I've been looking at the graphs available in the WOD and have created some examples of how information about FUNcube can be derived from the graphs.
For example, the following fits into the UK yr9 - yr 11 (14 - 16yrs) Physics/maths. - Sorry to copy and paste, but this was part of a note sent to a colleague who wants to organise a 'gifted and talented' satellite activity.
Interestingly, there was more interest in the internet availability of data, rather than direct reception in the school field. - Seems that direct reception may need to be done in a small group
whereas there are no limits on numbers for a classroom activity with a demo (globe and model) some video and audio clips and a laptop based investigation. (personally I think the cold British winter put them off)
.............The live data changes as you watch, but it stores the last 104 minutes as whole orbit data. –As below.
It could be possible, perhaps with a model of the satellite and a globe to visualise what’s going on 600km above us, to create a class exercise. We could give them a graph or CSV file and after a suitable introduction, get them to interpret the data from a graph. It would show how telecommunications and graph plotting is invaluable to ‘space science’. We do have recordings of the satellites telemetry downlink to listen to and can put that into a program that displays the data in the recording. Also, we can put text messages into the satellite on request, so if the students look at the live data they could see. “Hello Mr / Miss xxxxx and the students at xxxxx school, London UK” or something similar (200 character limit per message)
For example: This is a graph of data received in South Africa this morning, but available a few seconds later via the internet.
This graph shows total current from the solar panels (black) and the temperature of a chrome plated bar of Aluminium mounted on the outside
of the satellite (Red). The X axis is time from 0 to 104 minutes. New data would appear at the right end of the graph.
The graphs are available as downloadable jpg or pdf fles. The data is also available as a CSV file, so the students can plot on Excel.
What can you learn from the graphs?
1) From the ‘photo current’ graph in black, can you see when the satellite is in the Earth’s shadow (eclipse)?
The satellite is in the sun from t= 0 – 42 minutes. It then passes into eclipse. - Solar panel current drops to zero
2) How long is the satellite in eclipse? - Data is taken at 1 min intervals. There are 35 samples of current at zero, so 35 minutes is spent in eclipse.
3) When the satellite is illuminated its temperature increases. Does it reach equilibrium? – Yes. At t= 30 minutes the temp settles around 26 degrees.
At this point, the amount of energy absorbed by the spacecraft equals the amount of energy radiated into space.
4) Why is the temperature curve in eclipse smooth, but ‘jagged’ when illuminated by the sun?
In eclipse heat energy is lost by radiation in all directions. But when the satellite Is in the sun (most visible between t=30 and 40 minutes) the
metal bar has many small variations in temperature. This is due to satellite rotation. When facing the sun, energy is absorbed from the Sun’s radiation,
but as the satellite rotates and the bar faces The Earth or deep space, it loses some of its energy and the temperature drops a few degrees.
5) Assuming this graph is repeated each orbit, can you estimate the time taken for the FUNcube satellite to complete one orbit of the Earth?
The red plot of temperature has a value of 10 degrees near the start. It also has 2 samples at 10 degrees near the end of the graph. By counting the
samples between repeating points we can measure the period of the waveform which equals the time of one orbit. – The 10 degree samples are at
3 minutes and then at 98 and 100 minutes. If we take 3 minutes and then find the average of 98 and 100 to get 99, we can determine that one orbit takes
99 – 3 = 96 minutes.
6) Given the radius of the Earth is 6371km and the altitude of FUNcube is 630km, can you calculate the velocity of the satellite?
As the satellite is orbiting 630km above the surface of the Earth, the orbital radius is 6371 + 630 = 7001km
The total distance travelled in one orbit is equal to the circumference of the orbit. Circumference = 2 x Pi x radius.
Therefore circumference = 2 x 3.142 x 7001km or 43,989km
So, velocity = 43,989 / 96 = 458km per minute
or Velocity = 43,989 / 96 x 60 = 27,480 km/hr
Or perhaps most dramatically……………. 43,989 / 96 / 60 = 7.63 kilometres every second.
(there’s probably more than the just the above if we think about it)