IGCSEPhysicsCIERevision Notes6. Space PhysicsEarth & The Solar SystemOrbital Speed Equation

Orbital Speed Equation (CIE IGCSE Physics)

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Lindsay Gilmour

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Orbital speed equation

Extended tier only

  • When planets orbit around the Sun, or a moon moves around a planet, they move in circular orbits
  • In one complete orbit, a planet travels a distance equal to the circumference of a circle
    • This is equal to 2 straight pi r, where r is the radius of the circular path
  • The relationship between speed, distance and time is:

speed space equals space distance over time space equals space fraction numerator circumference space of space orbit over denominator orbital space period end fraction

  • The average orbital speed of an object can be defined by the equation:

v space equals space fraction numerator 2 straight pi r over denominator T end fraction

  • Where:
    • v = orbital speed in metres per second (m/s)
    • r = average radius of the orbit in metres (m)
    • T = orbital period in seconds (s)
  • This orbital period (or time period) is defined as:

The time taken for an object to complete one orbit

  • The orbital radius r is always taken from the centre of the object being orbited to the object orbiting

Orbital speed of a planet

Orbital Period, downloadable IGCSE & GCSE Physics revision notes

Orbital radius and orbital speed of a planet moving around a Sun

Worked example

The Hubble Space Telescope moves in a circular orbit. Its height above the Earth’s surface is 560 km and the radius of the Earth is 6400 km. It completes one orbit in 96 minutes.

6-1-2-worked-example-question-cie-igcse-23-rn

Calculate its orbital speed in m/s.

Answer:

Step 1: List the known quantities

  • Radius of the Earth, R = 6400 km
  • Height of the telescope above the Earth's surface, h = 560 km
  • Time period, T = 96 minutes

Step 2: Write the relevant equation

v space equals space fraction numerator 2 straight pi r over denominator T end fraction

Step 3: Calculate the orbital radius, r

  • The orbital radius is the distance from the centre of the Earth to the telescope

6-1-2-worked-example-solution-cie-igcse-23-rn

r space equals space R space plus space h

r space equals space 6400 space plus space 560 space equals space 6960 space km

Step 4: Convert any units

  • The time period needs to be in seconds

1 space minute space equals space 60 space seconds

96 space minutes space equals space 60 space cross times space 96 space equals space 5760 space straight s

  • The radius needs to be in metres

1 space km space equals space 1000 space straight m

6960 space km space equals space 6 space 960 space 000 space straight m

Step 5: Substitute values into the orbital speed equation

v space equals space fraction numerator 2 straight pi space cross times space 6 space 960 space 000 over denominator 5760 end fraction space equals space 7592.18 space equals space 7590 space straight m divided by straight s

Exam Tip

Remember to check that the orbital radius r given is the distance from the centre of the Sun (if a planet is orbiting a Sun) or the planet (if a moon is orbiting a planet) and not just from the surface. If the distance is a height above the surface you must add the radius of the body, to get the height above the centre of mass of the body.

This is because orbits are caused by the mass, which can be assumed to act at the centre, rather than the surface.

Don't forget to check your units and convert any if required!

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    Lindsay Gilmour

    Author: Lindsay Gilmour

    Lindsay graduated with First Class Honours from the University of Greenwich and earned her Science Communication MSc at Imperial College London. Now with many years’ experience as a Head of Physics and Examiner for A Level and IGCSE Physics (and Biology!), her love of communicating, educating and Physics has brought her to Save My Exams where she hopes to help as many students as possible on their next steps.