**Predicting the Path of a Satellite Activity**** **

(**This activity is optio****nal****,** but the mathematics are very helpful in this activity.)

Resource - Adapted from: ** http://octopus.gma.org/surfing/satellites/orbit.html (NASA Learning from Satellites)**

__Supplies Needed for this activity: __

**12-inch or 30.48 cm globe**** **(this is the globe diameter)**length of #10 wire (20 in to 40 in or 51 cm to 102 cm) to project orbital plane****clear plastic strip to project width of Earth image****data for a NOAA-class satellite (suggestion: use N2YO.com)****ruler****protractor****compass**

Satellite orbits lie in planes that bisect the orbited body. If the Earth were not rotating, each orbiting satellite would

pass over the same point on the Earth with each orbit, crossing the equator repeatedly at the same longitude.

Because the Earth is constantly rotating, each orbital pass of the satellite (as indicated by the model described in this activity) appears to be to the west of the previous one. In reality, the Earth is rotating eastward as the orbital plane remains fixed.

**You can examine the factors determining the length of a satellite's orbit around Earth**

Recognize that the Earth rotates 360 degrees in 24 hours, or:

__60 minutes__ X __24 hours__ = __1440 minutes__

1 hour 1 day 1 day

Dividing 360 degrees by 1440 minutes shows that the Earth is rotating 0.25 degrees every minute. Here's the math:

__360 degrees__ = __0__**.** 25 degrees

1440 minutes 1 minute

In this example, the satellite that we track travel around the Earth has an orbit period of 102 minutes. Thus, we can see that if the satellite crossed the equator at 0 degrees longitude on one orbit, it would move over on the globe by 25.5 degrees longitude 102 minutes later.

__0__**. **25 degrees = __25.5 degrees__

1 minute 102 minutes

The extremely large size of the Earth in relation to the small thickness of the atmosphere leads to frequent, intentional distortions of scale in map projections. A worthwhile activity is to construct a true scale physical model of an orbiting satellite's path around a classroom earth globe to understand the physical characteristics of a satellite's orbit.

Given the following information answer the questions that follow.

If you are tracking the NOAA series of weather satellites, the following figures will be a close approximation.

** Km. Miles**

**Mean orbital altitude** 860 534

**Width of field of view** 2900 1800

**Orbital period.. ** 102 minutes

Determine the scale of your globe by measuring either its diameter or circumference and comparing that to the Earth's actual diameter or circumference. **(Earth diameter: 12,756 km or 7,926 miles and circumference: 40,030 km or 24,874 miles**)

What follows is a series of questions to test for understanding.

**Questions:**

**What is the Earth's diameter? What is the diameter of the globe?**

**The sample calculations given here are based on the use of a 12-inch (30.48 cm) diameter globe and should be proportionally adjusted for use of other materials. The diameter of a globe can be determined by first finding the circumference using a tape measure.**

**A ratio is set up to determine the scale model of the globe:**

__Globe Size__ = __1 cm__ or __1 in__

**Earth Size x km x mi**

** ****Earth ** **Model**

** **

**Diameter 7,926 miles or 12,756 km ** 12 inches or 30.48 cm

** **

**Circumference 24,874 miles or 40,030 km **37.75 inches or 95.76 cm ** **

** **

**Path width 1,800 miles or 2,897 km ** 2.5 inches or 6.93 cm

** **

**Orbit altitude 534 miles or 860 km ** 0 .80 inches or 2.05 cm

**2.****What is the ratio of the model diameter compared to the Earth's diameter? This is your scale measure.**

** **

**Using the circumference of the globe compared to the circumference of the Earth, a ratio can be established as follows:**

**C=2 π r π=3.14 (to determine the globe's circumference) C= circumference π = pi r= radius of the globe**

__C of Globe__ =** ** 37.68__ inches__ = __1 inch __ or __ 95.7____ cm____ __ = __1 cm __

**C of Earth 24,874 miles 660 miles 40,234 km 418 km**

** 3. Just how high should you position a piece of wire above the classroom globe to represent the satellite path?**

Using a piece of wire (#10 - #12 works well), position the wire in such a way as to center the wire over your location on the globe. Let's assume the NOAA satellite is in a polar orbit (many NOAA satellites are in a geostationary orbit).

