500 Km

Example 9,Page No:601

In [14]:
import math

# variable Declaration
farth   = 30000;        # farthest point in satellite elliptic eccentric orbit
closest = 200;          # closest point in satellite elliptic eccentric orbit
Re      = 6370;         # Radius of earth in km

# Calculations
Apogee  = farth + Re;   # Apogee in km
Perigee = closest + Re; # perigee in km
a       = (Apogee + Perigee)/float((2));       # semi-major axis
e       = (Apogee - Perigee)/float((2*a));     # orbit eccentricity

# Result
print'Apogee = %d km\n'%Apogee;
print'Perigee = %d km\n'%Perigee;
print'orbit eccentricity = %3.3f'%e;
Apogee = 36370 km

Perigee = 6570 km

orbit eccentricity = 0.694

Example 10,Page No:604

In [15]:
import math
# Given data
e       = 0.5;      # orbit eccentricity
ae      = 14000;    # from fig. the distance from center of ellipse to the centre of earth

# Calculations
a       = ae/float(e);         # semi major axis
apogee  = a*(1 + e);    # Apogee in km
perige  = a*(1 - e);    # perigee in km

# output
print'Apogee = %d km\n'%apogee,'Perigee = %d km'%perige;
Apogee = 42000 km
Perigee = 14000 km

Example 11,Page No:605

In [15]:
import math

# variable Declaration
G       = 6.67*10**-11;      # Gravitational constant
M       = 5.98*10**24;       # mass of the earth in kg
Re      = 6370*10**3;        # radius of earth in m

# Calculations
u       = G*M
Vesc    = math.sqrt(2*u/float(Re));
Ves     = Vesc/float(1000);    # escape velocity in km/s

# Result
print'Escape velocity = %3.1f km/s'%Ves;
Escape velocity = 11.2 km/s

Example 12,Page No:605

In [16]:
import math

# Variable Declaration
a       = 25000*10**3;      # semimajor axis in m from fig
G       = 6.67*10**-11;     # Gravitational constant
M       = 5.98*10**24;      # mass of the earth in kg
h = 0

# Calculations
u       = G*M;
T       = 2*math.pi*math.sqrt((a**3)/float(u));
hr          = T/float(3600);                        # conv. from sec to hrs and min
t           = (T%3600);               # conv. from sec to hrs and min
mi          = t/float(60);                          # conv. from sec to hrs and min

# Result
print'Orbital time period = %d'%hr,'Hours %d minutes'%mi;
Orbital time period = 10 Hours 55 minutes

Example 13,Page No:605

In [11]:
import math

# variable Declaration
apogee  = 35000;        # farthest point in kms
perigee = 500;          # closest point in kms
r       = float(6360);         # radius of earth in kms
G       = 6.67*10**-11   # gravitational constant
M       = 5.98*10**24;   # mass of earth in kgs
# calculations
#funcprot(0)
apogee_dist = apogee + r       # apogee distance in kms
perigee_dist= perigee+r ;      # perigee distance in kms
a           = (apogee_dist + perigee_dist)/2;   # semi-major axis of elliptical orbit
T           = (2*math.pi)*math.sqrt((a*10**3)**3/(G*M));          # orbital time period
hr          = T/float(3600)                        # conv. from sec to hrs and min
t           = T%3600               # conv. from sec to hrs and min
mi          = t/60                          # conv. from sec to hrs and min
u           = G*M
Vapogee     = math.sqrt(u*((2/(apogee_dist*10**3)) - (1/(a*10**3))));      # velocity at apogee point
Vperigee    = math.sqrt((G*M)*((2/(perigee_dist*10**3)-(1/(a*10**3)))))    # velocity at perigee point

