OxfordAQA International AS and A-level Physics: Gravitational potential and orbits of planets and satellites (9630) – Revision Notes

These notes condense sections 3.7.3 (Gravitational potential) and 3.7.4 (Orbits of planets and satellites) of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, exams from May/June 2020 onwards. Both sections are marked International A-level only. For full explanations and longer worked examples, read the study guide. Test yourself afterwards with the practice questions, find gaps with the free diagnostics, and tick off topics on the checklist. The course hub lists the rest of the course.

Data for the reminders, as a question would give them: G = 6.67 × 10⁻¹¹ N m² kg⁻²; the Earth has mass 5.97 × 10²⁴ kg and radius 6.37 × 10⁶ m.

Definitions

• Gravitational potential, V: the work done per unit mass by an external agent in moving a small mass from infinity to the point. Zero at infinity, so negative everywhere else. Unit J kg⁻¹.
• Gravitational potential difference, ΔV: the work done per unit mass in moving a mass between two points.
• Equipotential surface: a surface on which V is constant. The field is at right angles to it.
• Geosynchronous orbit: an orbit whose period equals the rotation period of the planet (for the Earth, 23 h 56 min).
• Geostationary orbit: a geosynchronous orbit that is above the equator and in the same direction as the rotation, so the satellite stays above one point.

Formulas

Quantity	Formula	Notes
Work done	ΔW = mΔV	ΔV = V(final) − V(initial). Positive if the mass moves away.
Potential, radial field	V = −GM/r	r from the centre of the body. A scalar.
Field strength from potential	g = −ΔV/Δr	The magnitude is the gradient of the V–r graph.
Field strength, radial	g = GM/r²	From 3.7.2.
Orbital speed	v = √(GM/r)	Independent of the satellite’s mass.
Orbital period	T = 2πr/v, T² = 4π²r³/(GM)	T² ∝ r³.
Kinetic energy	Ek = GMm/(2r)
Potential energy	Ep = −GMm/r
Total energy	E = −GMm/(2r)	E = −Ek and Ep = −2Ek.

Method in steps: work done between two radii

1. Write down GM (or G and M separately).
2. Convert both distances to metres and measure them from the centre of the body.
3. Find V = −GM/r at each radius, keeping the minus sign.
4. Subtract: ΔV = V(final) − V(initial).
5. Multiply by m. Quote the sign if the question asks for work done by the field, and the magnitude if it asks for work needed.

Reminder. Work to lift 1200 kg from the Earth’s surface to a radius of 3 × 6.37 × 10⁶ m:

GM = 3.982e14
ΔV = GM (1/R - 1/3R) = 3.982e14 x (2/3) / 6.37e6 = 4.167 x 10^7 J/kg
ΔW = 1200 x 4.167e7 = 5.00 x 10^10 J

Graphs

	g against r (r larger than the body radius)	V against r
Shape	Falls steeply then flattens; inverse square	Rises from a negative value towards zero; inverse
Halving the distance	g is four times larger	V is twice as negative
Doubling the distance	g becomes one quarter	V becomes half as negative
Never reaches	Zero	Zero

To find g from a V–r graph, draw a tangent at the point, find its gradient and take the magnitude. To find the potential difference from a g–r graph, the area under the curve gives the magnitude of ΔV.

Method in steps: orbit problems

1. Gravity gives the centripetal force: GMm/r² = mv²/r.
2. Get v = √(GM/r), then T = 2πr/v.
3. For comparisons, use ratios: T² ∝ r³ and v ∝ r^(−1/2).
4. For energy, use Ek = GMm/(2r), Ep = −GMm/r and E = −GMm/(2r).
5. For a change of orbit, subtract the totals: ΔE = E(final) − E(initial).

Reminder. Orbit radius 2.00 × 10⁷ m about the Earth:

v = sqrt(3.982e14 / 2.00e7) = 4.46 x 10^3 m/s
T = 2 pi x 2.00e7 / 4462 = 2.82 x 10^4 s = 7.82 hours

Reminder. Energy to move a 250 kg satellite from a circular orbit of radius 7.00 × 10⁶ m to one of radius 1.00 × 10⁷ m about the Earth:

E(7.00e6) = -3.982e14 x 250 / (2 x 7.00e6) = -7.11 x 10^9 J
E(1.00e7) = -3.982e14 x 250 / (2 x 1.00e7) = -4.98 x 10^9 J
ΔE = +2.13 x 10^9 J

The energy needed is the change in total energy, not the change in potential energy alone, because the kinetic energy also changes.

