A synchronous satellite
- Round
- Theory
- Points
- 10
- Level
- Silver
- Taught in
- Section 1.7
Adapted from IOAA 2017 (Phuket), theory problem 5. We have restated it in our own words and kept the original data, adding a value for Earth's radius. The solution was last checked on 15 October 2026.
A synchronous satellite goes round Earth in exactly the time Earth takes to turn once, at a height of 35 786 km above the surface. One such satellite is placed in an orbit tilted at 6.69° to the plane of the equator. Find the greatest altitude the satellite can have for an observer at latitude 51.49°. Ignore refraction, and take Earth's radius to be 6378 km.
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Seen from Earth's centre, the satellite stays within 6.69° of the plane of the equator. It is highest for the observer when it crosses the observer's meridian at its furthest north, 6.69° north of the equatorial plane.
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In the plane of that meridian, the angle at Earth's centre C between the observer O and the satellite S is
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Dropping a perpendicular from O onto the line CS splits the triangle COS into two right-angled triangles. The perpendicular has length
so
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The zenith distance of the satellite is the outside angle of the triangle at O, which equals the sum of the angles at C and S,
AnswerThe greatest altitude is 38.4°. Treating the satellite as if it were infinitely far away would give
This problem also appears among the olympiad problems of Section 1.7, Problems for Chapter 1, which teaches the ideas it uses. Equation and figure numbers refer to the course.