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```					                       ISS Earth Circumference

Teachers’ Notes

Graham Colman, Wymondham College, Norfolk, UK.
www.wymondhamcollege.co.uk/maths
For more space & astronomy resources see www.colmanweb.co.uk

Email: grahamcolman@hotmail.com

This resource consists of 3 files:

   Teachers’ Notes & Answers (this file)
   ISSEarthCircumference.pps (PowerPoint slideshow)
   ISSEarthCircumference Print.ppt (PowerPoint slide for printing)

Key Learning Outcome

Pupils will consolidate their understanding of circle formulae and
distance/time/speed formulae via the medium of space and space exploration.

This is a nice starter activity. It should take around 15 minutes or so.

From just a few facts it is possible to calculate the circumference of the Earth to
within 2% accuracy.

Objective

Pupils will work out the circumference of the Earth from facts about the
International Space Station

Show the pupils the slideshow and give out the printable version or just draw a
rough sketch up.

Tell the pupils as little as possible. Try to let them work everything out, but
obviously if they are struggling give them something to go on.
To extend the problem ask the pupils to calculate Earth’s total surface area,
Earth’s total land area (⅓ total surface area) and its volume. Otherwise calculate
percentage error, convert to kilometres…

Solution

Via method here = 24,680 miles (39,488 km for explanation see further below)

Accurate answer = 25,046 miles (40,074 km)

http://pds.jpl.nasa.gov/planets/special/earth.htm
http://encarta.msn.com/encnet/refpages/search.aspx?q=earth+circumference&Submit2=Go

Method

Using ‘distance = speed x time’:

17500mph 1.5 hours
= 26250 miles

c
This is the circumference of the ISS orbit. Using c  2r  r        :
2

26250
r          4179.936... miles
6.28

This is the distance of ISS from the Earth’s centre, which is 250 miles more than
its distance from the Earth’s surface. So the radius of the Earth is given by:

4179.936  250  3929.936...miles

and hence using c  2r , Earth’s circumference is:

6.28 3929.936
 24680 miles

In kilometres

Using ‘distance = speed x time’:

28000 km/h 1.5 hours
= 42000 km

c
This is the circumference of the ISS orbit. Using c  2r  r        :
2

42000
r          6687.898... km
6.28
This is the distance of ISS from the Earth’s centre, which is 400 km more than its
distance from the Earth’s surface. So the radius of the Earth is given by:

6687.898  400  6287.898... km

and hence using c  2r , Earth’s circumference is:

6.28 6287.898
 39488km

Note: Pupils can use  or 3.14 and still get the same answer because they divide
and then multiply by  or 3.14 thereby cancelling the effect of each.

Extensions:

Earth’s radius = 3929.936 miles (using  = 3.14)
Earth’s total surface area = 4r 2 = 193,981,626 miles2
Earth’s total land area = surface = 64,660,542 miles2
3

Earth’s volume = 4 r 3 =254,111,791,785 miles3 (this is 2.54 1011 cubic miles)
3

In kilometres:

Earth’s radius = 6287.898 km (using  = 3.14)
Earth’s total surface area = 4r 2 = 496,844,918 km2
Earth’s total land area = surface = 165,614,972 km2
3

Earth’s volume = 4 r 3 = 1,041,370,000,000 km3 (this is 1.04 1012 cubic
3

kilometres)

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