9478 / 2027
Gravitational Fields overview

Lesson 2 of 5

Potential and potential energy

  • I can define gravitational potential as external work per unit test mass in bringing it from infinity to a point.Syllabus

    Syllabus 9478, 8(e). Define gravitational potential as external work per unit test mass in bringing it from infinity to a point.

  • I can use phi = -GM/r for point-mass gravitational potential.Syllabus

    Syllabus 9478, 8(f). Use phi = -GM/r for point-mass gravitational potential.

  • I can explain the gravitational potential energy U_G = -GMm/r of a two-point-mass system.Syllabus

    Syllabus 9478, 8(g). Explain the gravitational potential energy U_G = -GMm/r of a two-point-mass system.

Make a guess. You are not marked.

A satellite is moved to a higher circular orbit. What happens to the gravitational potential energy of the Earth-satellite system?

  1. It decreases, because the field is weaker further from the Earth.
  2. It increases, becoming less negative as the separation grows.
  3. It becomes positive once the satellite is above the atmosphere.
Show the answer

It increases, becoming less negative as the separation grows.

Gravitational potential energy is -GMm / r, taken as zero at infinity. A larger r makes it less negative, which is an increase. Work had to be done to lift the satellite.

Gravitational potential is an energy-per-mass quantity at a point. Gravitational potential energy belongs to a specified interacting system and also depends on the test body's mass.

Choose the zero at infinity

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, without a change in kinetic energy. Its unit is J/kg.

Set the potential to zero at infinite separation. This is an energy reference. Bringing the test mass inward slowly lets gravity do positive work. To prevent a kinetic-energy increase, the external agent does negative work, so the potential at the finite point is negative.

φ = -GM/r

This is the point-source potential and the exterior potential of a spherical source with zero at infinity. The radius is centre distance. Negative potential does not mean negative mass or an outward force: it describes the work relative to the chosen zero.

Multiply by mass to obtain system energy

For source mass M and test mass m separated by r, the gravitational potential energy is:

UG = mφ = -GMm/r

UG names the gravitational potential energy of the two-mass interaction, in joules. It is the gravitational case of the potential energy Ep used earlier, so Ep = UG in this account. It is not an additional energy store to add again.

Changing the small test mass at a fixed position leaves φ unchanged in this source-field model, but changes UG in proportion to that mass. Keep potential in J/kg distinct from potential energy in J.

Estimate the potential and energy scales

Using the rough inputs GM about 4 × 1014 m3/s2 and R about 6 × 106 m gives |φ| about 7 × 107 J/kg. For a roughly 400 kg test body, |UG| is of order 3 × 1010 J. Both signed values are negative with the infinity reference.

Worked outward transfer

A less negative potential energy is an increase

Use GM = 4.0 × 1014 m3/s2, R = 6.4 × 106 m and a 400 kg body. At the surface:

φ1 = -(4.0 × 1014)/(6.4 × 106)
= -6.25 × 107 J/kg
UG,1 = 400φ1 = -2.50 × 1010 J

At the larger centre distance r2 = 8.0 × 106 m:

φ2 = -(4.0 × 1014)/(8.0 × 106)
= -5.00 × 107 J/kg
UG,2 = 400φ2 = -2.00 × 1010 J

Moving outward raises the system's gravitational energy

For the 400 kg body, these are energy levels, not positions on a distance axis. The reference UG = 0 is the limit of indefinitely large separation.

Negative gravitational potential energy becomes less negative on an outward moveAn energy-level axis is labelled in ten to the ten joules, increasing upward. The infinity reference is zero; it is an energy reference rather than a finite radius drawn in space. The 400-kilogram system has gravitational potential energy negative 2.50 times ten to the ten joules at the surface, radius 6.4 million metres, and negative 2.00 times ten to the ten joules at radius eight million metres. Their vertical positions share an energy scale of one hundred drawing units per ten to the ten joules. An upward brown arrow between these levels shows the positive 0.50 times ten to the ten joule change. Gravity does the negative of that work. The diagram does not include final orbital kinetic energy.UG / 1010 J0Zero reference:r tends to infinity-2.0-2.5r = 8.0 × 106 mR = 6.4 × 106 m+0.50

ΔUG = +5.00 × 109 J; work by gravity is -5.00 × 109 J. This is a potential-energy change, not the complete energy needed to enter a circular orbit.