See the link below for information on types of orbits:

http://6. Understanding satellite orbits.

The polar orbit would take the satellite in a north-south path and it would travel close to both poles. If a Chicago, IL location is chosen, the wire should cross the equator at approximately 90 degrees west longitude, and continue up to the left of the north pole by about a few degrees. Experiment with different ways of supporting the wire slightly above the globe. You may be able to use the support bar already found on your globe for support of the wire. The globe should be able to rotate under the wire. You may want to add a piece of plastic transparency material to the wire. A stiff sheet of overhead transparency is a good material to use. This transparency will represent the width of the Earth that the NOAA satellite will image on a typical pass. Because this satellite width is approximately 1,800 miles or 2,099 km.

(The scale plastic strip will be approximately 2.5 in or 6.93 cm wide. See calculations below.)

By the scale established above, the orbital height of 534 miles or 860 km is represented by a scale distance of:

__ 1 inch__ = __0__**.**80 inch or __1 cm __ = __ 2__**.**05__ cm__

#### 660 miles 534 miles 418 km 860 km

Thus, on this model of the Earth, using a 12-inch diameter globe, the wire representing the orbital plane should be placed approximately 3/4" above the surface of the globe. (Or 30.48-cm globe, the wire will be 2.05 cm above the globe surface.)

**How many miles from your location are the points on Earth over which the satellite will come into or leave receiving range?**

On a large sheet of paper, draw a circle of the same diameter as your globe.

This circle will represent the surface of the Earth. Write the scale of measure in the lower right-hand corner as a legend. Draw a circle having the same center as the first, but with a radius of 534 miles or 860 km more than your first circle. This will represent the orbit of the satellite over the Earth's surface. Label point H on the inner circle (Earth's surface) as your town or city. Draw a straight line through this point. That will represent the horizon as it appears from your location.

The satellite we wish to examine can only be received while in an unobstructed straight line from the antenna; thus, it can only be received while above the horizon. The point at which we first receive a satellite's signal is known as Acquisition of Signal (AOS), and the point at which we lose the signal is referred to as Loss of Signal (LOS). A good analogy would be to think of sunrise as AOS and sunset as LOS.

On your drawing, label two points that lie directly under the points at which the satellite will come into (Acquisition of Signal, AOS) and go out of (Loss of Signal, LOS) receiving range.

You may wish to draw a circle on your globe to represent this range, known as the acquisition circle.

Assume the satellite pass is directly overhead of point H (your home location). R equals the radius of our reception area (acquisition circle) and D equals the diameter of the reception area.

Having established the AOS and LOS points on the orbital curve, project lines from both the AOS and LOS points down to the center of the Earth. Label this point G (Earth's Geocenter). With these lines in place, go back and label the two points where these two lines intersect the Earth's surface. Appropriately label these Points A and L. If we measure the angle formed by points AGL, we find it to be 55 degrees. This is angle D (diameter of acquisition circle). Angle R is half of angle D (R = radius of acquisition circle). Why is it important to know this angle? We are interested in knowing the size of our acquisition circle, that is, the distance from our home location (point H) that we can expect to receive the satellite signal.

We've previously determined that the circumference of the Earth is approximately 24,874 miles or 40,030 km. Set up a ratio to determine "D" distance which is the miles between A (AOS) and B (LOS), given that angle is 55 degrees:

__360 degrees__ = __55 degrees __ or __360 degrees__ = __55 degrees __

24,874 mi 3,800 mi 40,030 km 6,116 km

##### To determine the distance that the signal will be seen and the location, take a look at the angle HGA or HGL in the diagram above.

This is known as "R" in the diagram and represents how far on each side of the horizon the signal will be received. This amount will be 1/2 the

"D" distance. So the the 'Acquisition Distance' will be 1,900 miles or 3,058 km. We then look on a map north and south of Chicago and see that

signal will be seen north of the Hudson Bay in Hudson Straig and all the way south to Nicaragua in Central America.

