AQA 7408 · Sections 3.7.1 and 3.7.2 · Paper 2
Gravitational fields, potential and orbits
Move from universal attraction and radial fields to negative potential, circular-orbit energy, escape speed and geostationary satellites. Diagnose the higher-orbit misconception, test it in the simulator, then complete the original assessment.
Content diagnostic
A satellite moves to a higher circular orbit
Active revision
Use a three-pass radial-field routine
Measure from the centre
Convert altitude h to radius r = R + h before using any inverse-square, potential or orbit equation.
Separate vectors and scalars
Field strength is a vector directed inward. Potential is a negative scalar when zero is defined at infinity.
Check energy signs
For a bound circular orbit, Eₖ is positive, Eₚ is negative with twice its magnitude, and total energy is negative.
Force fields and gravity
Describe the interaction before calculating it
Force fields
A force field is a region in which a body experiences a non-contact force. Mass produces gravitational fields; static charge produces electric fields; moving charge produces magnetic fields. A vector field assigns a field vector to every point.
Newton’s law applies to point masses and spherically symmetric bodies treated as if their mass were concentrated at the centre. The two masses exert equal and opposite attractive forces.
Field strength
Gravitational field strength is force per unit mass, measured in N kg⁻¹. Around a spherical mass, field lines are radial, point inward and become less dense as radius increases.
Gravitational and electrostatic fields compared
| Feature | Gravitational | Electrostatic |
|---|---|---|
| Source | Mass | Positive or negative charge |
| Force between point sources | F = Gm₁m₂/r² | F = Q₁Q₂/(4πε₀r²) in vacuum |
| Interaction | Always attractive | Attractive or repulsive |
| Potential zero | Usually infinity; V is negative around an isolated mass | Usually infinity; sign depends on source charge |
| Field and potential | Field lines cross equipotentials at right angles | Same geometrical relationship |
Potential and potential energy
Use potential to track energy per unit mass
Gravitational potential is external work done per unit mass in bringing a small test mass slowly from infinity to the point, with no change in kinetic energy. Zero potential is defined at infinity.
Moving away from a mass increases V towards zero and increases gravitational potential energy. External work is required for a slow outward move.
With outward radius positive, the radial field component is negative. The potential graph has a positive slope, so the field points inward.
The signed area under a gᵣ–r graph is the negative potential difference. If a graph shows |g| instead, state that it displays magnitude.
Every point on an equipotential surface has the same V. Moving along it gives ΔV = 0 and therefore no gravitational work. Field lines cross equipotentials normally.
Useful simulation
Choose a sideways launch and read its orbit
Orbital Motion Lab controls Central body, Starting distance r and Sideways launch speed. Values are reported per kilogram, so satellite mass has no control and does not alter the path.
Circular orbit
Choose Earth, set Starting distance to 1.35R and select Circular to launch. Pause and compare gravity with v²/r in the At launch panel; they match. Read speed, field strength, period and specific energies. Total specific energy is negative; potential energy has twice the magnitude of kinetic energy.
A higher circular orbit
Keep Earth and Circular selected, then increase Starting distance to 2.00R and choose Launch with these settings. Compare launch speed, gravity, period and total specific energy: speed and field strength decrease, period increases and total energy becomes less negative. Radius and speed sliders describe the next launch; choose Launch with these settings to apply them. Scenario buttons launch immediately.
Speed selects the shape
At the same radius, compare Too slow, Circular, Ellipse and Escape. At circular speed gravity gives exactly v²/r. A greater speed requires more turning acceleration than gravity supplies for a circle at that launch radius, so the path moves outward. Escape occurs when the specific energy is non-negative.
Field and potential graphs
Open Field Graphs for signed gravitational g–r and V–r graphs, tangents and areas. The orbit lab shows launch paths and energies. Use the geostationary equations below to calculate its radius and period; the orbit lab has no named LEO or geostationary preset.
Planets and satellites
Derive motion and energy from gravity
Circular speed and period
- Gravity supplies the resultant centripetal force: GMm/r² = mv²/r.
- Cancel satellite mass m to obtain v² = GM/r.
- Use v = 2πr/T.
