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EnglandPhysicsSyllabus dot point

How much energy does it take to move a mass within a gravitational field?

Gravitational potential and potential energy in a radial field, the potential gradient, equipotential surfaces, and the work done moving a mass between points.

A focused answer to AQA A-Level Physics 3.7.2.3, covering gravitational potential in a radial field, gravitational potential energy, the potential gradient and its link to field strength, equipotential surfaces, and the work done moving a mass.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. Gravitational potential
  3. Gravitational potential energy
  4. Potential gradient and field strength
  5. Equipotentials
  6. Try this

What this dot point is asking

AQA specification point 3.7.2.3 wants you to define gravitational potential, calculate it in a radial field, relate the potential gradient to field strength, describe equipotential surfaces, and calculate the work done moving a mass between two points.

Gravitational potential

The potential is negative everywhere and rises towards zero at infinity, because gravity is attractive and we define infinity as the zero of potential. The closer to the mass, the deeper (more negative) the potential well, which is why escaping a planet requires a large input of energy. Unlike electric potential, gravitational potential can never be positive, because there is no repulsive gravity.

Gravitational potential energy

The potential energy of a mass mm at a point of potential VV in the field is the work needed to bring it there from infinity:

The work done moving a mass between two points is W=mΔV=m(V2−V1)W = m\Delta V = m(V_2 - V_1), which depends only on the start and end positions because gravity is a conservative force. Lifting a mass to a higher (less negative) potential requires positive work to be done on it.

Potential gradient and field strength

Field strength is the negative of the potential gradient. The minus sign shows the field points from high to low potential, that is, towards the mass producing the field, where the potential is most negative. A steep potential gradient means a strong field.

Equipotentials

Try this

Q1. Explain why gravitational potential is always negative. [2 marks]

  • Cue. The field is attractive and the zero is at infinity, so energy is released as a mass moves inward, giving negative potential.

Q2. State the relationship between gravitational field strength and gravitational potential. [1 mark]

  • Cue. Field strength is the negative potential gradient, g=−ΔVΔrg = -\dfrac{\Delta V}{\Delta r}.

Q3. State the value the gravitational potential approaches as rr tends to infinity. [1 mark]

  • Cue. Zero.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20185 marksA satellite of mass 1200 kg1200 \text{ kg} is raised from a circular orbit at 7.0×106 m7.0 \times 10^6 \text{ m} from the Earth's centre to one at 1.0×107 m1.0 \times 10^7 \text{ m}. Taking GM=3.99×1014 m3 s−2GM = 3.99 \times 10^{14} \text{ m}^3 \text{ s}^{-2}, calculate the change in gravitational potential energy of the satellite.
Show worked answer →

Find the potential at each radius using V=−GMrV = -\dfrac{GM}{r}.

V1=−3.99×10147.0×106=−5.70×107 J kg−1V_1 = -\dfrac{3.99 \times 10^{14}}{7.0 \times 10^6} = -5.70 \times 10^7 \text{ J kg}^{-1}.

V2=−3.99×10141.0×107=−3.99×107 J kg−1V_2 = -\dfrac{3.99 \times 10^{14}}{1.0 \times 10^7} = -3.99 \times 10^7 \text{ J kg}^{-1}.

Change in potential energy ΔEp=m(V2−V1)=1200×(−3.99×107−(−5.70×107))=1200×1.71×107=2.1×1010 J\Delta E_p = m(V_2 - V_1) = 1200 \times (-3.99 \times 10^7 - (-5.70 \times 10^7)) = 1200 \times 1.71 \times 10^7 = 2.1 \times 10^{10} \text{ J}.

Markers reward the negative potentials, finding the difference correctly (the energy increases as the satellite rises), and multiplying by the mass.

AQA 20213 marksExplain why gravitational potential is always negative, and state what happens to the potential as the distance from a mass tends to infinity.
Show worked answer →

The zero of gravitational potential is defined at infinity, where a test mass is beyond the field. Because gravity is attractive, work is done by the field (energy is released) as a mass moves inward from infinity, so the potential at any finite point is less than zero, that is, negative.

As the distance tends to infinity, V=−GMrV = -\dfrac{GM}{r} tends to zero from below, so the potential rises towards zero.

Markers reward defining the zero at infinity, linking the negative sign to the attractive field and energy released on approach, and stating V→0V \to 0 as r→∞r \to \infty.

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