The two positions have different potentials and interaction energies. Moving outward makes the negative potential energy less negative. The zero at infinite separation is a limiting reference, not another finite position.
ΔUG = UG,2 - UG,1
= -2.00 × 1010 - (-2.50 × 1010)
= +5.00 × 109 J

The potential-energy increase is positive. It is the energy needed to raise this interaction energy between the two positions, before accounting for a kinetic-energy change or losses.

Try it step by step

Work it out in steps: Potential phi = -GM / r in J/kg, then Potential energy U = m phi in J.

  1. Problem 1. A 500 kg satellite is 7.0 x 106 m from the centre of the Earth. Take GM = 3.98 x 1014 N m2/kg. Find the gravitational potential there and the potential energy of the Earth-satellite system.
  2. Problem 2. A 200 kg probe is 1.0 x 107 m from the centre of the Earth. Use the same GM. Find the potential energy of the Earth-probe system.
  3. Problem 3. A 1000 kg satellite is 4.2 x 107 m from the centre of the Earth. Use the same GM. Find the potential and the potential energy.
Show the answer

Problem 1.

Potential phi = -GM / r in J/kg: phi = -3.98 x 1014 / 7.0 x 106 = -5.69 x 107 J/kg

Potential energy U = m phi in J: U = 500 x (-5.69 x 107) = -2.84 x 1010 J

Problem 2.

Potential phi = -GM / r in J/kg: phi = -3.98 x 1014 / 1.0 x 107 = -3.98 x 107 J/kg

Potential energy U = m phi in J: U = 200 x (-3.98 x 107) = -7.96 x 109 J

Problem 3.

Potential phi = -GM / r in J/kg: phi = -3.98 x 1014 / 4.2 x 107 = -9.48 x 106 J/kg

Potential energy U = m phi in J: U = 1000 x (-9.48 x 106) = -9.48 x 109 J

Check your understanding

At a point with gravitational potential -5.00 x 107 J/kg, a 400 kg test body is placed in the source field. With zero potential energy at infinite separation, which statement is correct?

  1. Potential remains -5.00 x 107 J/kg; the source-body potential energy is -2.00 x 1010 J.
  2. The potential becomes -2.00 x 1010 J/kg because the test mass is 400 kg.
  3. Potential energy is +2.00 x 1010 J because energy cannot be negative.
Show the answer

Potential remains -5.00 x 107 J/kg; the source-body potential energy is -2.00 x 1010 J.

U_G = m phi = 400(-5.00 x 107) = -2.00 x 1010 J. Distinguish potential per unit mass from the energy of the two-mass interaction.

Give work its correct sign and recipient

For movement from r1 to r2, subtract the two potential energies:

ΔUG = GMm(1/r1 - 1/r2)
Wgravity = -ΔUG

During the outward transfer above, gravity does -5.00 × 109 J of work because its force opposes the outward displacement. If an external agent moves the body with no kinetic-energy change and no other transfer, its work is +5.00 × 109 J.

If kinetic energy changes, the external work must account for that too. With no dissipative transfer, Wexternal = ΔEk + ΔUG. Include any relevant losses or other stores in a more complete account. The field-work relationship does not say every applied work equals ΔUG under all conditions.

In particular, raising the body to a satellite's radius is not enough to put it into a circular orbit. That final state also needs the orbital kinetic energy.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

Define the gravitational potential at a point. Explain why it is negative everywhere near the Earth. [2]

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