Orbital Period (Edexcel IGCSE Physics (Modular))

Revision Note

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Written by: Ashika

Reviewed by: Caroline Carroll

Calculating orbital period

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:

$s p e e d = \frac{d i s t a n c e}{t i m e}$

The average orbital speed of an object is defined by the equation:

$v = \frac{2 π r}{T}$

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 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

$v = \frac{2 π r}{T}$

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

$v = \frac{2 π \times (6 960 000)}{5760}$

$v = 7592.18 = 7590 m / s$

Examiner Tips and Tricks

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

Last updated: 28 August 2024

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