Potential, potential energy and field gradient
Key idea: H2 Physics lessons on inverse-square gravitation, field strength, gravitational potential, escape speed and satellite orbits.
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The core idea
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Big question: Why is gravitational potential negative around an isolated mass?
Taking zero potential at infinity, a point mass gives φ = −GM/r and a test mass has Ep = mφ. The negative sign represents a bound system: energy must be supplied to reach infinity. Field strength is the negative potential gradient, so a steeper potential–distance graph means a stronger inward field.
Choose a zero that makes an isolated field useful
Gravitational potential at a point is the work done per unit mass by an external force in bringing a small test mass from infinity to that point. Equivalently, φ is potential energy per unit mass. For an isolated spherical mass, choosing φ = 0 at infinity gives φ = −GM/r outside the mass and Ep = mφ for a test mass.
The negative value means a mass in the field is bound: positive work must be done by an external agent to take it to infinity with no final kinetic energy. Moving farther away makes potential increase towards zero—it becomes less negative.
Check your understanding: Which is the higher potential, −20 MJ kg⁻¹ or −5 MJ kg⁻¹?
−5 MJ kg⁻¹ is higher because it is closer to the chosen zero at infinity.
Read field direction from a potential graph
Field strength is the negative potential gradient: g = −dφ/dr in one radial dimension. Around a positive source mass, φ rises with r, so dφ/dr is positive and g is negative relative to an outward-positive axis—directed inward.
Potential is scalar, so contributions from several masses add algebraically. Field strength is vector, so equal potentials do not imply equal fields and a point of zero field need not have zero potential.
Check your understanding: What does a steep potential–distance graph show?
A large magnitude of potential gradient and therefore a strong gravitational field.
Key ideas to keep
- Potential is energy per unit mass and is scalar; field strength is force per unit mass and is vector.
- Add potentials algebraically but add field strengths vectorially.
- Potential becomes less negative, not smaller in magnitude, as r increases.
See the reasoning
Worked example
Find work from a potential change
Question: A 500 kg satellite moves from radius 7.0 × 10⁶ m to 1.4 × 10⁷ m around Earth. Use GM = 3.99 × 10¹⁴ m³ s⁻². Find its gravitational potential-energy change.
Step 1: Find both potentials
Why: Potential uses centre distance and retains its negative sign.
Working: φi = −3.99×10¹⁴/(7.0×10⁶) = −5.70×10⁷ J kg⁻¹; φf = −2.85×10⁷ J kg⁻¹.
Step 2: Take final minus initial
Why: Energy change follows ΔEp = m(φf − φi).
Working: Δφ = (−2.85 − (−5.70))×10⁷ = +2.85×10⁷ J kg⁻¹.
Step 3: Multiply by satellite mass
Why: Potential is energy per unit mass.
Working: ΔEp = 500(2.85×10⁷) = 1.43×10¹⁰ J.
Answer: Gravitational potential energy increases by 1.43 × 10¹⁰ J.
Check: Moving outward makes potential less negative, so a positive energy change is physically sensible.
Another worked model
Question
A 600 kg spacecraft moves slowly from Earth's surface at 6.37 × 10⁶ m to r = 7.00 × 10⁶ m. Find the minimum external work, using GM = 3.99 × 10¹⁴ m³ s⁻².
Check the worked solution
φ₁ = −GM/r₁ = −6.26 × 10⁷ J kg⁻¹ and φ₂ = −5.70 × 10⁷ J kg⁻¹. For negligible kinetic-energy change, external work = ΔU = m(φ₂ − φ₁) = 600(5.64 × 10⁶) = 3.38 × 10⁹ J. Potential becomes less negative.
Use a hint if needed
Practise with support
Try this
Potential rises from −6.00 × 10⁷ to −5.80 × 10⁷ J kg⁻¹ over an outward radial interval of 2.00 × 10⁵ m. Estimate the radial field strength.
Hint: Calculate dφ/dr, then apply the negative-gradient sign.
Check your answer
dφ/dr ≈ [2.00 × 10⁶]/[2.00 × 10⁵] = +10.0 J kg⁻¹ m⁻¹. Hence g = −dφ/dr = −10.0 N kg⁻¹, where the negative sign means inward.
Now work without the hint
Practise independently
Your turn
For GM = 3.20 × 10¹⁴ m³ s⁻², find φ at r = 8.00 × 10⁶ m and the energy needed to move a 250 kg mass slowly from there to infinity.
Check your answer
φ = −GM/r = −4.00 × 10⁷ J kg⁻¹ and U = mφ = −1.00 × 10¹⁰ J. The minimum external work to reach zero potential energy at infinity is +1.00 × 10¹⁰ J.
Avoid these traps
Common mistakes
Common mistake
Gravitational potential is positive because gravity is attractive.
What is wrong with this reasoning?
Show better thinking
With zero at infinity, φ = −GM/r. A bound position has negative potential because external work is required to move a unit mass to infinity.
Common mistake
Field strength equals the positive slope of a potential–distance graph.
What is wrong with this reasoning?
Show better thinking
Radially, g = −dφ/dr. Potential rises towards zero as r increases, so its positive outward slope corresponds to an inward field.
Write for the examiner
Exam guidance
State the zero at infinity and preserve negative signs until the final energy interpretation.
Exam-style practice [7 marks]
Two point masses M and 4M are separated by distance 3d. Find the gravitational potential at the midpoint and explain why the gravitational field there is not zero.
Plan before you answer
- Add the two scalar potentials.
- Calculate or compare the two field vectors.
- Keep potential and field reasoning separate.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
At the midpoint each source is 1.5d away, so φ = −GM/(1.5d) − 4GM/(1.5d) = −10GM/(3d). The two field vectors point in opposite directions, but the field due to 4M is four times as large at the same distance. The resultant therefore points towards 4M and is not zero, even though potential was found by scalar addition.
Come back in three days
Check what stayed with you
Recall question 1
Where is gravitational potential conventionally zero for an isolated mass?
Check the answer
At infinity.
Recall question 2
Why is φ negative at finite r?
Check the answer
The mass is bound and external work is needed to move it to the zero at infinity.
Recall question 3
State the field–potential relationship.
Check the answer
Field strength is the negative potential gradient.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. The official topic states no explicit exclusions. Models use centre-to-centre distance, zero potential at infinity, point masses or spherical bodies viewed externally, and ideal circular orbits unless a question states otherwise.
- GCE A-Level H2 PhysicsTopic 8(e) / Topic 8(f) / Topic 8(g) / Topic 8(h) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027