**Knowing the period of a complete orbit, find a way to calculate the amount of time that the satellite will be in range if it passes directly overhead as you've illustrated.**

Rotate the globe to position the wire so that the northbound orbit will cross the equator at 0 degrees longitude. If the Earth were not rotating, the satellite would always follow the path illustrated by the wire. Would this path ever bring the satellite over your school?

Polar orbiting weather satellites have an orbital period of about 102 minutes. This means they complete a trip around the world in approximately 1 hour and 42 minutes.

The Earth rotates 360 degrees in 24 hours.

**How many degrees does the Earth rotate in 1 minute?**

** ** The Earth rotates 360 degrees in 1440 minutes. Thus:

360 / 1440 = 0.25 degrees/minute

**7.****How many degrees does the Earth rotate in 1 hour? **

If the Earth rotates 0.25 degrees each minute, then:

0 .25 x 60 = 15 degrees/ hour

**8.****How many degrees does the Earth rotate in 102 minutes? **

If the Earth rotates 0.25 degrees each minute, then:

0.25 x 102 min = 25.5 degrees/orbit

9. **If the satellite crossed the equator at 0 degrees longitude at 0000 UTC, at what longitude would it cross the equator 102 minutes later? **

(UTC is Coordinated Universal *Time* (abbreviated to *UTC*) is the primary *time* standard by which the world regulates clocks and *time*. It is established in Greenwich England and all time zones around the world use UTC as a reference.) ** **

**The satellite crosses the equator at 0 degrees west longitude. The next equator crossing will be 102 minutes later, and located 25.5 degrees west. ****204 minutes later? **

**Thus, the next equator crossing 102 minutes after the second pass,will take place at 25.5 degrees further west , which would place this crossing at 51 degrees west longitude (25.5 + 25.5). **This would be 51 degrees further west than the first pass.** **

** **

10. **How many orbits will the satellite complete in one day?**

#### Each orbit for this satellite takes 102 minutes, there are 1,440 minutes in a 24 hour period, so 1,440 ÷ 102 = 14.11 orbits in a day.

#### 11.**How many miles does the satellite travel during one orbit?**

Earth's circumference is 24,478 miles, but the satellite is has an altitude of 534 miles, and the diameter of Earth is 7,926 mi so one orbit is now C_{satellite}= (7,926 mi + 534 mi) x **π** = 26,564 mi, not counting for the bulge at Earth's equator and taper at the poles.

or

C_{satellite}= (**12,756 km + 860 km**) x **π** = 42,754 km, not counting for the bulge at Earth's equator and taper at the poles.

12. **How many miles does the satellite travel during each day? **

**Our calculations show that the satellite will make 14.11 orbits/day x 26,564 mi/ orbit = 375,818 miles/day !!**

**or**

**Our calculations show that the satellite will make 14.11 orbits/day x 42,754 km/ orbit = 603,259 km/day !!**

**13.**** How many times will your location be viewed by this satellite in one day?**

**At least four times/day. The satellite will usually come within range on two consecutive orbits, sometimes three. Usually figure on a pass to the east of your location, nearly overhead, and then to the west. Remember that at one part of the day the satellite will be on an ascending pass (crossing the equator going north) and at another time of the day the satellite will be on a descending pass (crossing the equator going south).**

**Here is one final activity which will test your understanding of orbital parameters. Assume you are tracking a NOAA-class satellite with an orbital period of 102 minutes, and a longitudinal increment of 25.5 degrees west/orbit.**

**Complete the chart for the remaining five orbits.**

**Orbit # Equator Crossing (EQX) Time**

** **

** 1 0 degrees 1200 UTC**

** **

** 2 **

** **

** 3 **

** **

** 4 **

** **

** 5**

** **

** 6**

**Here's what you should have:**

__ __ 2 25.5 degrees 1342 UTC

** **

** 3 51.0 degrees 1524 UTC**

** **

** 4 76.5 degrees 1706 UTC**

** **

** 5 102 degrees 1848 UTC**

** **

** 6 127.5 degrees 2030 UTC**