- Substitution gives T² = 4π²r³/(GM), so T² ∝ r³ for a fixed central mass.
Energy of a circular orbit
Moving to a larger circular orbit requires a net energy input. Total energy increases towards zero even though final orbital speed is lower.
Escape speed
This follows by setting total energy at launch equal to zero at infinity. In the ideal model, escape speed is independent of the escaping mass.
Orbit applications
Distinguish synchronous, geostationary and low Earth orbits
Synchronous
The orbital period equals the rotation period of the body. A synchronous orbit need not be circular or equatorial, so it need not remain above one longitude.
Geostationary
Circular, in Earth’s equatorial plane, same direction as Earth’s rotation, with the sidereal-day period and radius about 4.22 × 10⁷ m. It is useful for communications and continuous weather monitoring.
Low Earth orbit
Much smaller radius and shorter period. Useful for high-resolution Earth observation, crewed missions and low-latency communications, but one satellite does not remain above one point.
Application trade-off
Geostationary satellites offer continuous regional coverage but large signal delay and weaker received signals. LEO systems need tracking or constellations but reduce path length and latency.
Data and graph skills
Determine a central mass from orbital data
Linearise Kepler’s relationship
- Measure or obtain orbital radius r from the centre and period T for several near-circular orbits around the same body.
- Calculate r³ and T² with consistent SI units.
- Plot T² vertically against r³ horizontally.
- For a line through the origin, gradient = 4π²/(GM).
- Calculate M = 4π²/(G × gradient).
Evaluate the model
- Use centre-to-centre radius, not altitude.
- Check that the central mass is much larger than each orbiting mass.
- Use near-circular orbits or interpret the stated orbital radius carefully.
- Include uncertainty in period and radius; cubing r magnifies fractional uncertainty by approximately a factor of three.
- Look for an intercept or curvature that may indicate systematic error or a broken assumption.
Misconception repair
Eight ideas that commonly lose marks
- “Use altitude in the inverse-square equation.” Use radius from the centre: r = R + h.
- “There is no gravity in orbit.” Gravity supplies the centripetal acceleration; astronauts feel weightless because they are in free fall.
- “Potential is a vector.” Potential is a scalar; gravitational field strength is a vector.
- “Negative potential means negative force.” The negative potential records the bound energy reference. Force direction comes from the field vector.
- “A higher orbit has a faster satellite.” Circular speed v = √(GM/r) decreases as radius increases.
- “A higher orbit has lower total energy.” E = −GMm/(2r) becomes less negative, so total energy increases.
- “A synchronous satellite is automatically geostationary.” Geostationary also requires a circular equatorial orbit in Earth’s rotation direction.
- “A heavier satellite falls faster in the same orbit.” Satellite mass cancels from g, circular speed and period.
Original exam-style assessment
37 marks with coded mark schemes
Attempt every question before opening its mark scheme. All constants required for a calculation are supplied. No response requires drawing or annotating; numerical dependencies explicitly allow error carried forward.
1. Define a force field. State the source of a gravitational field and one difference between gravitational and electrostatic interactions.[3 marks]
Mark scheme
- (B1) A force field is a region in which a body experiences a non-contact force.
- (B1) Mass is the source of a gravitational field.
- (B1) Masses always attract, whereas charges may attract or repel.
Related PhysicsUK checklist points: 3.7.1(a), 3.7.1(c), 3.7.1(f)
2. A satellite of mass 850 kg is 8.2 × 10⁶ m from the centre of a planet of mass 7.5 × 10²⁴ kg. Calculate the gravitational force on the satellite. Use G = 6.67 × 10⁻¹¹ N m² kg⁻².[3 marks]
Mark scheme
- (M1) Uses F = GMm/r² with r measured from the centre.
- (M1) Substitutes F = (6.67 × 10⁻¹¹ × 7.5 × 10²⁴ × 850)/(8.2 × 10⁶)².
- (A1) F = 6.32 × 10³ N, directed towards the planet.
Related PhysicsUK checklist points: 3.7.2.1(b)
3. A point is 3.00 × 10⁶ m above Earth’s surface. Earth has radius 6.37 × 10⁶ m and mass 5.97 × 10²⁴ kg. Calculate the gravitational field strength and state its direction. Use G = 6.67 × 10⁻¹¹ N m² kg⁻².[3 marks]
Mark scheme
- (M1) Uses centre-to-centre radius r = 6.37 × 10⁶ + 3.00 × 10⁶ = 9.37 × 10⁶ m.
