Gravitational potential (A-level only) — AQA A-Level Physics
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Gravitational potential (V) at a point is the work done per unit mass to move a small test mass from infinity to that point, defined as V = -GM/r. It is a scalar quantity, always negative, and its gradient gives the gravitational field strength g = -dV/dr.
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Gravitational potential (A-level only) explained
Gravitational potential (A-level only) covers 5 specification points, including Understanding of definition of gravitational… and Understanding of gravitational potential….
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Open a point below for its explanation and revision guidance.
Understanding of definition of gravitational potential, including zero value at infinity.
Gravitational potential, V, at a point is the work done per unit mass in bringing a small test mass from infinity to that point. Because the gravitational force is attractive, the work done by the field is positive when moving from infinity, so the potential is negative near a mass and increases towards zero at infinity. The defining equation is V = -GM/r for a point mass or uniform sphere, where G is the gravitational constant, M is the mass producing the field, and r is the distance from the centre. The zero value is chosen at infinity, where the field strength is zero. For example, at the surface of the Earth, using M = 5.97 × 10²⁴ kg and r = 6.37 × 10⁶ m gives V ≈ -6.25 × 10⁷ J kg⁻¹. Potential is a scalar, so values add algebraically.
Understanding of gravitational potential difference.
Gravitational potential difference, ΔV, between two points is the difference in gravitational potential between them, equal to the work done per unit mass in moving a small test mass between the two points. It is calculated as ΔV = V_final - V_initial, where each V is found from V = -GM/r. Because potential is a scalar, the difference is simply the algebraic subtraction of the two potential values. For example, moving from r = 2R to r = R near a planet gives ΔV = (-GM/R) - (-GM/2R) = -GM/2R, which is negative, meaning the potential decreases as the test mass moves closer. The magnitude of ΔV is the energy transferred per unit mass, and the sign indicates the direction of potential change. This concept is essential for understanding changes in potential energy and orbital motion.
Work done in moving mass m given by ∆W = m∆V Equipotential surfaces.
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.
Idea that no work is done when moving along an equipotential surface.
An equipotential surface is a surface on which the gravitational potential V is the same at every point. If a mass m moves from one point on the surface to another, the potential difference ∆V is zero, so the change in potential energy m∆V is zero. The gravitational force acts perpendicular to the equipotential surface, while the displacement along the surface is tangential, so the angle between force and displacement is 90°. Since work done is force × displacement × cos θ and cos 90° = 0, the gravitational field does no work on the mass. This is why no energy is transferred when a satellite moves in a circular orbit, because its path lies along an equipotential. It also explains why equipotential surfaces are always perpendicular to field lines.
Graphical representations of variations of g and V with r.
This topic requires you to sketch and interpret graphs of gravitational field strength g and gravitational potential V against distance r from a point mass or uniform sphere. For a point mass, the magnitude of g is GM/r² and V = −GM/r. The radial component of g is negative, but its magnitude falls as 1/r², while V is always negative and rises towards zero as 1/r. Inside a uniform sphere, g's magnitude increases linearly with r from zero at the centre, whereas V becomes more negative towards the centre (e.g. V at centre is −3GM/2R). You should identify asymptotes, signs, gradients and discontinuities, and use the graphs to compare fields and potentials at different distances.
Your focus
- State the definition of gravitational potential, including the zero reference at infinity.
- Calculate gravitational potential at a point using V = -GM/r.
- Explain why gravitational potential is negative near a mass and how it changes with distance.
Show all 15 objectives
- Define gravitational potential difference between two points.
- Calculate gravitational potential difference using V = -GM/r for each point.
- Interpret the sign and magnitude of gravitational potential difference in terms of energy transfer per unit mass.
- Calculate the work done in moving a mass between two points using ∆W = m∆V and interpret the sign of the result.
- Derive and use ∆V = GM(1/r₁ − 1/r₂) for movement in a radial gravitational field.
- Describe the shape and spacing of equipotential surfaces around a point mass and relate them to field lines.
