University Physics I Project - Geostationary Orbits

Sharr Boshnjakaj

The Objective:

Calculate whether two points on Earth are visible by a satellite in a geostationary orbit and how long the signal takes to reach the satellite from both points.

Introduction:

Geostationary orbits are defined to be satellites that orbit the Earth at the same speed as the rotation of the Earth which results in the satellite "hanging" over the same point. This is mainly used for communcation and meteorological purposes.

Theory:

If three communications satellites are evenly placed on the geostationary orbit, global communications can be achieved excluding the north and south poles. When the satellite receives a weak radio signal from a ground station, it will automatically turn it into a high-power signal, and then send it to the other ground station, or to another communication satellite, and then send it to the earth. Satellites or man-made satellites orbiting the earth in this orbit are always located at the same position on the surface of the earth. Its orbital eccentricity and orbital inclination are both zero. The movement period is 23 hours, 56 minutes and 04 seconds, which coincides with the rotation period of the earth, and the orbital radius is 42164.169km.

Since the trajectory of the sub-satellite point of a satellite moving in a geostationary orbit is a point, observers on the ground can always observe the satellite at the same position in the sky at any time and will find that the satellite is stationary in the sky, so many artificial Satellites, especially communication satellites, mostly use geostationary orbit. Geostationary orbit service technology is of great significance in promoting the development of economy, national defense and space technology. The artificial satellite is relatively stationary with the ground and is fixed above the equator. The direct circular orbit is 35,786 kilometers above the ground, and the speed of the satellite is 3.07 kilometers per second. A satellite can cover about 40% of the earth’s area. Meteorological satellites, communication satellites, and broadcasting satellites often use this orbit. There are around 400 geostationary satellites in space orbitting Earth right now.

Statement of the Problem:

Attempting to send a signal via a station on Earth to a geostationary satellite. The first equation shown below calculates the distance between the center of the planet and the satellite. The following equation computes the "boundaries" with which the satellite can and cannot connect to. The great circle distances takes two coordinates ($\lambda$ and $\phi$) and calculates the distance between them with trigonometric functions. Depending on the location (coordinates) of the station on Earth, there will be a time delay when the satellite receives the signal. The time delay may be caused by poor connection, thw location of the station itself, among other realistic reasons. If the coordinates reach outside the connection boundaries of the satellite, the delay increases. In this case, we increased the delay by 0.01 second although it may vary in real-time.

Time delay is determined by calculating the great circle distance and taking the square root of it and finally dividing it by the speed of light. It is clear when the coordinates reach outside the connection boundaries of the satellite because the time delay increases so much. This is visible on plot graphs below, where Mercury, Venus, Earth, and Mars are plotted to show the differences. The planets have different masses and different sidereal rotational periods which are taken into account when calculating time delays for the signal of a geostationary satellite. The faster and lighter a planet is, the closer the satellite is which results in a significantly shorter time delay. This is seen in planets such as Earth and Mars. Although Mercury has a lighter mass, it also rotates much slower which is why the time delay is larger. Venus is both heavier and slower than the previous three planets. Because the proportions of Venus are so different, it must be graphed separately to properly provide a clear representation of time delay. This is all also very stringent on the fact that the location must be found on the prime meridian, meaning the longitude is zero.

Equations:

$m\omega^2r = \frac{GMm}{r^2}$

$arccos(\frac{r}{r+h})$

$d\sigma = arccos(sin(\phi1)sin(\phi2)+cos(\phi1)cos(\phi2)cos(\delta\lambda))$

Analysis:

import equations     # Importing a text file with defined functions to calculate what is necessary
import numpy as np   # Importing a python library that will help calculate mathematical equations

Defining some variables:

Defining the speed of light to calculate the period of time it takes for the signal to travel from the surface of Earth to geostationary orbit

c = 3e8                                               # Defining the speed of light
h = equations.geo_radius(5.972e24, 24*3600) - 6.4e6   # Using the equation to calculate Earth's distance
h/c                                                   # Dividing the radius by speed of light to find the period of time

0.11943882962308378

Defining signal delay:

How long it takes for a signal to reach the satellite

def signal_delay(m, T, l2, phi2, R=6.4e6, deltat = 0.01):
    h = equations.geo_radius(m, T) - R                                # Hard-coding an equation
    accessible, gc_dist = equations.is_accessible(m, T, 0, phi2, r=R)
    if accessible:                                                    # If-statement to calculate if the connection is
        t = ((h**2+(gc_dist)**2)**(1/2)) / c                          # in the boundary of the visibility cone angle
    else: 
        t = (gc_dist/c) + (h/c) + deltat                              # or if the boundary is not visible
    return t

Plots:

angles = np.linspace(0, np.pi/2, 10000)                    # Finding the angle for 10,000 points in increments of 2 degrees
delay_times1 = []
delay_times2 = []
delay_times3 = []
for angle in angles:
    delay_times1.append(signal_delay(5.972e24, 24*3600, 0, angle))
    delay_times2.append(signal_delay(6.390e23, 1.026*24*3600, 0, angle))
    delay_times3.append(signal_delay(3.285e23, 1407.5*3600, 0, angle))

import pylab as pl
pl.figure(figsize=(12,10))
pl.plot(angles*180/np.pi, delay_times3, 'r', label = "Mercury")
pl.plot(angles*180/np.pi, delay_times1, 'k',  label = "Earth")
pl.plot(angles*180/np.pi, delay_times2, 'g',  label = "Mars")
pl.legend()
pl.rcParams.update({'font.size': 18})
pl.xlabel('Angle (Degrees)')
pl.ylabel('Time (Hours)')
pl.title('Signal Delay for Mercury, Earth, and Mars (Angle vs. Time Graph)')

Text(0.5, 1.0, 'Signal Delay for Mercury, Earth, and Mars (Angle vs. Time Graph)')

Because Venus rotates so slowly, the signal delay's "jump" is at a higher angle. If Venus were to be graphed with the other three planets, it would disproprotionate the graph, which is why the planet is graphed separately.

delay_times4 = []
for angle in angles:
    delay_times4.append(signal_delay(4.867e24, 2802*3600, 0, angle))

import pylab as pl
pl.figure(figsize=(12,10))
pl.plot(angles*180/np.pi, delay_times4, 'b', label = "Venus")
pl.legend()
pl.rcParams.update({'font.size': 18})
pl.xlabel('Angle (Degrees)')
pl.ylabel('Time (Hours)')
pl.title('Signal Delay for Venus (Angle vs. Time Graph)')

Text(0.5, 1.0, 'Signal Delay for Venus (Angle vs. Time Graph)')