Sinusoidal Function of the Orbit of a Satellite

Since the orbit's vertical height above, or below the Equator is the only thing being compared to the core of the Earth, a sinusoidal function will be perfect. The orbit of the satellite is from Equator, to South, to Equator, to North, and back to Equator; this is a perfect example of a Sine curve reflected in the x axis because the satellite is going South (downwards) from the midline (equator, which is in level with the Earth's core), to the minimum (South - Antartica), back North first to the midline (equator) and then to the maximum (North Pole) and finally back South to the midline (equator). This is the curve of a sine function reflected over the x axis. (refer to rough diagram below) The orbit's vertical height in comparison to the Equator, which is parallel to the core of the Earth is a smooth repetetive oscillation. This orbit will be continuous, and this will thus make the choice of a Sin graph reflected over the x axis perfect, since it is continuous as well. The satellite travels around Earth in the path of a circle at a constant speed.

• Orbit goes from Equator to South to Equator to North back to Equator
• a Sin function with reflection over the x axis (refer to previous page)
• Distance above Earth's surface = 900 km = distance of satellite to earth's surface
• Earth's mean radius = 6371 km = distance from Earth's core to Earth's surface
• Amplitude = distance from satellite to Earth's core = satellite's distance above the Earth's surface + Earth's mean radius = 900 + 6371 = 7271 Km = 7.271 Mm
• half of full curve
• This would also be the maximum height
• Vertical stretch = 7.271
• since amplitude is 7.271 and vertical stretch = amplitude
• Time for one orbit = 90 minutes
• Period = time for one cycle = time for one orbit = 90 minutes
• Horizontal compression = 2pi (which is one cycle of the parent function, which makes the period 1) divided by 90, to stretch one cycle to 90 minutes = 2pi/90 = pi/45
• will make the period fit the time for one orbit, 90 minutes
• horizontal shift = reciprocal of pi/45 = 45/pi
• Height at core of the Earth = height parallel to the equator = 0
• Vertical shift = 0 = we are measuring vertical height relative to the CORE OF THE EARTH/equator and since the satellite begins its orbit at the equator, which is parallel to the core of the Earth, it will be 0
• Time for orbit begins at 0 minutes = time when satellite is launched into orbit is 0, graph begins at 0 minutes
• Horizontal Translation = 0

f(x) = -7.271 sin (pix/45)

Sinusoidal Function of the Vertical Height of a Polar Orbit Satellite

Revision

The sinusoidal graph created to model the vertical height of a satellite orbit (height above or below the Equator), relative to the core of the Earth was almost as precise as possible. However there are certain areas of improvement that could have made the graph more accurate or appealing. (listed below, described furthermore in following pages)

• Instead of accounting for the vertical height of the satellite relative to the core, the distance from the satellite to the surface of the Earth could have been calculated and graphed.
• Also, accounting for the time during the launch using a piecewise function could also make this model more accurate.
• Phrasing the description of the graph differently could also make the understanding of this graph better.
• Changing the point of relativity could also have made this project better. We could have tracked the satellite's distance from a certain city in the world. This would give us a vertical shift, as long as it was not on the equator.

Why were Radians Used?

Change of Distance between Orbit and Earth

There are multiple different ways that this situation could have been modelled. For example, instead of modelling the distance from the satellite to the core of the Earth, the distance from the satellite to the surface of the Earth could have been modelled. This could have been done by ignoring the Earth's radius, and just using the distance from the orbit to the Earth's surface. This would have resulted in a lower vertical stretch factor and amplitude; vertical stretch would have been just 900 km, instead of 7271 km. On top of the original assumptions, the assumption that the mean earth radius is ongoing throughout all of Earth's surface has to be assumed for this model to be more precise (will not account for changes in earth's radius like mountains, which will increase the radius, and beaches which will be a lower radius) (rough sketch below)

Radians were used because:

• radians are the natural unit of trigonometric functions
• f(x) = -7.271 sin (pix/45) is a trig sinusoidal function
• radians for trigonometric functions (this one which is modelling the vertical height of a satellite's orbit relative to the Earth's core) are convenient and easy to use.
• if we were to use angles or other units, the numbers given would be ugly while radians are neat and simple
• the parent function of a sin function would be 180 degrees or 2pi - 2pi is much cleaner
• Radians are pure numbers
• no need for units - makes graphing easier and more convenient
• since angles are in base 60, the numbers would be very messy, but when using radians, it is much more pure and precise
• no need for rounding numbers or using 5 decimal places

Information on Satellite Orbit

Accounting for Time taken to Launch Satellite

The original model of the vertical height between the satellite and the core of the Earth began only when the satellite went into orbit (it had a period of 90 minutes); it did not take into account the time taken to launch the satellite into space. This is something that can be revised. If we were to add in a function to represent the launch time, it will be a straight horizontal line There will be no increasing slope as this graph is only modelling the vertical height in comparison to the core, not distance from surface of the Earth in general. That would be completely different. Since the satellite is launched to be parallel with the equator, which is parallel to the core, the satellite will not have any vertical height, as it will be in line with the point of relativity (core/equator) as well. A piecewise function will have to be used to model this situation.

f(x) = { = 0(x) ; 0 <= x <= 10 }

{ = -7.271 sin [pi/45(x-10)] ; 10<= x <= 100 }

The first part of the piecewise function represents the launching of the satellite. There will be no slope (it is just horizontal) because the vertical height between the satellite and the core of the Earth is not changing (it is just 'coming out'); the launch is parallel to the equator and core (vertical height relative to the Equator does not change). There is a limit of 0 to 10 because the launch only takes 10 minutes. The second part of the piecewise function represents the original function but with a couple of changes. There is a horizontal shift of 10 towards the right since the function is starting 10 minutes later due to the added launch time. The limits are also increased by 10 due to this. The period of one cycle of the orbit is still 90 minutes, but the period of the entire piecewise function, for one cycle will be 100 minutes now. If the satellite was to orbit forever, the limit 10 <= x <= 100, would be changed to 10 <= x <= infinity. The launch would stay the same in this case as it is only launched up once.

