Calculate average orbital speed with consistent units
Use v = 2πr/T for the stated circular or average-radius orbital model.
Section 6.1.1 · The Earth
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028. Use supplied distances, periods and stated orbital models. Drawings are not to scale. Written practice is self-checked. Human subject-teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Use circumference of a circle.
- Work with powers of ten and time conversions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the circumference of a circular orbit of radius r?
What you will learn
- Use v = 2πr/T for the stated circular or average-radius orbital model.
- Calculate an orbital average from centre-referenced radius and a consistently converted period.
Divide a complete path by its period
Average speed is total path length divided by elapsed time. For a circular orbit the circumference is 2πr, so average orbital speed v = 2πr/T. In this syllabus r is the average orbital radius and T the orbital period. Use the stated circular or average-radius approximation when applying this expression.
Radius is measured from the central body to the orbit, not the orbit’s diameter and not an altitude above a planet’s surface. If only altitude is supplied, add the supplied planetary radius before using the orbit radius. For strongly elliptical paths, 2πr is a model approximation rather than an exact ellipse circumference.
Choose one consistent unit system
For metres per second, express r in metres and T in seconds. A kilometre is 1000 metres, a day is 24 × 3600 = 86400 seconds, and a supplied 365-day year is 31536000 seconds. Convert before division and round the final result after calculation.
Using a radius in metres and a period in days produces metres per day, not metres per second. An average over the entire orbit also does not specify the instantaneous speed at one point on an elliptical path; that distinction matters in the Solar System energy lessons.
Pause and explain
With r in metres and T in seconds, what unit does 2πr/T have?
Worked example
Use r = 1.50 × 10¹¹ m and T = 365 days to estimate Earth’s average orbital speed.
Show the worked solution
- Convert T: 365 × 86400 = 31536000 s.
- Calculate 2πr/T = 2π × 1.50 × 10¹¹ ÷ 31536000.
- Obtain 29885.8 m/s and round appropriately.
Answer Approximately 2.99 × 10⁴ m/s, or 29.9 km/s.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find average orbital speed for a supplied circular orbit with r = 7.0 × 10⁶ m and T = 6000 s.
Give me a hint
Use the full circumference and the supplied seconds.
Compare my reasoning
- Write v = 2πr/T.
- Substitute 2π × 7.0 × 10⁶ ÷ 6000.
- Calculate and round 7330.38 m/s.
Approximately 7.33 × 10³ m/s.
Look for these in your work
- Use radius rather than diameter.
- Report the speed unit.
An ideal circular orbital model has r = 4.0 × 10⁸ m and T = 25 days. Calculate average speed.
Give me a hint
Convert the supplied period into seconds before substituting.
Compare my reasoning
- T = 25 × 86400 = 2160000 s.
- v = 2π × 4.0 × 10⁸ ÷ 2160000.
- The result is 1163.55 m/s.
Approximately 1.16 × 10³ m/s.
Look for these in your work
- Convert the whole period.
- Round only after division.
A satellite’s supplied altitude is 4.0 × 10⁵ m above a planet of radius 6.0 × 10⁶ m; T = 5500 s. Find its average circular orbital speed.
Give me a hint
The radius in the equation starts at the planet’s centre.
Compare my reasoning
- Add radius and altitude: r = 6.4 × 10⁶ m.
- Use v = 2π × 6.4 × 10⁶ ÷ 5500.
- Calculate 7311.34 m/s.
Approximately 7.31 × 10³ m/s.
Look for these in your work
- Add altitude to planetary radius.
- State the circular-orbit assumption.
Common mistakes
- Do not use diameter as radius.
- Convert days before reporting metres per second.
- Whole-orbit average is not local ellipse speed.
What can you explain now?
With r in metres and T in seconds, what unit does 2πr/T have?
Compare with the explanation
Metres per second
The numerator is a distance and the denominator a time, giving metres per second.
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Make it stick
Calculate an orbital average from centre-referenced radius and a consistently converted period.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.