Orbital Period (Edexcel IGCSE Physics: Double Science)

Revision Note

Test yourself
Ashika

Author

Ashika

Last updated

Did this video help you?

Orbital period equation

  • When planets move around the Sun, or a moon moves around a planet, they orbit in circular motion
    • This means that in one orbit, a planet travels a distance equal to the circumference of a circle (the shape of the orbit)
    • This is equal to 2πr, where r is the radius a circle 

  • The relationship between speed, distance and time is:

s p e e d space equals space fraction numerator d i s t a n c e over denominator t i m e end fraction

  • The average orbital speed of an object is 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 is always taken from the centre of the object being orbited to the object orbiting

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 (HST) moves in a circular orbit around the Earth. 

The HST orbits at a height of 560 km above the Earth’s surface and completes one orbit in 96 minutes. The radius of the Earth is 6400 km.

Calculate the orbital speed of the HST in m/s.

Answer:

Step 1: List the known quantities

  • Radius of the Earth = 6400 km
  • Height of the HST above the Earth's surface = 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

r = radius of the Earth + height of the HST above the Earth's surface

r = 6400 + 560 = 6960 km

Step 4: Convert any units

  • The time period needs to be in seconds

1 minute = 60 seconds

T = 96 minutes = 60 × 96 = 5760 s

  • The radius needs to be in metres

1 km = 1000 m

r = 6960 km = 6 960 000 m

Step 5: Substitute values into the orbital speed equation

v space equals space fraction numerator 2 straight pi cross times open parentheses 6 space 960 space 000 close parentheses over denominator 5760 end fraction

v space equals space 7592.18 space equals space 7590 space straight m divided by straight s

Examiner Tip

Remember to always check that the orbital period 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

You've read 0 of your 5 free revision notes this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Ashika

Author: Ashika

Expertise: Physics Project Lead

Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.