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Gravitational Potential & Gravitational Potential Energy

Key idea: Define gravitational potential φ as work done per unit mass from infinity, use φ = −GM/r and U = mφ, and apply g = −dφ/dr (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Relate gravitational potential, potential energy and field gradient.

1. Definitions (Must Know)

A. Gravitational potential, φ

Gravitational potential, φ, 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 (slowly, so its kinetic energy does not change).

For a point mass M:

φ = -GM/r

Unit: J kg⁻¹.

B. Reference level (zero at infinity)

By convention, φ = 0 at infinity. Since gravity is attractive, φ is negative everywhere in the field of an isolated mass.

Gravitational potential and its radial gradientA graph of gravitational potential phi against outward radial distance r starts negative near a mass and rises towards zero. Its positive slope corresponds to an inward, negative radial field component through g sub r equals minus d phi by d r.φ = 0φrdφ/dr > 0gᵣ < 0: inwardφ = −GM/r approaches zero
Scroll diagram horizontally to read all labels.
With outward chosen positive, φ rises towards zero as r increases, so dφ/dr is positive and the radial field component gᵣ = −dφ/dr is negative (inward).

Gravitational potential vs distance (Earth example)

Potential is negative and approaches zero as distance increases.

Scroll across the graph to read all labels.

Potential is negative and approaches zero as distance increases.Potential is negative and approaches zero as distance increases.
Using φ = -GM/r with GM_E ≈ 3.99 × 10¹⁴ m³ s⁻²: as r increases, φ becomes less negative and approaches 0.
Open full-size graph
View figure data
Values for Gravitational potential vs distance (Earth example)
Distance from Earth's centre, r (10⁷ m)φ = −GM/r (scaled)
0.637-6.26
1-3.99
2-2
4.22-0.945

C. Gravitational potential energy, U

Gravitational potential energy of a mass m at a point is:

U = mφ = -GMm/r

It is the energy of the two-body system (M and m) relative to zero at infinity.

D. Potential difference and work done

The potential difference between points A and B is:

Δφ = φ_B-φ_A

For slow movement (no change in kinetic energy), the work done by the external force is:

Wₑₓₜ = mΔφ = Δ U

2. Key Ideas (What Earns Marks)

  • φ is energy per unit mass (J kg⁻¹); U is energy (J).
  • For an isolated mass with φ = 0 at infinity, φ is negative at finite r; “higher potential” means “less negative”.
  • For a point mass: φ = -GM/r, U = -GMm/r
  • Field strength and potential are linked (syllabus): gᵣ = -dφ/dr
  • Moving to a higher orbit requires energy: r increases, φ increases (becomes less negative), so Δ U > 0.
Sign trap

Don’t panic when the numbers are negative. A “gain in potential” usually means the value becomes less negative (e.g. from -5 × 10⁷ to -3 × 10⁷).

3. Detailed Explanations

A. Why φ is negative

Using the definition with φ = 0 at infinity:

  • Gravity pulls the test mass inward.
  • To move the test mass slowly inward, the external force must act outward (opposite the displacement).
  • So the external force does negative work. Therefore φ is negative.

Equivalently: a bound system has U < 0 relative to infinity, so φ = U/m < 0.

B. Using potential to calculate energy changes

  1. Find φ at each position using φ = -GM/r.
  2. Compute Δφ = φ_B-φ_A.
  3. Multiply by mass to get energy change: Δ U = mΔφ.

If the object is moved slowly, Wₑₓₜ = Δ U.

Choose the outward radial direction as positive. The radial component of gravitational field strength is then

gᵣ = -dφ/dr.

If r is measured positive outward from the centre, then vector g points inward, so it has a negative radial sign.

From φ = -GM/r:

dφ/dr = +GM/r² ⇒ gᵣ = -GM/r².

So the magnitude of the field strength is:

| vector g| = GM/r²

Near Earth’s surface (uniform field model), the potential change over a small height change Δ h is approximately:

Δφ ≈ g Δ h

so Δ U ≈ mg Δ h.

D. Equipotential surfaces

An equipotential surface is a surface where φ is constant.

  • Moving along an equipotential: Δφ = 0 ⇒ Δ U = 0 ⇒ Wₑₓₜ = 0.
  • Gravitational field lines are perpendicular to equipotential surfaces.
  • Work done by gravity is W_g = -Δ U = -mΔφ.

4. Common Mistakes

  • Missing the negative sign in φ = -GM/r or U = -GMm/r.
  • Using radius/altitude wrongly: r is centre-to-centre; for altitude h, r = R_E + h.
  • Confusing φ (per unit mass) with U (total energy).
  • Forgetting that “increase in potential” can mean “less negative”.

5. Exam Tips

  • State the reference: “φ = 0 at infinity”.
  • If the question asks for energy required to raise orbit: use Δ U = m(φ₂-φ₁) and expect Δ U > 0.
  • If you use gᵣ = -dφ/dr, define outward as positive so the negative radial component means inward.

6. Worked Examples

Modelled example 1

Gravitational potential at a given r

Core

Problem

Find the gravitational potential at 1.00 × 10⁷ m from Earth’s centre. Take M_E = 5.97 × 10²⁴ kg and zero potential at infinity.
Study the worked solution

  1. Choose the potential model

    Method

    Use φ = -GM/r.

    Reason

    Outside a spherical Earth, its mass may be treated as concentrated at its centre.

    Working

    φ = -GM/r

  2. Substitute

    Method

    Keep the reference-dependent negative sign.

    Reason

    With zero at infinity, a finite point in an attractive field has negative potential.

    Working

    φ = -(6.67 × 10⁻¹¹)(5.97 × 10²⁴)/(1.00 × 10⁷)

  3. Evaluate

    Method

    Obtain -3.98 × 10⁷ J kg⁻¹.

