chapter 2: Satellite Orbits and Trajectories

Example 2.1, page no-36

In [16]:
import math

#Variable Declaration
r1=6370.0        # Earth's Orbit in km
r2=630.0         # Height of satellite from surface in km
G=6.67*10**-11   # Gravitational constant inNm^2/kg^2
M=5.98*10**24    # Mass of earth in kg

#Calculation
R=r1+r2
v=math.sqrt(G*M/(R*10**3))

#Result
print("The velocity of sattelite %.2fkm/s"%(math.floor(v/10)*10**-2))

The velocity of sattelite 7.54km/s

Example 2.2, page no-37

In [17]:
import math

#Variable Declaration
A=45000.0    #Apogee in km
P=7000.0     #Perigee in km


#Calculation
#(a)
a=(A+P)/2
#(b)
e=(A-P)/(2*a)
#(c)
e=(math.floor(e*100))/100
d=a*e

#Result
print("(a)\nSemi-major axis of elliptical orbit is %d km"%a)
print("\n(b)\nEccentricity = %.2f"%e)
print("\n(c)\nThe distance between centre of earth and centre of ellipse is %d km "%d)

(a)
Semi-major axis of elliptical orbit is 26000 km

(b)
Eccentricity = 0.73

(c)
The distance between centre of earth and centre of ellipse is 18980 km

Example 2.3, page no-37

In [18]:
#Variable Declaration
ma=42000.0        # Major axis distance in Km
P=8000.0          # Perigee distance in Km


#Calculation
A=ma-P
e=(A-P)/ma

#Result
print("Apogee=%dkm\n Eccentricity=%.2f"%(A,e))

Apogee=34000km
 Eccentricity=0.62

Example 2.4, page no-37

In [19]:
#Variable Declaration
e=0.6                #Eccentricity
d=18000.0            #distance between earth's centre and centre of ellipse


#Calculation
a=d/e
A=a*(1+e)
P=a*(1-e)


#Result
print("Semi-major axis of elliptical orbit is %d km\n Apogee distance=%dkm\n Perigee distance=%dkm"%(a,A,P))

Semi-major axis of elliptical orbit is 30000 km
 Apogee distance=48000km
 Perigee distance=12000km

Example 2.5, page no-38

In [20]:
#Variable Declaration
AP_diff=30000.0   #difference between apogee and perigee in km
AP_sum=62800.0    #Apogee+perigee


#Calculation
E=AP_diff/AP_sum


#Result
print("Orbit Eccentricity= %.3f"%E)

Orbit Eccentricity= 0.478

Example 2.6, page no-38

In [21]:
import math
#Variable Declaration
R=7000.0*10**3       # sattelite orbit in m
mu=39.8*10**13       # constant G*M in Nm^2/kg
A=47000.0*10**3      # appogee distance in m
P=7000.0*10**3       # perigee distance in m


#Calculation
v=math.sqrt(mu/R)
a=(A+P)/2
v1=math.sqrt(mu*((2/R)-(1/a)))


#Result
print("Velocity of satellite A at point X is v=%.2fkm/s\nVelocity of satellite B at point X is V=%.3fkm/s"%(v/1000,v1/1000))
#value in book is different at 3rd decimal place.

Velocity of satellite A at point X is v=7.54km/s
Velocity of satellite B at point X is V=9.949km/s

Example 2.7, page no-39

In [22]:
import math

#Variable Declaration
R=42000.0*10**3       #sattelite orbit in m
mu=39.8*10**13        #constant G*M in Nm^2/kg
A=42000.0*10**3       #appogee distance in m
P=7000.0*10**3        #perigee distance in m

#Calculation
v=math.sqrt(mu/R)
a=(A+P)/2
v1=math.sqrt(mu*((2/R)-(1/a)))

#Result
print("Velocity of satellite A at point X is v=%.3fkm/s\n Velocity of satellite B at point X is V=%.3fkm/s"%(v/1000,v1/1000))

Velocity of satellite A at point X is v=3.078km/s
 Velocity of satellite B at point X is V=1.645km/s

Example 2.8, page no-40

In [23]:
import math
#Variable Declaration
R=25000.0*10**3       #sattelite orbit in m
mu=39.8*10**13        #constant G*M in Nm^2/kg
A=43000.0*10**3       #appogee distance in m
P=7000.0*10**3        #perigee distance in m

#Calculation
v=math.sqrt(mu/R)
a=(A+P)/2
v1=math.sqrt(mu*((2/R)-(1/a)))

#Result
print("Velocity of satellite A at point X is v=%.3fkm/s\n Velocity of satellite B at point X is V=%.3fkm/s"%(v/1000,v1/1000))
#value in book is different at 3rd decimal place.

Velocity of satellite A at point X is v=3.990km/s
 Velocity of satellite B at point X is V=3.990km/s

Example 2.9, page no-40

In [24]:
import math
#Variable Declaration
a=(50000.0/2)*10**3           #Semi-major axis in m
mu=39.8*10**13                #constant G*M in Nm^2/kg


#Calculation
T=2*math.pi*math.sqrt((a**3)/mu) #math.pi gives variation in answer
h=T/(60*60)
x=T%3600
m=x/60
s=x%60

#Result
print("Orbital time period is given by, T = %dsec\n\t\t\t\t    = %dh %dm %ds"%(T,math.floor(h),math.floor(m),math.floor(s)))
#value in book is different for seconds.

