Extract the Inclination from line 2.

Assigned data

1 XX445U YYNNNPPP 22314.79993809  .00000153  00000-0  00000-0 0  999Z


2 XX445   0.0687  62.6476 0004015 195.9237 105.3447  1.00272850 5257Z


Assignment 1 version 1.0

The objectives of this assignment are:

  • To process orbital element data for a satellite to determine its position at the data epoch.
  • To use the position data to identify the satellite (and any co-located satellites).

Q5. Inclination


Extract the Inclination from line 2.  You should be able to characterize the satellite’s orbital plane based on the Inclination value, for example an inclination of approximately 90° is characterized as a polar orbit (since the satellite passes over [or close to] the North and South Poles in each orbit).


Characterize the orbital plane as precisely as possible based solely on the Inclination value  – do not consider any other orbital elements for this answer.


Show the Inclination from your TLE set, adding a trailing degree symbol (  °  ),  and state your characterization of the orbital plane, as in this example:


Inclination: 89.8989°  
This satellite’s orbit is polar.


Note: If you are using Microsoft® Word you should select

Insert, Symbol, More Symbols …, and find the Degree Sign symbol.


(7 points)

Q6. Right Ascension of the Ascending Node


Extract the Right Ascension of the Ascending Node from line 2 of your TLE set and show it, adding a trailing degree symbol (  °  ), as in this example:


Right Ascension of the Ascending Node: 302.2222°


(1 point)

Q7.      Eccentricity


Extract the Eccentricity from line 2.  You should be able to make an observation  about the satellite’s orbital path from the Eccentricity value.


Show the Eccentricity from your TLE set, adding a leading zero and decimal point (  0.  ),  and state your observation about the orbital path, as in this example:


  Eccentricity: 0.999999    
  The orbital path of this satellite is almost a radial trajectory.  

(7 points)

Q8. Argument of Perigee 


Extract the Argument of Perigee from line 2 of your TLE set and show it, adding a trailing degree symbol (  °  ), as in this example:


Argument of Perigee: 324.4444°


(1 point)

Q9. True Anomaly 


In this context the anomaly of the satellite is the angle  between a line joining its perigee to the center of the Earth  and a line joining the satellite’s current position to the center of the Earth.   This angle is one way to express how far the satellite has progressed  along its elliptical path in the current orbit, with the perigee as the starting point.


Your TLE set shows the Mean Anomaly, which would be the anomaly  if the satellite’s speed was constant.  Because the satellite’s speed is always changing  we need to determine the True Anomaly (the anomaly at the epoch).   We can do this by first calculating another angle, the Eccentric Anomaly,  and then using trigonometry to calculate the True Anomaly from the Eccentric Anomaly.


Kepler’s equation  M = E − ε sin E  shows the relationship between  the Mean Anomaly (M), the Eccentricity (ε), and the Eccentric Anomaly (E):    This equation cannot be solved algebraically, but iterative techniques can be used  to quickly find a very good approximation for E given M and ε.


To find an close approximation of the True Anomaly (θ):

  • browse to http://www.jgiesen.de/kepler/kepler1.html ;
  • copy-and-paste click the Eccentricity and Mean Anomaly values into the appropriate boxes; – click “calculate Newton” to calculate an approximation using the Newton-Raphson method;  – round the result to 4 decimal places – this is accurate enough for this assignment.       Be sure to round, not just truncate.


The convention is to show anomalies in the range [0°, 360°),

i.e. greater than or equal to 0°, and less than 360°.

If your result is negative then add 360° as in the second example below.


[If you are interested you can click “calculate series” to calculate an approximation using the   series expansion shown above the table.  It will calculate the same value (within 8 decimal   places for a small value of e) except that for a non-negative Mean anomaly it will always be   non-negative in the  range [0°, 360°), while the Newton-Raphson method will be in the range   (–180°, +180°].   The negative and non-negative values are mathematically identical and either   one may be used in later calculations without affecting the result,   but for this assignment you are required to use the non-negative form.]


continued …


Show the Mean Anomaly from your TLE set

and your calculated and rounded value for the True Anomaly,  adding a trailing degree symbol (  °  ) to each,  as in this example where the Newton-Raphson result was positive:


Mean Anomaly: 65.4321°

True Anomaly:        66.6666°


OR as in this example where the Newton-Raphson result was negative                                                 and 360° was added to it:


Mean Anomaly: 65.4321°  
True Anomaly: −293.3334° = 66.6666


(5 points)

Q10. Orbital period


A conventional method to describe the amount of time it takes for a satellite to complete  one orbit around the Earth is the Mean Motion (N) – the number of revolutions [orbits]  around the Earth that the satellite completes in one solar [24-hour] day.   The TLE set shows the Mean Motion using this convention.


A more useful measure is the Orbital Period (Psat) – the amount of time it takes  to complete one orbit.  The Orbital Period is conventionally measured in minutes.