Graph Analysis

Thank You!

The End!

Kennedy Space Center, located in Cape Canaveral in Florida, is a launch site for spacecraft. When a satellite is launched from Cape Canaveral, it is usually at approximately the Equator. The site has been used by NASA for every human space flight in it's history, and a countless number of satellite launches. Most satellites launched into space orbit the Earth in a low earth orbit. A LEO is when the satellite orbits below 2000 km from the surface of the Earth.

A satellite was launched from Cape Canaveral into a low earth orbit (refer to diagram below) where the satellite is always 900 km above the Earth's surface, except for during the satellite launch. This satellite will be launched to be parallel with the equator, and will then head South towards Antartica, where it will then orbit North towards the North Pole (passing the Equator again), and finally orbit South to the equator, finishing the cycle. This type of orbit is called a polar orbit. (refer to diagram) This orbit will take 90 minutes. Model a sinusoidal equation, where x is the time in minutes taken for the orbit, and f is the vertical height in Megameters from the core to the satellite, for this satellite's vertical height above or below the Equator, relative to the core of the Earth. Assume the vertical height at the core of the Earth, which is parallel to the equator is 0. Since the surface of the Earth varies from place to place, the mean radius of the Earth, 6371 km, will be used for the function. Assume that time begins when the satellite gets into orbit. Do not take into account the Earth's axial tilt, shifts in the Earth's orbit around the sun, or rotation. It also has to be assumed that the satellite's orbit is a perfect circle.

1 Megameter = 1000 Kilometers

• X intercepts are at (0, 0) (45, 0) and (90, 0)
• X intercepts are where the satellite comes into the same vertical height (or level) as the Equator or earth's core (at height 0, y=0)
• Earth's core has a height of 0 (point of relativity; point everything else is compared to), it is parallel to the Equator
• It will take 45 minutes for the satellite to complete half the orbit and reach the Equator for the second time
• will reach the equator 3 times - at 0 minutes, 45 minutes and 90 minutes
• Period is 90 minutes - 90 minutes to complete one full cycle
• maximum height is 7.271 Megameters = 7271 Kilometers
• the highest the satellite will orbit above the core of the Earth is 7.271 Megameters - or 7.271 Mm above the Equator (same thing)
• minimum height is -7.271 Megameters = 7271 Kilometers
• the lowest the satellite will orbit is 7.271 Megameters below the core of the Earth/below the Equator
• There was a limit ; 0 <= x <= 90 set on the graph so it would just show one cycle of the curve; show the data from just one orbit
• The orbit's vertical height, relative to the earth's core and Equator would be a smooth repetetive oscillation if not for the limit set on the graph ~ ; 0 <= x <= 90
• satellite would continue to orbit over and over again

Usage

Accounting for Assumptions

My original function could have been used to find the vertical height of the satellite in comparison to the core at a certain time. This can be done by simply plugging in the minutes you would like and calculating. Oppositely, the time when the satellite would be at a vertical height could be found by reversing the function and solving.

The original model had a couple of assumptions that allowed the graph to remain straightforward and in a sinusoidal curve. The assumptions were the Earth's axial tilt, orbit around the sun, and rotation not being taken into account. If these assumptions were to be accounted for, the graph modelling the vertical height of the satelite relative to the Earth's core (North Pole being the max, and Antartica being the min) would be completely different. There would be many other factors to take into account, and it would be very hard to keep track of everything with all the different factors affecting the Earth and it's orbit.

Graph

Which Sinusoidal Function?

Different Description

f(x) = -7.271 sin (pix/45) ; 0 <= x <= 90

Vertical Height of a Satellite's Orbit over Earth

Relative to the Earth's Core

Transformations + Equation

The original statement for this trig model was:

Model a sinusoidal equation, where x is the time in minutes taken for the orbit, for this satellite's vertical height above or below the Equator, relative to the core of the Earth. Assume the vertical height at the core of the Earth, which is parallel to the equator is 0. Since the surface of the Earth varies from place to place, the mean radius of the Earth, 6371 km, will be used for the function. Assume that time begins when the satellite gets into orbit. Do not take into account the Earth's axial tilt, shifts in the Earth's orbit around the sun, or rotation.

There can be confusion regarding the concept of the vertical height of the satellite relative to the core of the Earth. This can be rewritten to be vertical height of the satellite relative to the Equator. This states the same thing except the readers will not be confused by the idea of height and the core of the Earth. The Equator provides a good point of relativity as it is known as the 'middle' or 'half' of the Earth. Changing of this statement will result in the exact same graph and calculations, except it may clear up some of the confusion behind vertical height. It is measuring the height above or below to the Equator, where the Equator is the half, which is 0, not height away from the surface of the Earth - that would be completely different.

Trig Modelling Assignment

Louie Li