Work done in moving mass m given by ∆W = m∆V Equipotential surfaces.

The idea

Gravitational potential V at a point is the work done per unit mass in bringing a small test mass from infinity to that point, so V is negative near a mass. Moving a mass m between two points with potentials V₁ and V₂ changes its potential energy by m∆V, where ∆V = V₂ − V₁. The work done by the gravitational field equals −m∆V; the external work needed is +m∆V. For a radial field, V = −GM/r, so ∆V = GM(1/r₁ − 1/r₂). An equipotential surface joins points of equal V; around a point mass these are concentric spheres. Because V is constant on such a surface, m∆V = 0, so no work is done moving along it. This links potential difference to energy transfer and explains why equipotentials are always perpendicular to field lines.

What earns the marks

• State that gravitational potential V is the work done per unit mass in bringing a test mass from infinity to the point, giving V = −GM/r for a radial field.
• Use ∆W = m∆V with ∆V = V₂ − V₁, recognising that the sign of ∆W depends on the direction of movement relative to the field.
• Explain that the gravitational force is conservative, so work done depends only on the initial and final potentials, not on the path taken.
• Describe an equipotential surface as a surface on which every point has the same gravitational potential, so ∆V = 0 between any two points on it.
• Show that moving mass m along an equipotential surface gives ∆W = m × 0 = 0, meaning no work is done against or by the gravitational field.
• Relate equipotential surfaces to field lines: they are perpendicular, and the field does no work along a surface because displacement is perpendicular to the force.
• Apply V = −GM/r to calculate ∆V between two radii and hence the energy needed to move a satellite between orbits.

Common mistakes and how to put them right

• Treating gravitational potential V as always positive.

  Put it right by stressing that V is negative for a bound mass because the zero is at infinity and the field attracts.

• Using ∆W = m∆V but ignoring the sign, so the direction of energy transfer is wrong.

  Put it right by defining ∆V = V_final − V_initial and interpreting a negative ∆W as work done by the field.

• Confusing gravitational potential V with gravitational potential energy.

  Put it right by stating that V is energy per unit mass (J kg⁻¹) while potential energy is mV (J).

• Assuming work is done whenever a mass moves.

  Put it right by checking whether the displacement has a component along the field; along an equipotential there is none.

Exam tips

• Always write the defining equation ∆W = m∆V before substituting values, and state clearly whether the work is done by the field or by an external agent.
• When calculating ∆V for a radial field, use ∆V = GM(1/r₁ − 1/r₂) and keep the negative sign convention consistent throughout.
• Sketch equipotential surfaces as dashed circles or spheres around a point mass and draw field lines crossing them at 90°, labelling the potential values.
• In explanation questions, link the zero work along an equipotential to the perpendicularity of force and displacement, not just to equal potential values.