- (A1) g = GM/r² = 4.54 N kg⁻¹.
- (B1) The field is directed radially towards Earth’s centre, as shown by inward field lines.
Related PhysicsUK checklist points: 3.7.2.2(a), 3.7.2.2(b), 3.7.2.2(c)
4. A 120 kg satellite is moved slowly from radius 7.0 × 10⁶ m to 1.4 × 10⁷ m from Earth’s centre. Calculate the increase in gravitational potential and the work done on the satellite. Use Earth mass 5.97 × 10²⁴ kg and G = 6.67 × 10⁻¹¹ N m² kg⁻².[4 marks]
Mark scheme
- (M1) Uses V = −GM/r at both radii.
- (A1) ΔV = V₂ − V₁ = +2.84 × 10⁷ J kg⁻¹.
- (M1) Uses ΔW = mΔV.
- (A1) Work done on the satellite = +3.41 × 10⁹ J.
Related PhysicsUK checklist points: 3.7.2.3(a), 3.7.2.3(b), 3.7.2.3(c)
5. Outward radial direction is defined as positive. The signed area under a gᵣ–r graph between radii A and B is −1.80 × 10⁷ J kg⁻¹. Determine Vᴮ − Vᴬ and the change in potential energy of a 50 kg satellite moving from A to B.[4 marks]
Mark scheme
- (B1) Uses gᵣ = −dV/dr, so ∫gᵣdr = −ΔV.
- (M1) Therefore ΔV is the negative of the signed area.
- (A1) Vᴮ − Vᴬ = +1.80 × 10⁷ J kg⁻¹.
- (A1) ΔEₚ = mΔV = +9.0 × 10⁸ J.
Related PhysicsUK checklist points: 3.7.2.3(g), 3.7.2.3(h)
6. Derive the relationship T² ∝ r³ for a satellite of mass m in a circular orbit of radius r around a much more massive body of mass M.[4 marks]
Mark scheme
- (M1) Equates gravitational force to centripetal force: GMm/r² = mv²/r.
- (A1) Cancels m and obtains v² = GM/r.
- (M1) Uses v = 2πr/T and substitutes into the force relationship.
- (A1) Obtains T² = 4π²r³/(GM), hence T² ∝ r³ for fixed M.
Related PhysicsUK checklist points: 3.7.2.4(a), 3.7.2.4(b)
7. A satellite is in a circular Earth orbit of radius 7.20 × 10⁶ m. Calculate its orbital speed and period. Use Earth mass 5.97 × 10²⁴ kg and G = 6.67 × 10⁻¹¹ N m² kg⁻².[4 marks]
Mark scheme
- (M1) Uses v = √(GM/r).
- (A1) v = 7.44 × 10³ m s⁻¹.
- (M1) Uses T = 2πr/v or T² = 4π²r³/(GM).
- (A1) T = 6.08 × 10³ s, approximately 101 min. Allow valid error carried forward from v.
Related PhysicsUK checklist points: 3.7.2.4(a)
8. A 1200 kg satellite orbits Earth at radius 8.0 × 10⁶ m. Calculate its kinetic, gravitational potential and total energies. Then calculate the energy that must be supplied to transfer it to a circular orbit of radius 1.2 × 10⁷ m. Use Earth mass 5.97 × 10²⁴ kg and G = 6.67 × 10⁻¹¹ N m² kg⁻².[4 marks]
Mark scheme
- (A1) Eₖ = GMm/(2r) = +2.99 × 10¹⁰ J.
- (A1) Eₚ = −GMm/r = −5.97 × 10¹⁰ J.
- (A1) E = −GMm/(2r) = −2.99 × 10¹⁰ J.
- (A1) Required increase in total energy = E₂ − E₁ = +9.95 × 10⁹ J. Allow error carried forward from a consistently calculated E₁.