Writing good explanations

• Equipotential question. State that the gravitational force is perpendicular to the surface, so the force has no component along the path. Then say the work done is zero because ΔV = 0.
• Why a bigger orbit means a lower speed. Gravity provides the centripetal force, GMm/r² = mv²/r, so v² = GM/r. A larger r gives a smaller v.
• Drag paradox. Drag removes total energy. The satellite falls to a smaller radius, where the potential energy is lower by twice the kinetic energy gained, so the speed rises.
• Geosynchronous. State the period (the Earth’s rotation period), the equatorial plane and the direction of rotation if the answer must be geostationary. Then give the use: a fixed ground antenna can stay pointed at it.

Must-know distinctions

• Potential (J kg⁻¹, scalar, negative) versus potential energy (J, mV) versus field strength (N kg⁻¹, vector).
• Total energy of an orbit becomes more negative as the orbit gets smaller, yet the speed becomes larger.
• A satellite loses mechanical energy to drag and falls to a lower orbit. Its kinetic energy rises.
• Geosynchronous means the period matches the planet’s rotation. Geostationary needs the equatorial plane as well.
• Zero work along an equipotential means no change in potential energy. It does not mean no force acts: the force is perpendicular to the motion.

Quick self-test

1. State what is meant by the gravitational potential at a point.
2. Find V at the Earth’s surface.
3. Find V at a distance of 2 × 6.37 × 10⁶ m from the Earth’s centre.
4. Find the work to move 1200 kg from the Earth’s surface to a radius of 3 × 6.37 × 10⁶ m.
5. A planet has V = −2.4 × 10⁷ J kg⁻¹ at r = 4.0 × 10⁶ m, and V is proportional to −1/r. Find g and M.
6. A satellite’s orbital radius doubles. By what factors do v and T change?
7. A satellite has a kinetic energy of 6.0 × 10⁹ J in a circular orbit. State its potential energy and total energy.
8. Find the orbital radius for a geosynchronous orbit of the Earth (T = 86 164 s).
9. Find the extra energy needed to free a 450 kg satellite from a 7.00 × 10⁶ m circular orbit to infinity.
10. One planet orbits a star at four times the radius of another. By what factor is its period larger?
11. Why is no work done moving a mass along an equipotential?
12. Explain why the total energy of a satellite in a circular orbit is negative.

Answers

1. The work done per unit mass, by an external agent, in moving a small mass from infinity to the point.
2. V = −3.982e14/6.37e6 = −6.25 × 10⁷ J kg⁻¹.
3. V = −3.982e14/1.274e7 = −3.13 × 10⁷ J kg⁻¹.
4. 5.00 × 10¹⁰ J (see the reminder above).
5. g = −ΔV/Δr gives g = V/r in magnitude for an inverse relationship, so g = 2.4e7/4.0e6 = 6.0 N kg⁻¹. M = gr²/G = 6.0 × (4.0e6)²/6.67e-11 = 1.44 × 10²⁴ kg.
6. v falls by a factor of 1/√2 (to 0.707 of its value). T increases by 2^1.5 = 2.83.
7. Ep = −2Ek = −1.2 × 10¹⁰ J. E = −Ek = −6.0 × 10⁹ J.
8. r = (GMT²/4π²)^(1/3) = 4.21 × 10⁷ m.
9. Extra energy = −E = GMm/(2r) = 3.982e14 × 450/(2 × 7.00e6) = 1.28 × 10¹⁰ J.
10. T² ∝ r³, so T increases by 4^1.5 = 8.
11. The field is perpendicular to the surface, so the force has no component along the displacement. Equivalently, ΔV = 0, so ΔW = mΔV = 0.
12. Ek = +GMm/(2r) and Ep = −GMm/r, so E = −GMm/(2r), which is less than zero. A negative total means the satellite is bound: it would need an input of energy equal to |E| to reach infinity with zero speed.

Where marks are usually lost

• Dropping the sign of V, or taking initial minus final.
• Using height above the surface instead of the distance from the centre.
• Leaving r in kilometres.
• Quoting an energy change as the total energy of the orbit.
• Stating that the speed of a satellite depends on its mass.
• Writing “the satellite speeds up because energy is added” when drag is the cause. Explain the transfer from potential to kinetic energy.
• Defining potential without saying “per unit mass” and “from infinity”.
• Calling any orbit with a 24 hour period geostationary, or giving 24 hours instead of 23 h 56 min for the Earth’s rotation period.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for exams from May/June 2020 onwards, published by OxfordAQA. Sections 3.7.3 and 3.7.4.