    Reason

    Potential is energy per unit mass.

    Working

    φ ≈ -3.98 × 10⁷ J kg⁻¹

Guided practice 2

Potential energy from U = mφ

About 4 min

Problem

A 1000 kg satellite is where φ = -3.98 × 10⁷ J kg⁻¹. Find the Earth–satellite system’s gravitational potential energy relative to infinity.

Try this before viewing the solution

Unit: J

Hints

Hint 1: distinguish phi from U
Use U = mφ; the result is in joules.
View solution step by step

  1. Link potential and energy

    Method

    Use U = mφ.

    Reason

    Potential is gravitational potential energy per unit test mass.

    Working

    U = mφ

  2. Evaluate

    Method

    Obtain -3.98 × 10¹⁰ J.

    Reason

    The negative sign identifies a bound two-body system relative to infinite separation.

    Working

    U = (1000)(-3.98 × 10⁷) = -3.98 × 10¹⁰ J

Common misconception 3

Energy needed to raise a satellite’s orbit

Find and correct the mistake

Learner claim

A 700 kg satellite is moved slowly from altitude 200 km to 1000 km. Take R_E = 6.37 × 10⁶ m and GM_E = 3.99 × 10¹⁴ m³ s⁻². A learner says the external work is negative because both endpoint potentials are negative. Diagnose and calculate.

Try this before viewing the solution

Sign of external work

View solution step by step

  1. Convert both altitudes

    Method

    Use centre distances r₁ = 6.57 × 10⁶ m and r₂ = 7.37 × 10⁶ m.

    Reason

    Potential depends on distance from Earth’s centre.

    Working

    r₁ = R_E + 2.00 × 10⁵, r₂ = R_E + 1.00 × 10⁶

  2. Find the potential increase

    Method

    Obtain Δφ ≈ +6.6 × 10⁶ J kg⁻¹.

    Reason

    The final potential is less negative, so φ₂-φ₁ is positive.

    Working

    Δφ = GM(1/r₁-1/r₂) ≈ 6.6 × 10⁶ J kg⁻¹

  3. Find slow-lift work

    Method

    Obtain Wₑₓₜ ≈ 4.6 × 10⁹ J.

    Reason

    With negligible kinetic-energy change, external work equals Δ U = mΔφ.

    Working

    Wₑₓₜ = (700)(6.6 × 10⁶) ≈ 4.6 × 10⁹ J

Examiner practice 4

Using gᵣ = -dφ/dr

3 marks

Examination question

For φ = -GM/r, use gᵣ = -dφ/dr to determine the radial field component at r = 4.20 × 10⁷ m, where GM_E = 3.99 × 10¹⁴ m³ s⁻². Take outward as positive. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Differentiate potential

    1 mark

    Method

    dφ/dr = +GM/r².

    Reason

    Differentiating -GM r⁻¹ produces a positive derivative.

    Working

    dφ/dr = +GM/r²

  2. Apply the gradient sign

    1 mark

    Method

    gᵣ = -GM/r².

    Reason

    The negative-gradient relation makes the component negative when outward is positive.

    Working

    gᵣ = -(3.99 × 10¹⁴)/((4.20 × 10⁷)²)

  3. Evaluate and interpret

    1 mark

    Method

    gᵣ ≈ -0.23 N kg⁻¹.

    Reason

    The negative component means the field points inward; its magnitude is 0.23 N kg⁻¹.

    Working

    gᵣ ≈ -0.23 N kg⁻¹

Challenge 5

Potential difference and energy change between two radii

Minimal support

Independent transfer

A satellite of mass m moves from radius r₁ to r₂ around a planet of mass M. Derive expressions for Δφ = φ₂-φ₁ and Δ U, then state the sign when r₂ > r₁.

Try this before viewing the solution

Hints

Hint 1: write both endpoint potentials
Begin with φ₁ = -GM/r₁ and φ₂ = -GM/r₂.
View solution step by step

  1. Subtract endpoint potentials

    Method

    Obtain Δφ = GM(1/r₁-1/r₂).

    Reason

    Potential change is final minus initial.

    Working

    Δφ = -GM/r₂ + GM/r₁ = GM(1/r₁-1/r₂)

  2. Convert to system energy

    Method

    Multiply by satellite mass.

    Reason

    U = mφ at each point.

    Working

    Δ U = mGM(1/r₁-1/r₂)

  3. Interpret outward motion

    Method

    If r₂ > r₁, both changes are positive.

    Reason

    1/r₁ > 1/r₂, so the bracket is positive.

    Working

    The potential and potential energy increase by becoming less negative.

7. Mind Stretchers

Mind stretcher 1: Radius when potential is halvedExtension

At what radius is the gravitational potential half (same sign) of its value at radius r? (Express your answer in terms of r.)

Show answer

Since φ = -GM/r:

φ'/φ = (-GM/r')/(-GM/r) = r/r'

For φ' = (1/2)φ, we need r/r' = 1/2, so r' = 2r.

Mind stretcher 2: Why U stays constant in a circular orbitExtension

Explain why an object in circular orbit has constant gravitational potential energy even though a gravitational force acts on it.

Show answer

For a circular orbit, r is constant, so φ = -GM/r and U = mφ are constant.

Also, gravity points towards the centre while the instantaneous displacement is tangential, so the gravitational force does no work on the object in uniform circular motion.

Mind stretcher 3: Optional (Enrichment)Extension

A. Force as a potential gradient (calculus form)

Sometimes you will see:

Fᵣ = -dU/dr

This is the radial component of the gravitational force derived from how U changes with r. For A Level, the key examinable gradient relationship is gᵣ = -dφ/dr when outward is positive.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027