Related PhysicsUK checklist points: 3.7.2.4(c), 3.7.2.4(d)
9. Use energy conservation to calculate the minimum escape speed from Earth’s surface, ignoring atmosphere and Earth’s rotation. Use Earth mass 5.97 × 10²⁴ kg, radius 6.37 × 10⁶ m and G = 6.67 × 10⁻¹¹ N m² kg⁻².[3 marks]
Mark scheme
- (M1) Sets initial total energy equal to zero at infinity: ½mv² − GMm/R = 0.
- (M1) Obtains v = √(2GM/R).
- (A1) v = 1.12 × 10⁴ m s⁻¹, or 11.2 km s⁻¹.
Related PhysicsUK checklist points: 3.7.2.4(e)
10. State four requirements for a geostationary satellite. Then give one application for which geostationary orbit is preferred over low Earth orbit and explain why.[5 marks]
Mark scheme
- (B1) The orbit is circular.
- (B1) The orbit lies in Earth’s equatorial plane.
- (B1) The satellite travels in the same direction as Earth’s rotation.
- (B1) Its period equals Earth’s sidereal rotation period, giving radius about 4.22 × 10⁷ m from Earth’s centre.
- (B1) Valid application with reason, for example communications or weather monitoring because the satellite remains above one longitude and a fixed ground antenna can be used.
Related PhysicsUK checklist points: 3.7.2.4(f), 3.7.2.4(g), 3.7.2.4(h)
Specification coverage
AQA 7408 sections 3.7.1 and 3.7.2
Letter suffixes in this checklist are PhysicsUK subdivisions for navigation. The official AQA references are the numbered sections above.
- 3.7.1(a) Define a force field as a region where a body experiences a non-contact force.
- 3.7.1(b) Represent a force field as a vector field with direction determined by inspection.
- 3.7.1(c) Recognise fields arising from interactions of mass, static charge and moving charges.
- 3.7.1(d) Compare gravitational and electrostatic inverse-square force laws.
- 3.7.1(e) Compare field lines, potentials and equipotential surfaces in gravitational and electrostatic fields.
- 3.7.1(f) State that masses always attract but charges can attract or repel.
- 3.7.2.1(a) Describe gravity as a universal attractive force between all matter.
- 3.7.2.1(b) Use Newton's law of gravitation F = Gm1m2/r² for point masses.
- 3.7.2.2(a) Represent gravitational fields using gravitational field lines.
- 3.7.2.2(b) Define gravitational field strength g = F/m.
- 3.7.2.2(c) Use g = GM/r² for a radial gravitational field.
- 3.7.2.3(a) Define gravitational potential with zero at infinity.
- 3.7.2.3(b) Define and use gravitational potential difference.
- 3.7.2.3(c) Use ΔW = mΔV for work done moving mass in a gravitational field.
- 3.7.2.3(d) Use equipotential surfaces and state that no work is done moving along an equipotential.
- 3.7.2.3(e) Use V = −GM/r for gravitational potential in a radial field.
- 3.7.2.3(f) Explain the significance of the negative sign in gravitational potential.
- 3.7.2.3(g) Interpret graphs of gravitational field strength g and potential V against radius r.
- 3.7.2.3(h) Use g = −ΔV/Δr and use area under a g-r graph to find ΔV.
- 3.7.2.4(a) Relate orbital period and speed to radius for circular orbits.
- 3.7.2.4(b) Derive T² proportional to r³ for circular orbits.
- 3.7.2.4(c) Use energy considerations for an orbiting satellite.
- 3.7.2.4(d) Calculate total energy of an orbiting satellite.
- 3.7.2.4(e) Use escape velocity.
- 3.7.2.4(f) Describe synchronous orbits.
- 3.7.2.4(g) Describe low Earth orbit and geostationary orbit applications.
- 3.7.2.4(h) State the plane and radius requirements for a geostationary orbit.
Continue through AQA Fields and Paper 2
Return to the AQA 3.7 fields module hub, then connect this topic with capacitors, energy and exponential graphs. Revisit circular motion for centripetal acceleration and mechanics and materials for force and energy. Continue through the AQA Paper 2 revision hub, problem-solving practice and MCQ practice.
Written against the AQA Physics 7408 content in sections 3.7.1 and 3.7.2. Questions and values are original.