- Explain why the work done on a mass moving along an equipotential surface is zero, using ∆W = m∆V and the perpendicularity of force and displacement.
- Deduce that a satellite in a circular orbit has constant speed because gravity does no work on it.
- Sketch equipotential surfaces perpendicular to gravitational field lines and interpret their spacing in terms of field strength.
- Sketch graphs of g and V against r for a point mass and for a uniform sphere, showing correct signs and asymptotic behaviour.
- Relate the gradient of a V–r graph to gravitational field strength and use it to compare values.
- Describe how g and V vary inside a uniform sphere and contrast this with the behaviour outside the sphere.
Which command word is this?
Give a precise, formal definition. For gravitational potential, you must state 'work done per unit mass' and 'from infinity to the point'. Include the equation if appropriate.
Gravitational potential (A-level only) exam tips
Topic Overview
Gravitational potential is a fundamental concept in A-level Physics that describes the energy per unit mass in a gravitational field. It allows you to calculate the work done in moving masses and understand orbits, escape velocity, and energy changes without dealing with forces directly. The topic builds on Newton's law of gravitation and introduces the idea of potential fields, which is essential for later study of electric fields.
In the AQA A-level specification, gravitational potential is a key part of the 'Fields' topic. You must be able to define potential, use the formula V = -GM/r, relate it to potential energy, and interpret potential-distance graphs. This concept also explains why gravitational potential is negative and why the gradient of a potential-distance graph gives the gravitational field strength.
Key Concepts
- →Gravitational potential V at a point is the work done per unit mass to bring a small test mass from infinity to that point: V = -GM/r.
- →Gravitational potential is a scalar quantity and is always negative for a point mass or spherical mass, with zero at infinity.
- →Gravitational potential energy U of a mass m in a field is U = mV = -GMm/r.
- →The gravitational field strength g is the negative gradient of the potential: g = -dV/dr.
- →The potential difference between two points is the work done per unit mass to move between them: ΔV = V_final - V_initial.
Marking Points
- Defines gravitational potential as work done per unit mass in moving a small test mass from infinity to the point.
- States that the zero of gravitational potential is at infinity.
- Explains that gravitational potential is negative near a mass because the gravitational force is attractive and the field does positive work as the test mass moves from infinity.
- Uses V = -GM/r correctly for a point mass or uniform sphere, with r measured from the centre.
- Recognises that gravitational potential is a scalar quantity and that potentials from multiple masses add algebraically.
- Calculates potential at a point using given values of G, M and r, including correct substitution and units J kg⁻¹.
- Defines gravitational potential difference as the difference in gravitational potential between two points.
- Calculates potential difference using ΔV = V_final - V_initial, with each V obtained from V = -GM/r.
- Interprets the sign of ΔV in terms of whether potential increases or decreases between the two points.
- Relates gravitational potential difference to the work done per unit mass in moving between the two points.
- Applies the concept to problems involving changes in distance from a mass, such as moving between two orbital radii.
- Uses the scalar nature of potential to add or subtract potentials algebraically without vector resolution.
- 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.
- Define an equipotential surface as a surface where the gravitational potential V is constant, so ∆V = 0 between any two points on it.
- Use ∆W = m∆V to show that if ∆V = 0 then ∆W = 0, so no work is done by or against the gravitational field.
- Explain that the gravitational force is perpendicular to the equipotential surface, so the component of force along the displacement is zero.
- Apply the work formula W = Fs cos θ and state that θ = 90° for motion along an equipotential, giving cos 90° = 0.
- Link the zero work to a circular orbit, where the gravitational force provides the centripetal force and the satellite's speed remains constant.
- Describe how equipotential surfaces are drawn perpendicular to field lines, with closer spacing where the field is stronger.
- Correctly sketches the magnitude of g against r outside a point mass as a positive curve proportional to 1/r², tending to zero as r tends to infinity.
- Correctly sketches V against r outside a point mass as a negative curve proportional to −1/r, approaching zero from below as r tends to infinity.