Use a calculator to determine the Orbital Period from the Mean Motion for your TLE set,  rounded to 5 decimal places, as in this example:


Psat = [1 / (N rev/day)] × (24 hr/day × 60 min/h)

Psat = [1 / (2.34567890 rev/day)] × (24 hr/day × 60 min/hr)

Psat ≈ 613.89477 min

Show your calculations as in the example.


To check your calculation multiply the Mean Motion by the Orbital Period  – the result should be the number of minutes in one solar day.

Divide that result by the number of minutes in one hour:  – this result should be the number of hours in one solar day    (within a very small rounding error).

Show your calculations for this check.


Show the Mean Motion from your TLE set,  and your calculated Orbital Period rounded to 5 decimal places,  as in this fictitious example:


Mean Motion: 2.34567890 revs/day
Orbital Period: 613.89477 minutes  


WARNING: Round only your final answer to 5 decimal places.

Do not round intermediate values.


(5 points)


Q11. Earth’s rotational period


The Earth rotates around its axis (a line through the North and South poles) once a day.   A day is typically measured as the time taken for the sun to re-appear in a given position  in the sky.  This is known as a solar day and the mean value is very close to 24 hours.


The Earth rotates about its axis in the same direction as it orbits around the Sun.

When the Sun re-appears in the same position the Earth has moved almost 1° further in its orbit since the orbital period of the Earth (PEarth) is approximately 365.25636 solar days.

(There are 360 degrees in a circle; 1/365.24363 is very close to 1/360,    so in one day the Earth moves almost 1 degree through a complete orbit.


The convention we will use for this assignment is to measure the Earth’s rotation against the mean vernal equinox.  The equinox is precessing (changing very slowly) so the length of a sidereal day (measured against the mean vernal equinox) is not quite the same as the length of a stellar day (measured against the fixed stars).  At present the sidereal day is 8.4 ms shorter than the stellar day – the difference is so small that we will ignore it.


Calculate the approximate length of a sidereal day as follows:

  • A solar day is 24 hours.
  • PEarth is approximately 365.25636 solar days.
  • In one solar day the Earth completes one rotation plus (1 / PEarth) of a rotation.
  • The sidereal day length is (PEarth ) / (PEarth + 1) solar days.

Hint: The length of a sidereal day is within 5 minutes of the length of a solar day.


Periods are conventionally expressed in minutes

so convert your answer from solar days (24 hour units) to minutes  by multiplying by the number of hours in a solar day  and by the number of minutes in one hour.


Show your calculation, and your answer, as in this incorrect example:  (Round all values to 5 decimal places, as in the example.)


Sidereal day length:          365.25636 / (365.25636 + 1) days
 ≈ 10.00274 days  
 ≈ 10.00274 days × 24 hours/day  
 ≈ 240.06576 hours  
 ≈ 240.06576 hours × 60 minutes/hour  
 ≈ 14403.94560 minutes  


(5 points)

Q12. Relative periods


Show the Orbital Period of your satellite (from Q10 )  and the rotational period of the Earth (the sidereal day length from Q11),  both rounded to 1 decimal place.


Use appropriate terminology (where possible)  to describe the relationship between the two periods,  as in this fictitious and incorrect example:


Satellite orbital period:  600 minutes
Earth rotational period: 6000 minutes


The orbital period of this satellite is approximately one-tenth
the rotational period of the Earth  
so the satellite’s orbit may be described as deciduous.  


For this answer consider only the two periods

– do not consider any other orbital elements at this point.


(5 points)

Q13. Type of orbit


Combine your characterizations of the Inclination (from Q5), the Eccentricity (from Q7),  and the Orbital Period (from Q10) of the satellite to determine the type of orbit of the satellite.


Browse to https://www.n2yo.com/satellites/ .

Click the link that corresponds to the type of satellite you determined above.


State the type of orbit you determined  and show the URL of the category page,  as in this fictitious and incorrect example:


The orbit of this satellite is decadaily.  
The category page I chose is https://www.n2yo.com/?c=999999


(5 points)

Q14. Relative position


Describe the position of the satellite at some point in its orbit in relation to a point on the surface of the Earth, as in this fictitious example:


The satellite passes over the North Pole once every 2.4 hours approximately.


The example above is for a polar orbit and uses the North Pole as its point  – you should choose a point relevant to your satellite’s orbit type.


Depending on your satellite’s orbit you might not be able to identify a specific point  but you should be able to describe the relationship of the satellite’s position to some point even if you do not know what that point is.


(5 points)


Q15. Rotation of the Earth


In order to determine the position of the satellite as seen from the Earth it is necessary to determine how much the Earth has rotated within the reference frame relative to the fixed point (the mean vernal equinox).  The conventional reference line on the Earth is the meridian (North-South line) through the Royal Observatory in Greenwich, England.   The angle between this meridian and the mean vernal equinox can be calculated  from the Greenwich mean sidereal time (GMST).


