Explain changing speed in an elliptical orbit through energy
Trace gravitational potential and kinetic energy changes as one object approaches and leaves the Sun.
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 kinetic energy and energy conservation.
- Compare signed quantities and square-root ratios.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
At fixed mass, which expression relates kinetic energy to speed?
What you will learn
- Trace gravitational potential and kinetic energy changes as one object approaches and leaves the Sun.
- Explain local elliptical-orbit speed through signed energy conservation rather than an orbital-average formula.
Follow one object along its path
For one object in an elliptical orbit, speed is greater when it is closer to the Sun and smaller when farther away. During approach, gravitational potential energy of the Sun–object system decreases and kinetic energy increases. During departure, kinetic energy decreases while gravitational potential energy increases.
This explanation uses conservation of total mechanical energy when other transfers are neglected. The Sun’s attraction does not need a fuel-burning engine to speed the object on approach. It also does not create new total energy; it changes how the conserved total is shared.
Use a stated energy reference
Gravitational potential energy can be negative when zero is chosen at infinite separation. A change from −80 MJ to −110 MJ is a decrease of 30 MJ, not an increase because the numerical magnitude is larger. Kinetic energy must then increase by 30 MJ in the isolated model.
For constant object mass, greater kinetic energy means greater speed because Ek = ½mv². A numerical speed ratio requires the square root of the kinetic-energy ratio. The whole-orbit average 2πr/T cannot specify the speed at the nearest and farthest points independently.
Pause and explain
As one elliptical-orbit object moves closer to the Sun, how do its energies change in the isolated model?
Worked example
A supplied model gives far-point energy Ep = −80 MJ, Ek = 20 MJ, and near-point Ep = −110 MJ. Find near-point Ek.
Show the worked solution
- Total mechanical energy is −80 + 20 = −60 MJ.
- Conservation gives −110 + Ek = −60 MJ.
- Ek = 50 MJ, a 30 MJ increase.
Answer Near-point kinetic energy is 50 MJ.
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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.
Explain why an orbiting comet speeds up as it approaches the Sun.
Give me a hint
Trace energy stores with other transfers neglected.
Compare my reasoning
- Gravitational potential energy decreases.
- Kinetic energy increases by the corresponding amount.
- For unchanged mass, greater kinetic energy means greater speed.
Potential energy becomes kinetic energy during approach.
Look for these in your work
- State both energy changes.
- Retain conservation and constant-mass conditions.
A supplied orbit has Ep = −120 MJ and Ek = 30 MJ far away. At a nearer point Ep = −160 MJ. Find Ek there.
Give me a hint
Keep the same total mechanical energy at both points.
Compare my reasoning
- The initial total is −90 MJ.
- Solve −160 + Ek = −90.
- Ek = 70 MJ.
70 MJ.
Look for these in your work
- Keep signed potential energy.
- Conserve the total.
For the same object mass, kinetic energy rises from 20 MJ to 80 MJ. Find the speed ratio.
Give me a hint
Speed is proportional to the square root of kinetic energy at fixed mass.
Compare my reasoning
- The kinetic-energy ratio is 80/20 = 4.
- Take its square root.
- The speed ratio is 2, not 4.
The final speed is twice the initial speed.
Look for these in your work
- Use the squared-speed relationship.
- Keep the mass fixed.
Common mistakes
- More negative potential energy is lower energy.
- Gravity redistributes conserved energy rather than creating it.
- A kinetic-energy ratio is not the same as a speed ratio.
What can you explain now?
As one elliptical-orbit object moves closer to the Sun, how do its energies change in the isolated model?
Compare with the explanation
Potential energy decreases and kinetic energy increases
The decrease in gravitational potential is balanced by increased kinetic energy, giving greater speed.
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Make it stick
Explain local elliptical-orbit speed through signed energy conservation rather than an orbital-average formula.
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.