Edexcel A-Level Physics 9PH0
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Gravitational Fields
Orbital speed, period and geostationary satellites
- 1 thing to know
- Papers 2 and 3
Overview
Recall questions on what supplies the centripetal force, deriving orbital speed and period, the period-radius relationship, and the conditions for a geostationary orbit.
What you need to know
Answer each one before you reveal it.
Questions from this subtopic
Each one with the model answer underneath.
- 1
What provides the centripetal force that keeps a satellite in orbit?
Model answer
The gravitational attraction between the satellite and the body it orbits provides the centripetal force.
- 2
How is the expression for the orbital speed of a satellite in a circular orbit of radius r around a body of mass M obtained?
Model answer
Equating GMm/r² to mv²/r and cancelling m and one factor of r gives v² = GM/r, so v = √(GM/r).
- 3
How is the expression for the orbital period of a satellite in a circular orbit of radius r obtained?
Model answer
Substituting v = 2πr/T into v² = GM/r gives 4π²r²/T² = GM/r, so T² = 4π²r³/(GM).
- 4
What relationship between orbital period and orbital radius follows from that derivation?
Model answer
The square of the period is proportional to the cube of the orbital radius.
- 5
Does a heavier satellite need to travel faster in a given orbit than a lighter one?
Model answer
No, because the satellite's mass cancels from the equation, so orbital speed depends only on the central mass and the orbital radius.
- 6
How do the speed and the period of a satellite change if it is moved into a higher orbit?
Model answer
Its orbital speed decreases because v = √(GM/r), while its period increases because T² is proportional to r³.
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