To calculate GMST as a conventional time of day (in a 24 hour range):  – browse to https://neoprogrammics.com/siderealtimecalculator/ ;

  • copy-and-paste the year, month, and day (note the order) from the converted Epoch you calculated in Q4        into the form under “Local Date”;
  • copy-and-paste the hour, minute, and second (rounded to an integer) from the converted Epoch you calculated in Q4        into the form under “Local Time”;
  • set the Time Zone Difference” to “+”, “00”, “00”;
  • set the “ΔT (Delta T)” to “+”, “00”, “00”;
  • select “Calendar Mode” as “Gregorian”;
  • set “Longitude” to “0” (it will change to “00 00 00.0” when computation is performed); – click “Compute”.


In the computed results find the line

“Greenwich Mean Sidereal Time at Longitude 0.0°”.


On the next line, the first value is the GMST.

(It is shown in hours, minutes, and decimal seconds, and also in decimal hours.)


On the next line, the final value is the GMST converted to an angle  which is known as the Greenwich meridian angle (GMA).

(It is shown in degrees, minutes, and decimal seconds, and also in decimal degrees.)


Show the date and the rounded time derived from the Epoch,  the GMST (as shown in the computed result,                     in hours, minutes, and decimal seconds),

and the GMA (as shown in the computed result,                           in decimal degrees,                           rounded to 4 decimal places)  as in this example:


Epoch: 2015DEC31 23:59:59
GMST: 06h 40m 20.840s  
GMA: 100.0868 °


(10 points)

Q16. Satellite longitude


The position of the satellite over the Earth can be calculated from the satellite’s position within the reference frame relative to the Earth’s position in the same frame.  For this assignment we will calculate only the satellite’s longitude (its position due North/South of a point on the Earth’s equator, expressed as an angle relative to the Greenwich meridian at the equator).


Calculate the satellite’s longitude within the reference frame (LONGframe) as follows:

  • take the Right Ascension of the Ascending Node (RAAN) from Q6;
  • add the Argument of Perigee (AofP) from Q8;
  • add the True Anomaly (TA) from Q9;
  • if the result is greater than or equal to 360° then subtract 360°; repeat this step until the result is less than 360°.



LONGframe = RAAN + AofP + TA

LONGframe = 302.2222° + 324.4444° + 66.6666°

LONGframe = 693.3332°

This is more than 360° so we subtract 360°:

LONGframe = 693.3332° − 360°

LONGframe = 333.3332°


Calculate the satellite’s longitude relative to the Earth (LONGEarth) as follows:

  • take the satellite’s longitude within the reference frame (LONGframe) from above; – subtract the Greenwich meridian angle (GMA) from Q15; – if the result is negative then add 360°.



LONGEarth = LONGframe – GMA

LONGEarth = 333.3332° − 100.0872°

LONGEarth = 233.2460°

This result is not negative so no adjustment is needed.


By convention the Greenwich meridian is the reference longitude 0°  and the longitude of a point on the Earth is measured east or west of the reference value.   If your LONGEarth is less than or equal to 180° it is easterly and is your satellite longitude – show it with a suffix “E” as in this example (not related to the examples above):


LONGsat = LONGEarth  
LONGsat = 179.1234° E


otherwise subtract your LONGEarth from 360° to obtain a westerly satellite longitude  (in the range 0 to less than 180°) and show it with a suffix “W”, as in this example:


LONGsat = 360° − LONGEarth  LONGsat = 360° − 233.2460°

LONGsat = 126.7540° W


Your answer must be in the range 0 to 180°, and it must be “E” or “W” (except 0).


Show your calculations of LONGframe, LONGEarth, and LONGsat,  as in the three examples above, and your results rounded to 4 decimal places.


(10 points)

Q17. Satellite identification


You now have enough information to identify your satellite and any co-located satellites.


Browse to the category page you selected in Q13.

Click on the column heading “Period [minutes]” to sort by that value.


Round your Period (from Q10) to one decimal place,  your Inclination (from Q5) to one decimal place,  and your Longitude (from Q16) to one decimal place.   Use the information on the Web site to find all satellite/s  within ± 0.1 minutes of the rounded Period,  within ± 0.5° of the rounded Inclination,  and within ± 0.2° of the rounded Longitude of your satellite.

Each of the values from the Web site must be within the ± margin shown  of the corresponding rounded value you calculated for your satellite

If a Longitude is not shown with a suffix of “E” or “W”, treat a negative value as westerly.


Note: You TLE set is from an actual satellite.  You should find at least one                         and typically more than one satellite within your search ranges.

If you cannot find at least one satellite you have most likely made an error                           and should go back and check every step.


Full credit for this question will be given only if you find all co-located satellites.


Show your search parameters (including the plus/minus margins),  and for each satellite that matches your parameters (within the margins)  show its name, NORAD ID, Period, Inclination, and Longitude.   The following is a fictitious example for two co-located satellites:










Search parameters:

Period:               99.9 ± 0.1 minutes

Inclination:        89.9° ± 0.5°

Longitude:         179.9° W ± 0.2°


Satellite/s found:

Satellite name:  IT353 SAT-A

NORAD ID: 35301

Period:               99.9 minutes

Inclination:        89.9°

Longitude:          179.9° W


Satellite name:  IT353 SAT-B

NORAD ID: 35302

Period:               99.8 minutes

Inclination:        90.0°

Longitude:          179.7° W


(20 points)






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