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| The International Space Station in orbit (ISS) |
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Specs
Brightness |
approximately -4 (less than Venus) |
Launch Window |
5-10 min. |
Orbital Altitude |
361 km at perigee
437 km at apogee
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Mass |
approx. 420 000 kg |
Dimensions |
111.08 m by 89.2 m |
Speed approx. |
26720 km/h |
Orbital Inclination |
51.5947° |
Estimated Cost |
28 billions |
Orbital Period |
approx. 90 min. |
Observational Visibility |
between 60 N and 60 S |
Orbital Type |
elliptical |
Countries - Canada, Belgium, Brazil, Denmark, France, Germany, Italy, Japan, the
Netherlands, Norway, Russia, Spain, Sweden, Switzerland, the United Kingdom, the
United States
Definitions
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Altitude |
Distance of orbit above the Earth’s surface.
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Apogee |
An Earth - centred orbiting object’s farthest point from Earth.
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Brightness |
Measured in magnitudes (as stars and planets are). Brightness depends on
elevation above the horizon as well - objects appear dimmer when close to
the horizon because of atmospheric interference.
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Circular Orbit |
An orbit that has no perigee or apogee.
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Elliptical Orbit |
An Earth - centred orbit having a perigee (closest approach to Earth) and an
apogee (farthest distance from Earth). Most orbits are elliptical or “oval
shaped”.
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Equatorial |
Orbit which encompasses inclinations from approximately 0 – 70o N or
S (includes most satellites).
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Launch Window |
Best time of day for launch based on ISS position relative to launch site, among
other things.
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Orbital Inclination |
Inclination is measured in degrees parallel to latitude. It is used instead of
latitude to refer to objects in orbit, such as the ISS. It is an angular
measurement taken from the Earth’s equator.
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Orbital Period |
The time it takes for an object to complete one orbit.
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Perigee |
An Earth - centred orbiting object’s closest approach to Earth.
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Polar Orbit |
An object which orbits around the poles. The object’s orbital inclination is
90o. The orbit can be circular or elliptical (example: weather
satellites).
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Viewing Windows |
Variations in time of day a satellite, space station or ISS passes. The time ISS
passes occur each day will vary because our 24-hour day is not evenly divisible
by the ISS Æ 90-min. orbital period. The Space Station will appear in the sky a
little earlier (or later) on subsequent days and thus move from daytime to night
time passes cyclically.
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Orbital Elements
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Distances |
Perigee of orbiting element
Apogee of orbiting element
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Orbital Inclination |
Inclination is measured in degrees parallel to latitude. It is used instead of
latitude to refer to objects in orbit, such as the ISS. It is an angular
measurement taken from the Earth’s equator.
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Orbital Period |
The time it takes for an object to complete one orbit
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Orbital Types |
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Circular Orbit |
An orbit that has no perigee or apogee |
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Elliptical Orbit |
An Earth - centred orbit having a perigee (closest approach to Earth) and an
apogee (farthest distance from Earth). Most orbits are elliptical or “oval
shaped”.
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Special Cases |
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Polar Orbit |
An object which orbits around the poles. The object’s orbital inclination is
90o. The orbit can be circular or elliptical (example: weather
satellites).
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Equatorial |
Orbit which encompasses inclinations from approximately 0 - 70o N or
S (includes most satellites)
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Geostationary
or geosynchronous -
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Satellites orbiting at altitudes of approx. 36 000 km take one day to orbit the
Earth. Since they are moving at the speed the Earth rotates, they are positioned
over the same location at all times (example: communications and meteorological
satellites).
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Activities
Junior High or High School
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Have the students trace out a ground track of the ISS orbit (between
51.6o N and 51.6o S) on a map. An example of the track
on a Mercator Projection is given in the student handout. Because a map is not a
three-dimensional projection of the Earth, the track will be an S-curve. Also,
the altitude of the ISS will not be represented.
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Compare these two-dimensional representations to a track mapped out on a globe.
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Tie a string around the globe (again, between 51.6o N and
51.6o S). This will represent the inclination of the ISS as it orbits
around the Earth. Also, the width of the string could adequately represent the
altitude of the ISS (the scale is about right).
High School
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Have the students figure out why the ISS can be seen from higher latitudes if
its altitude is increased. This concept is illustrated in the student handout
(pgs. 2 and 3). In this illustration, we can see that the horizon is tangent to
your point of latitude. (You can scale your own Earth from a full Earth, a half,
a quadrant, etc.).
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Have the students draw lines from the centre of a scaled Earth at angles
representing their latitudes, the ISS inclination latitude, and other latitudes.
All lines (except for the ISS line) should end at the surface of the Earth.
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Have them extend the ISS line above the surface to a 400 km altitude (based on
your scale). The ISS, then, will be a point at the tip of this line.
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Next, have the students draw tangent lines to the points of all the angled lines,
as in the illustration. These lines will be the horizons for each latitude. If
the ISS (the tip of the point on the 51.6o line) falls below the
horizon lines, it will not be seen. If it is above, it will.
Expand on this activity
Having the students create an illustration like the one below will help them
ascertain the altitude the ISS would need to be seen at latitudes outside its
inclination limitations. The students can pick a latitude, follow the same
procedures of drawing tangent lines, and find the length the hypotenuse line “H”
(being the ISS line) would have to be to just meet with the horizon line for that
latitude. The altitude at which the ISS would need to be to be seen at that
latitude would have to be slightly greater than the length of “H”.
Example
If an observer is at 80o N, what altitude would the ISS have to be for
the observer to see it?
In this case, we are looking at the inner triangle B, D, H in the diagram below. We
know the length of B (Earth’s radius) and the angle of A (angle of B (80o)
- angle of line H (51.6o). We also know the angle of P (180o -
(A + M)). Given the angle of P and the length of B, we can find the length of D (tan
P =opposite (B) / adjacent (D)). Once we know the length of D, we can use it and B
to find the length of H (Pythagorean). The length of H minus the length B (Earth’s
radius) will equal E - the altitude at which the ISS would have to be to just meet
the 80o horizon. So, anything greater than this number would put the
altitude right over the horizon to be seen at 80o N or S.
EXAMPLE – ISS VISIBILITY AT 80° N (or S)
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