Gravitational potential and energy — OCR A-Level Physics
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Gravitational potential and energy explained
Gravitational potential at a point is defined as the work done per unit mass in bringing a small test mass from infinity to that point.
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It is a scalar quantity, measured in joules per kilogram (J kg⁻¹). Because the gravitational force is attractive, the work done by the gravitational field is positive when moving from infinity to a point, so the potential is negative. At infinity, the potential is defined as zero. For example, near Earth's surface, the potential is approximately -GM/r, where M is Earth's mass and r is the distance from Earth's centre. The negative sign indicates that energy is released as the mass moves closer. This concept is fundamental for calculating gravitational potential energy and escape velocity.
(b) gravitational potential V g r GM =- at a distance r from a point mass M ; changes in gravitational potential
Gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point, giving V = −GM/r for a point mass M at distance r. The negative sign shows that potential is zero at infinity and decreases as you approach the mass, so a test mass falls spontaneously towards M. For a spherical mass of uniform density, the same expression applies outside the sphere, with r measured from the centre. A change in potential ΔV = V_final − V_initial equals the work done per unit mass moving between two points without change in kinetic energy. For example, moving from r = 2R to r = R near a planet changes V from −GM/2R to −GM/R, a change of −GM/2R, meaning potential decreases and gravitational potential energy is released.
(c) force–distance graph for a point or spherical mass; work done is area under graph
For a point mass or a spherical mass treated as a point at its centre, the gravitational force on a test mass varies with distance r as F = GMm/r², an inverse-square relationship. A force–distance graph plots this force on the vertical axis against separation r on the horizontal axis, producing a curve that falls steeply as r increases. The work done by the gravitational field in moving the test mass between two separations equals the area under the force–distance graph between those limits, with the sign indicating whether the field does work or work is done against it. Because the curve is not linear, the area must be found by integration, by counting squares, or by using the potential difference: work done per unit mass equals the change in potential. For example, the area under F = GMm/r² from r = R to r = 2R gives the energy released as the mass falls inward.
(d) gravitational potential energy E mV r GMm g = =- at a distance r from a point mass M
Gravitational potential energy of a mass m at distance r from a point mass M is E = mV = −GMm/r, where V = −GM/r is the gravitational potential at that point. The expression applies outside a spherical mass, with r measured from the centre. The negative sign means the potential energy is zero at infinity and decreases as the mass moves closer, so energy is released when a mass falls towards M. Because E is inversely proportional to r, doubling the separation from r to 2r changes E from −GMm/r to −GMm/2r, an increase of GMm/2r, which equals the energy needed to move the mass outward. For example, a satellite of mass m in orbit at radius r has E = −GMm/r, and the energy required to move it to a higher orbit is the difference between the two values.
(e) escape velocity.
Escape velocity is the minimum speed a projectile needs at the surface of a mass M to escape its gravitational field and reach infinity with zero kinetic energy. At the surface, radius R, the total energy is ½mv² − GMm/R. Setting this equal to zero at infinity gives ½mv² = GMm/R, so v = √(2GM/R). The mass m cancels, so escape velocity depends only on the mass and radius of the body, not on the projectile. For example, for a planet of mass 6.0 × 10²⁴ kg and radius 6.4 × 10⁶ m, v = √(2 × 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ / 6.4 × 10⁶) ≈ 1.1 × 10⁴ m s⁻¹. Escape velocity is a scalar speed, and the direction of launch is ignored in this idealised treatment.
Your focus
- Define gravitational potential and state its unit.
- Explain why gravitational potential is zero at infinity and negative near a mass.
- Calculate gravitational potential using V = -GM/r.
Show all 15 objectives
- Recall and use V = −GM/r for a point mass or outside a spherical mass.
- Explain the physical meaning of gravitational potential and its zero reference at infinity.
- Calculate changes in gravitational potential between two distances from a mass.
- Sketch and interpret a force–distance graph for a point or spherical mass.
- Explain why work done equals the area under a force–distance graph.
- Relate the area under the graph to changes in gravitational potential energy.
- Recall and use E = mV = −GMm/r for a point mass or outside a spherical mass.
- Explain the meaning of the negative sign in gravitational potential energy.
- Calculate changes in gravitational potential energy between two separations.
- Derive and use v = √(2GM/R) for escape velocity.
- Explain why escape velocity is independent of the mass of the escaping object.
- Calculate escape velocity for a given mass and radius using SI units.
Gravitational potential and energy exam tips
Marking Points
- Define gravitational potential as work done per unit mass in bringing a mass from infinity to the point.
- State that gravitational potential is zero at infinity.
- Recognise that gravitational potential is negative for a point near a mass.
- Use the equation V = -GM/r for a point outside a spherical mass.
- Explain that gravitational potential is a scalar quantity with units J kg⁻¹.
- States V = −GM/r for a point mass M at distance r, with r measured from the centre of the mass.