- States that inside a uniform sphere the magnitude of g is proportional to r and reaches zero at the centre, while V is finite and becomes more negative towards the centre.
- Uses the gradient of the V–r graph to obtain the radial component of gravitational field strength, recognising that g_r = −dV/dr.
- Interprets the area under a g–r graph between two radii as the change in gravitational potential, with an appropriate sign.
Examiner Tips
- 💡When defining gravitational potential, always include the phrase 'work done per unit mass' and specify the path from infinity.
- 💡In calculations, write the equation V = -GM/r, substitute values with units, and give the answer in J kg⁻¹ with an appropriate number of significant figures.
- 💡If a question asks about the zero of potential, state clearly that it is at infinity and explain that this choice makes the potential negative near a mass.
- 💡When calculating potential difference, show the two potential values clearly before subtracting, and state the sign of the result.
- 💡If asked for the work done per unit mass, remember it is equal to the negative of the potential difference, or simply the magnitude of ΔV depending on context.
- 💡Use consistent units throughout, and express potential difference in J kg⁻¹.
- 💡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.
- 💡In explanations, explicitly state that ∆V = 0 on an equipotential and substitute into ∆W = m∆V to justify zero work.
- 💡Use the perpendicularity argument with a clear diagram showing the force arrow at 90° to the equipotential surface and the displacement along it.
- 💡When discussing circular orbits, link constant speed to zero work done by gravity, avoiding any claim that gravity is absent.
- 💡Define equipotential surfaces in words before using them in calculations or explanations, so the meaning of ∆V = 0 is unambiguous.
- 💡Label axes with the quantity and symbol, and mark the asymptotic behaviour towards the r-axis clearly.
- 💡Use dashed lines or a separate sketch to distinguish the region inside a uniform sphere from the region outside it.
- 💡When asked to compare graphs, refer explicitly to signs, powers of r and limiting values rather than describing shapes vaguely.
- 💡Always define gravitational potential with the key phrases 'work done per unit mass' and 'from infinity'. Missing either loses marks.
- 💡When calculating potential or potential energy, ensure you use the correct radius: for objects on or above a planet's surface, r is the distance from the centre, so add the planet's radius to the altitude.
- 💡In graph questions, remember that the gradient of a V-r graph gives -g, and the area under a g-r graph gives the change in potential.
Common Mistakes
- Treating gravitational potential as a vector and attempting to resolve it into components; the correction is to treat it as a scalar and add values algebraically.
- Omitting the negative sign in V = -GM/r; the correction is to include the negative sign because the potential is negative near a mass when zero is at infinity.
- Confusing gravitational potential with gravitational field strength; the correction is to remember that potential is work done per unit mass (J kg⁻¹) while field strength is force per unit mass (N kg⁻¹).
- Subtracting potentials in the wrong order, giving the wrong sign for ΔV; the correction is to use ΔV = V_final - V_initial consistently.
- Treating potential difference as a vector quantity; the correction is to treat it as a scalar and subtract algebraically.
- Forgetting that potential is negative, leading to sign errors when subtracting; the correction is to keep the negative signs in the calculation and simplify carefully.
- Treating gravitational potential V as always positive: correct this 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: correct this 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: correct this 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: correct this by checking whether the displacement has a component along the field; along an equipotential there is none.
- Claiming that no force acts on a mass on an equipotential surface: correct this by stating that the gravitational force still acts, but it is perpendicular to the displacement.
- Thinking that no work is done because the mass is stationary: correct this by explaining that the mass may move at constant speed along the surface while the field does zero work.
- Assuming potential energy is zero on an equipotential: correct this by stating that potential energy mV is constant but not necessarily zero, since V is measured relative to infinity.
- Confusing equipotential surfaces with surfaces of constant field strength: correct this by noting that equipotentials join points of equal potential, while field strength depends on the spacing of equipotentials.
- Drawing V as positive outside the mass; the correction is that gravitational potential is negative because the zero is defined at infinity.
- Treating g as proportional to 1/r rather than 1/r²; the correction is that the inverse-square law gives a steeper fall for g than for the magnitude of V.
