Calculate electromagnetic travel using the vacuum speed
Use 3.0 × 10⁸ m/s, convert prefixes and distinguish a complete relay path from one segment.
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028. Follow the selected spectrum or sound section. Diagrams illustrate stated ideal models and supplied data. 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 speed = distance/time and v = fλ.
- Convert kilometres, milliseconds and frequency prefixes.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does 600 km equal in metres?
What you will learn
- Recall the vacuum speed and its air approximation.
- Calculate travel time, distance and vacuum wavelength with consistent units.
Use the correct value and environment
Electromagnetic speed in vacuum is 3.0 × 10⁸ m/s in the syllabus approximation, and is approximately the same in air. Use metres and seconds with this value. For a 600 km path, 600 000/(3.0 × 10⁸) gives 0.0020 s or 2.0 ms.
All electromagnetic regions share this vacuum speed. A microwave frequency of 2.4 GHz therefore has vacuum wavelength 0.125 m. This uses the specified speed and does not claim that glass, water or a fibre carries the wave at the same speed.
Count the full path
For an ideal 78 000 km ground–satellite–ground route, total travel time is 78 000 000/(3.0 × 10⁸) = 0.26 s. The total must include the upward and downward paths. A return conversation over the same route would add another relay journey.
Calculated propagation time omits encoding, switching, buffering and processing unless the question supplies them. State this limit. A measured total delay may be longer without contradicting the vacuum-speed model.
Pause and explain
What is the vacuum wavelength for 2.4 GHz at 3.0 × 10⁸ m/s?
Worked example
Find ideal propagation time along a 78 000 km electromagnetic relay route.
Show the worked solution
- Convert the total route to 7.8 × 10⁷ m.
- Use t = d/c₀.
- t = 7.8 × 10⁷/(3.0 × 10⁸) = 0.26 s.
Answer 0.26 s, excluding processing delays.
Try a topic test
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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 the ideal travel time for 600 km in air using the vacuum-speed approximation.
Give me a hint
Convert the path to metres, then divide by speed.
Compare my reasoning
- d = 600 000 m.
- t = d/c₀ = 600 000/(3.0 × 10⁸).
- t = 0.0020 s = 2.0 ms.
2.0 ms.
Look for these in your work
- Convert kilometres.
- State the air approximation.
Calculate vacuum wavelength at 100 MHz.
Give me a hint
Use frequency in hertz and λ = c₀/f.
Compare my reasoning
- 100 MHz = 1.00 × 10⁸ Hz.
- λ = 3.0 × 10⁸/(1.00 × 10⁸).
- λ = 3.0 m.
3.0 m.
Look for these in your work
- Convert megahertz.
- Give a length unit.
The 78 000 km relay is traversed twice for an ideal outward-and-return exchange. Calculate propagation delay and explain a larger measured total.
Give me a hint
Count two complete relay journeys, then distinguish processing.
Compare my reasoning
- Two routes total 156 000 km.
- Propagation time = 156 000 000/(3.0 × 10⁸) = 0.52 s.
- Processing and buffering can add further measured delay.
0.52 s ideal propagation; additional processing can increase the total.
Look for these in your work
- Count both complete journeys.
- Separate propagation from total system delay.
Common mistakes
- Use the total path.
- Convert prefixes before division.
- Air is an approximation; other media need supplied speeds.
What can you explain now?
What is the specified electromagnetic vacuum speed?
Compare with the explanation
3.0 × 10⁸ m/s.
It applies to all electromagnetic regions and is approximately the same in air.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Recall the vacuum speed, use the air approximation and calculate complete-path times or wavelengths with prefixes.
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.