- Explains that gravitational potential is defined as work done per unit mass in bringing a test mass from infinity to the point.
- Interprets the negative sign: V is zero at infinity and becomes more negative closer to M.
- Calculates a change in potential as ΔV = V_final − V_initial, e.g. −GM/R − (−GM/2R) = −GM/2R.
- Applies the expression outside a spherical mass, treating it as a point mass at its centre.
- Sketchs or interprets a force–distance graph showing F = GMm/r² decreasing with increasing r.
- States that work done is represented by the area under the force–distance graph between two separations.
- Recognises that the area cannot be found by a simple triangle or rectangle rule because the curve is non-linear.
- Links the area under the graph to the change in gravitational potential energy, or per unit mass to the change in potential.
- Applies the result to a point mass or to a spherical mass outside its surface, using r from the centre.
- States E = mV = −GMm/r for a mass m at distance r from a point mass M.
- Explains that E is the product of the mass and the gravitational potential at that point.
- Interprets the negative sign: E is zero at infinity and more negative closer to M.
- Calculates changes in gravitational potential energy as ΔE = E_final − E_initial using signed values.
- Applies the expression outside a spherical mass, with r measured from the centre.
- States that escape velocity is the minimum speed for a projectile to reach infinity with zero kinetic energy.
- Derives v = √(2GM/R) by equating total energy at the surface to zero at infinity.
- Explains that the projectile mass m cancels, so escape velocity is independent of the projectile.
- Calculates escape velocity using G = 6.67 × 10⁻¹¹ N m² kg⁻² and consistent SI units.
- Recognises that escape velocity depends on the mass and radius of the body being escaped from.
Examiner Tips
- 💡Always include the negative sign when stating the value of gravitational potential.
- 💡Use the definition to derive the equation V = -GM/r if needed.
- 💡Remember that potential is a scalar, so no direction is required.
- 💡Write the defining equation with its negative sign before substituting numbers, so the sign is not lost in rearrangement.
- 💡When asked for a change in potential, show both potential values with their signs and subtract them explicitly.
- 💡Check that r is larger than the radius of any spherical mass before using the point-mass expression.
- 💡Label both axes with quantity and unit, and mark the two separation limits between which the area is required.
- 💡If a numerical area is needed, state the method used, such as integration of GMm/r² or counting squares.
- 💡Use the link work done = mΔV to check an area obtained graphically.
- 💡Write E = mV and substitute V = −GM/r explicitly to show where the negative sign comes from.
- 💡For energy changes, calculate both energies with signs and subtract, rather than working with magnitudes.
- 💡Check that the mass m is the object in the field, not the source mass M.
- 💡Start from the energy condition ½mv² − GMm/R = 0 and cancel m before substituting numbers.
- 💡Convert radii from km to m and use G = 6.67 × 10⁻¹¹ N m² kg⁻² in SI units.
- 💡Give the answer as a speed with the correct unit, and check the order of magnitude against typical values such as 1.1 × 10⁴ m s⁻¹ for Earth.
Common Mistakes
- Forgetting the negative sign in the definition or equation; correct by remembering that gravitational potential is always negative near a mass.
- Confusing gravitational potential with gravitational potential energy; correct by noting potential is per unit mass.
- Thinking potential is zero at the surface of a planet; correct by stating it is zero only at infinity.
- Omitting the negative sign and writing V = GM/r; the correction is that potential is negative because work is done by the field as the test mass moves in from infinity.
- Measuring r from the surface rather than the centre of the spherical mass; the correction is that r is the distance from the centre when the point lies outside the sphere.
- Treating a change in potential as a simple subtraction of magnitudes; the correction is to subtract signed values, V_final − V_initial, so the sign of ΔV is meaningful.
- Treating the force–distance graph as a straight line and using ½ × base × height; the correction is that F ∝ 1/r² gives a curve, so integration or square-counting is needed.
- Confusing the area under a force–distance graph with the gradient; the correction is that gradient gives force per unit distance, while area gives work done.
- Using the surface radius instead of the centre-to-centre separation for a spherical mass; the correction is that r is measured from the centre when outside the sphere.
- Dropping the negative sign and writing E = GMm/r; the correction is that gravitational potential energy is negative relative to the zero at infinity.
- Using r as the height above the surface rather than the distance from the centre; the correction is that r is the centre-to-centre separation outside the sphere.
- Adding magnitudes when finding a change in energy; the correction is to subtract signed values, E_final − E_initial, so the direction of energy transfer is clear.
- Including the projectile mass m in the final expression; the correction is that m cancels, leaving v = √(2GM/R).
- Using the orbital speed formula v = √(GM/r) instead of the escape formula; the correction is that escape requires √2 times the circular orbital speed at the same radius.
- Substituting the radius in kilometres without converting to metres; the correction is to convert all quantities to SI units before calculating.