- Stating V becomes less negative inside a uniform sphere; the correction is that V becomes more negative towards the centre, reaching a minimum at r=0.
- Students often think gravitational potential is positive because they confuse it with potential energy or forget the negative sign. Correction: Potential is negative because the gravitational force is attractive and zero potential is at infinity.
- Many believe that gravitational potential and gravitational field strength are the same thing. Correction: Potential is a scalar (J kg^-1) while field strength is a vector (N kg^-1); they are related by g = -dV/dr.
- Some think that potential energy is always negative, but it can be positive if the reference point is not infinity. Correction: In A-level, we always take zero potential at infinity, so U is negative for bound systems.
Revision Plan
- 1Day 1: Learn the definition of gravitational potential and the formula V = -GM/r. Practice simple calculations for a point mass.
- 2Day 2: Understand the relationship between potential and potential energy (U = mV). Solve problems involving both.
- 3Day 3: Study potential-distance graphs and how to determine field strength from the gradient. Sketch graphs for different configurations.
- 4Day 4: Apply concepts to orbits and escape velocity. Calculate changes in potential energy for moving between orbits.
- 5Day 5: Complete past paper questions on gravitational potential, focusing on definitions, calculations, and graph interpretation. Review mistakes.
Exam Question Types
- 📋Definition and explanation questions: 'Define gravitational potential and explain why it is negative.' Use precise wording and refer to infinity and attractive forces.
- 📋Calculation questions: 'Calculate the gravitational potential at a point' or 'Find the change in potential energy.' Show all steps and units.
- 📋Graph interpretation: 'Sketch a graph of potential against distance and explain its shape.' Label axes, show asymptotic behaviour, and mention the gradient.
- 📋Comparison questions: 'Compare gravitational potential and gravitational field strength.' Highlight scalar vs vector, units, and relationship.
Command Word Expectations (AQA)
Give a precise, formal definition. For gravitational potential, you must state 'work done per unit mass' and 'from infinity to the point'. Include the equation if appropriate.
Show the formula, substitute correct values, and give the answer with units and appropriate significant figures. Include negative signs where necessary.
Provide a reasoned account linking cause and effect. For example, explain why potential is negative due to the attractive nature of gravity and the zero reference at infinity.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: Calculate the gravitational potential at the surface of the Earth. Mass of Earth = 5.97e24 kg, radius = 6.37e6 m, G = 6.67e-11 N m^2 kg^-2.
- 1.Step 1: Identify given values: M = 5.97e24 kg, r = 6.37e6 m, G = 6.67e-11 N m^2 kg^-2.
- 2.Step 2: Use the formula V = -GM/r.
- 3.Step 3: Substitute: V = -(6.67e-11 * 5.97e24) / (6.37e6).
- 4.Step 4: Calculate numerator: 6.67e-11 * 5.97e24 = 3.98e14.
- 5.Step 5: Divide by r: 3.98e14 / 6.37e6 = 6.25e7 J kg^-1.
- 6.Step 6: Include negative sign: V = -6.25e7 J kg^-1.
Question: A satellite of mass 500 kg is in a circular orbit at an altitude of 400 km above the Earth's surface. Calculate the gravitational potential energy of the satellite. (Earth mass = 5.97e24 kg, Earth radius = 6.37e6 m, G = 6.67e-11 N m^2 kg^-2).
- 1.Step 1: Calculate orbital radius: r = Earth radius + altitude = 6.37e6 + 4.00e5 = 6.77e6 m.
- 2.Step 2: Use gravitational potential energy formula: U = -GMm/r.
- 3.Step 3: Substitute values: U = -(6.67e-11 * 5.97e24 * 500) / (6.77e6).
- 4.Step 4: Calculate numerator: 6.67e-11 * 5.97e24 * 500 = 1.99e17.
- 5.Step 5: Divide by r: 1.99e17 / 6.77e6 = 2.94e10 J.
- 6.Step 6: Include negative sign: U = -2.94e10 J.