Orbital time period is given by, T = 39368sec
				    = 10h 56m 8s

Example 2.10, page no-42

In [25]:
#Variable Declaration
a1=18000.0*10**3           #Semi-major axis for first satellite in m
a2=24000.0*10**3           #Semi-major axis f0r 2nd satellite in m

#Calculation
T2_by_T1=(a2/a1)**(3.0/2.0)

#Result
print("Orbital time period of sattelite 2 is %.2f times that of sattelite 1"%T2_by_T1)

Orbital time period of sattelite 2 is 1.54 times that of sattelite 1

Example 2.11, page no-42

In [26]:
import math
#Variable Declaration
a=25000.0*10**3       #appogee distance in m
b=18330.0*10**3       #perigee distance in m


#Calculation
e=(math.sqrt(a**2-b**2)/a)


#Result
print("Apogee distance  = a(1+e)= %dkm\n Perigee distance = a(1-e)= %dkm\n"%(a*(1+e)/1000,math.ceil(a*(1-e)/1000)))

Apogee distance  = a(1+e)= 42000km
 Perigee distance = a(1-e)= 8000km

Example 2.12, page no-43

In [27]:
#Variable Declaration
e=0.6                  # eccentricity of elliptical orbit
a=0.97                 # area of shaded region
b=2.17                 # Area of non-shaded region
t=3                    # time taken by satellite to move from pt B to A


#Calculation
x=b/a
y=x*t

#Result
print("Time taken by satellite to move from A to B is %.3f hours "%y)

Time taken by satellite to move from A to B is 6.711 hours

Example 2.13, page no-44

In [28]:
#Variable Declaration
A=42000.0            # Apogee in km
P=8000.0             # Perigee in km
v_p=9.142            # velocity at perigee point


#Calculation
v_a=v_p*P/A


#Result
print("Velocity at apogee = %.3f km/s"%v_a)

Velocity at apogee = 1.741 km/s

Example 2.14, page no-44

In [29]:
import math

#Variable Declaration
theta=56.245      #angle made by direction of satellite with local horizontal
d=16000.0         #distance of particular point
P=8000.0          #Perigee in m
v_p=9.142         #velocity at perigee point


#Calculation
v=(P*v_p)/(d*math.floor(math.cos(theta*math.pi/180)*1000)/1000)


#Result
print("The velocity of satellite at that particular point is %.3f km/s"%v)

The velocity of satellite at that particular point is 8.236 km/s

Example 2.16, page no-49

In [30]:
import math
#Variable Declaration
A1=12000.0                 # first Apogee distance
P=8000.0                   # Perigee distance
v1=1.0                     # assume v1 as 1
v2=1.2*v1                  # 20% higher than v1 


#Calculation
x=(v2/v1)**2
k=(((1+(P/A1))/x)-1)
k=math.floor(k*10**4)/10**4
A2=P/k


#Result
print("A2 = %.0fkm"%math.ceil(A2))

A2 = 50826km

Example 2.17, page no-50

In [31]:
import math
#Variable Declaration
vp=8.0         # horizontal velocity of satellite in km/s
r=1620.0       # distance from earth's surface in km
R=6380.0       # Earth's radius in km
d=10000.0      # distance of point at which velocity to be calculated
theta=30.0     # angle made by satellite with local horzon at that point


#Calculation
P=r+R
v=(vp*P)/(d*math.cos(theta*math.pi/180))

#Result
print("v = %.2f km/s"%v)

v = 7.39 km/s

Example 2.18, page no-50

In [32]:
import math

#Variable Declaration
r=620.0         # distance from earth's surface in km
vp=8.0          # horizontal velocity of satelliteat 9000km height in km/s
R=6380.0        # Earth's radius in km
d=9000.0        # distance of point at which velocity to be calculated
theta=30.0      # angle made by satellite with local horzon at that point
mu=39.8*10**13  # Nm**2/kg


#Calculation
P=r+R
m=vp*d*math.cos(theta*math.pi/180)/P    #m=sqrt((2mu/P)-[2mu/(A+P)])
m=(m*10**3)**2
x=(2*mu/(P*10**3))-m          #x=[2mu/(A+P)]
x=math.floor(x/10**4)*10**4
k=(2*mu)/x               #k=A+P
k=math.ceil(k/10**4)*10**4
A=k-(P*10**3)

#Result
print("A = %.0f km"%(A/1000))

A = 16170 km

Example 2.19, page no-58

In [33]:
import math
#variable declaration
R=6380                  #Earth's radius in km
T=86160                 #Orbital period of Geostationary satellite in km
mu=39.8*10**13          #in Nm^2/k

#calculations

r=(T*math.sqrt(mu)/(2*math.pi))**(2.0/3.0) # Answer matches to the answer given in the book if value of pi is taken as 3.14 

#Result
print('Radius of satellite is, r = %.0f km'%(r/1000))
print('Therefore, height of satellite orbit above earth surface is %.0f km '%((r/1000)-R))

Radius of satellite is, r = 42142 km
Therefore, height of satellite orbit above earth surface is 35762 km

Example 2.20, page no-59

In [34]:
import math
#variable declaration

R=6380                  #radius of earth in km
P=400                   #Perigee distance in km
A=40000                 #Apogee distance in km
mu=39.8*10**13          #in Nm^2/k

#calculation

a=(A+P+R+R)/2           #semi-major axis of the elliptical orbit

T=(2*math.pi*(a*10**3)**(3.0/2.0))/math.sqrt(mu)

h=T/(60*60)
x=T%3600
m=x/60
s=x%60

#Result
print('T = %dsec\n  = %dh %dm %ds\n\nThis approximately equal to 12 hour'%(T,math.floor(h),math.floor(m),math.floor(s)))

T = 43158sec
  = 11h 59m 18s

This approximately equal to 12 hour