Define gravitational potential and calculate its value
Define gravitational potential relative to infinity and use the negative point-mass potential −GM/r.
A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 13. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use Newton's laws, circular motion, vectors, scientific notation and SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Where is gravitational potential conventionally zero for an isolated point mass?
What you will learn
- Define gravitational potential as work done per unit mass from infinity
- Use φ = −GM/r for a point mass
- Interpret negative potential and the zero at infinity
Potential is energy transferred per unit mass
Gravitational potential φ at a point is the work done per unit mass in bringing a small test mass from infinity to that point, with the usual zero of potential chosen at infinity. Potential is a scalar, measured in J kg⁻¹.
Gravity attracts, so an external agent bringing a test mass slowly from infinity does negative work: the field transfers energy to the motion unless the agent removes it. This reference choice makes the potential around an isolated mass negative.
A point mass has potential −GM/r
For a point mass M, φ = −GM/r. The potential becomes less negative as r increases and approaches zero at infinity. A stronger source mass makes the potential more negative at the same radius.
Do not confuse potential φ, which is energy per unit mass, with potential energy Ep for a particular mass. Multiplying φ by a test mass gives its gravitational potential energy.
Pause and explain
What is the sign of φ = −GM/r for positive M and r?
Worked example
Find the gravitational potential at radius 6.0 × 10⁶ m from a point mass of 6.0 × 10²⁴ kg.
Show the worked solution
- Use φ = −GM/r.
- Substitute 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ / 6.0 × 10⁶.
- Retain the negative sign and give J kg⁻¹.
Answer φ ≈ −6.67 × 10⁷ J kg⁻¹.
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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.
Calculate φ at radius 2.0 × 10⁷ m from a mass 6.0 × 10²⁴ kg.
Give me a hint
Substitute into −GM/r and keep the sign through the calculation.
Compare my reasoning
- φ = −GM/r.
- φ = −(6.67 × 10⁻¹¹)(6.0 × 10²⁴)/(2.0 × 10⁷).
- Give J kg⁻¹.
φ ≈ −2.0 × 10⁷ J kg⁻¹.
Look for these in your work
- Potential equation used
- Negative sign retained
- Unit stated
At a fixed distance, a source mass is tripled. How does its gravitational potential change?
Give me a hint
Use direct proportionality of φ to source mass while holding r fixed.
Compare my reasoning
- φ = −GM/r.
- With M replaced by 3M, φ₂ = −3GM/r.
- The potential remains negative and its magnitude triples.
The potential becomes three times as negative; its magnitude triples.
Look for these in your work
- Distance held fixed
- Mass proportionality applied
- Sign and magnitude distinguished
Two points are at radii R and 2R from the same isolated mass. Compare their potentials.
Give me a hint
Apply the inverse-distance dependence, including the common negative sign.
Compare my reasoning
- φ(R) = −GM/R.
- φ(2R) = −GM/(2R) = φ(R)/2.
- The farther point has half the magnitude and is less negative.
At 2R the potential is half as negative as at R.
Look for these in your work
- Equation applied twice
- Radius ratio correct
- Negative values interpreted
Common mistakes
- Omitting the negative sign
- Giving potential in joules instead of joules per kilogram
- Setting potential to zero at the surface while using φ = −GM/r
What can you explain now?
Where is gravitational potential conventionally zero for an isolated point mass?
Compare with the explanation
At infinity
The potential approaches zero as the separation becomes very large.
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Make it stick
Why is gravitational potential negative when its zero is chosen at infinity?
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