# Result
print'Orbital Time Period = %d Hrs'%hr,'%d min' %mi 
print'Velocity at apogee = %3.3f Km/s' %(Vapogee/1000)
print 'Velocity at perigee = %3.3f Km/s'%(Vperigee/1000)
print'Note: Calculation mistake in textbook in finding velocity at apogee point'
Orbital Time Period = 10 Hrs 20 min
Velocity at apogee = 1.656 Km/s
Velocity at perigee = 9.987 Km/s
Note: Calculation mistake in textbook in finding velocity at apogee point

Example 14,Page No:606

In [17]:
import math

# Variable Declaration
ra_S_rp     = 50000;        # sum of apogee and perigee distance
ra_D_rp     = 30000;        # difference of apogee and perigee distances

# Calculations
e           = ra_D_rp/float(ra_S_rp);  # eccentricity

# Result
print'Target eccentricity = %3.1f'%e;
Target eccentricity = 0.6

Example 15,Page No:607

In [26]:
import math

# Given data
a       = 20000;            # semi major axis of elliptical sate. orbit in kms
b       = 16000;            # semi minor axis of elliptical sate. orbit in kms

# calculations
# a = (ra + rp)/2
# b = math.sqrt(ra*rp)
# let k = (ra + rp)
# let l = ra*rp
k       = 2*a;              # ra+ rp ----------------1
l       = b**2;              # ra*rp -----------------2
# ra**2 -40000ra + 256000000



Ap = (k - (math.sqrt(k**2-4*1*l)))/(2*1)
Pe = (k + (math.sqrt(k**2-4*1*l)))/(2*1)





print'Apogee distance = %d km\n'%Ap,'Perigee distance = %d km' %Pe;
Apogee distance = 8000 km
Perigee distance = 32000 km

Example 16,Page No:607

In [15]:
import math

# Variable delaration
H       = 35800;        # height of orbit in kms
re      = 6364;         # radius of earth in kms
i       = 2;            # angle of inclination in degrees

# Calculations
r       = H+re;         # radius of orbit in kms
lamdamax = i;           # max latitude deviation
long_dev = (i**2)/float(228);   # max. longitude deviation
disp_lamda = (r*i*math.pi/180) # max disp in km due to lamdamax
max_disp1  = disp_lamda*(long_dev/lamdamax) # max disp.due to max.longitude deviation

# Result
print'Maximum deviation in latitude = %d°'%lamdamax
print'Maximum deviation in longitude = %3.4f°'%long_dev
print'Maximum displacements due to latitude displacement = %d Km'%disp_lamda
print'Maximum displacements due to longitude displacement = %3.1f Km\n'%max_disp1 ;
    
Maximum deviation in latitude = 2°
Maximum deviation in longitude = 0.0175°
Maximum displacements due to latitude displacement = 1471 Km
Maximum displacements due to longitude displacement = 12.9 Km

Example 17,Page No:608

In [17]:
import math

# Variable Declaration
r       = 42164;        # orbital radius in kms
Dlamda_max = 500;       # max displacement due to latitude deviation

# Calculations
i   = Dlamda_max/float(r);     # angle of inclination in radians
i_deg = i*180/math.pi;      # rad to deg conv

# Result
print'Angle of inclination = %3.2f°'%i_deg;
Angle of inclination = 0.68°

Example 18,Page No:609

In [22]:
import math

# Variable Declaration
H   = 35786;        # ht of orbit from earth surface
Re  = float(6378)          # radius of earth in kms

# Calculations
# For theoretical max coverage angle,elevation angle E = 0 
E = 0
# max coverage angle = 2amax
# 2amax = 2asin(Re/(Re+H)cosE)
amax = 2*math.asin((Re/(Re+H))*math.cos(E))
amax_deg = amax*180/math.pi     # rad to deg conversion
D   = math.sqrt( Re**2 + (Re+H)**2 - 2*Re*(Re + H)*math.asin(E + math.asin((Re/(Re+H))*math.cos(E)))) # Max slant range

# Result
print'Maximum Coverage angle = %3.1f°'%amax_deg
print'Maximum slant Range = %d Km'%D;
Maximum Coverage angle = 17.4°
Maximum slant Range = 41671 Km