The key idea
Newton's inverse-square law, applied from lab bench to geostationary orbit. The algebra is short; the marks concentrate around one negative sign — gravitational potential — and one satellite: the geostationary one.
Understand it · 01
Measure distance from the centre
For a point mass or outside a spherically symmetric body, gravitational field strength has magnitude GM/r². The distance r is measured from the centre of the attracting body, not from its surface. At altitude h above a planet of radius R, use r = R + h. Doubling the centre-to-centre distance reduces the field strength to one quarter. The field is a vector directed towards the attracting mass.
The expression F = GMm/r² can be written as F = mg, identifying g as GM/r². This explains why a small object’s gravitational acceleration does not depend on its own mass in the ideal model: its gravitational force and inertia increase together. The model assumes the object does not significantly alter the main field and ignores effects such as atmospheric drag. Inside a planet, its entire mass cannot generally be treated as a point mass at the centre.
Understand it · 02
Understand why the potential is negative
Gravitational potential at a point is the potential energy per unit mass relative to a chosen zero. For an isolated mass, defining zero at infinite separation gives V = −GM/r. The negative sign means that work must be supplied to move a mass from that bound position to infinity. As r increases, potential rises towards zero, even though its magnitude decreases. Potential is a scalar and does not carry an inward or outward direction.
The energy change for moving a mass m between two radii is m(Vfinal − Vinitial). Near a planet’s surface, over a small height change, this approaches mgΔh with approximately constant g. Over a large altitude change, substituting the surface value of g into mgh overestimates the required energy because the field weakens with height. On a potential–distance graph, the negative gradient gives the radial component of field strength.
Understand it · 03
Connect gravity with circular motion
For a circular orbit where gravity provides the inward resultant, GMm/r² = mv²/r. Cancelling the orbiting mass gives v² = GM/r and, using v = 2πr/T, T² = 4π²r³/(GM). A higher circular orbit has a lower orbital speed and a longer period. These statements compare different circular orbits; they do not mean that simply slowing a satellite at one point moves it neatly into a higher circle.
An orbiting astronaut feels weightless because the astronaut and spacecraft fall freely together, rather than because gravity has vanished. Observing an orbital radius and period can estimate the central mass using the circular model. Use the full orbital radius and consistent SI units. A geostationary orbit has additional requirements beyond its period: it must be circular, above the equator, and move in the same rotational direction as Earth to remain above the same ground location.
Key definitions
- Newton's law of gravitation
- The gravitational force between two point masses is proportional to the product of the masses and inversely proportional to the square of their separation: F = GMm ÷ r².
- Gravitational field strength
- The gravitational force per unit mass at a point: g = F ÷ m. Units: N/kg.
- Gravitational potential (φ)
- Potential energy per unit mass, with zero at infinity: φ = −GM/r. Equivalently, the external work per unit mass for a slow transfer from infinity with no kinetic-energy change. It is negative for an isolated attracting mass.
- Geostationary orbit
- A circular equatorial orbit in Earth's rotational direction with a period equal to Earth's sidereal rotation, approximately 23 h 56 min (often rounded to 24 h in exam models). The satellite remains above the same ground point.
Equations, units & conditions
Gravitational force
F = G M m ÷ r²
Units: N
Field strength
g = G M ÷ r²
Units: N/kg
Potential
φ = −G M ÷ r
Units: J/kg
Orbit condition
G M m ÷ r² = m v² ÷ r
Worked examples, step by step
Example 1
Calculate the orbital speed of a satellite 400 km above Earth's surface. (M = 6.0 × 10²⁴ kg, R = 6.4 × 10⁶ m, G = 6.67 × 10⁻¹¹)
Follow the method
- r = R + h = 6.4 × 10⁶ + 4.0 × 10⁵ = 6.8 × 10⁶ m.
- Gravity provides centripetal force: GMm/r² = mv²/r, so v = √(GM/r).
- v = √(6.67 × 10⁻¹¹ × 6.0 × 10²⁴ ÷ 6.8 × 10⁶).
v ≈ 7.7 × 10³ m/s (7.7 km/s)
Practice questions with worked solutions
Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.
Build · 3 suggested marks
1. Altitude and field strength
A planet has radius R and surface field strength 12 N kg⁻¹. Find the field strength at altitude 2R above its surface.
Show solution and self-check: question 1
- The distance from the centre is R + 2R = 3R.
- g = 12 × (R/3R)² = 12/9.
1.33 N kg⁻¹ towards the planet.
Self-check · one mark per point
- Use centre-to-centre distance 3R.
- Apply the inverse-square relationship.
- Give the inward field direction.
Build · 3 suggested marks
2. Circular orbit
For a planet, GM = 3.60 × 10¹⁴ m³ s⁻². A satellite has a circular orbital radius of 9.0 × 10⁶ m. Find its speed and period.
Show solution and self-check: question 2
- v = √(GM/r) = √(3.60 × 10¹⁴/9.0 × 10⁶) = 6325 m s⁻¹.
- T = 2πr/v = 8941 s.
Speed 6.3 km s⁻¹; period 8.9 × 10³ s, approximately 2.5 h.
Self-check · one mark per point
- Equate gravity to the inward resultant.
- Obtain speed using the full orbital radius.
- Calculate period from circumference divided by speed.
Common mistakes & how to avoid them
- Dropping the minus sign on potential. φ = −GM/r: potential is always negative, rises towards zero at infinity, and 'higher potential' means less negative. Escape questions live and die on this sign.
- Using surface g for orbital altitude. g falls as 1/r² — and r is measured from the centre of the Earth, not from the surface. Adding the altitude to the radius is a required step.
- Geostationary answers missing a condition. State a circular orbit above the equator, in Earth's rotational direction, with the same period as Earth's rotation (approximately 24 h in the usual exam approximation).
- Confusing field strength (vector, N/kg, −dφ/dr) with potential (scalar, J/kg). The radial component of field strength is the negative gradient of potential; where the potential curve flattens, the field magnitude decreases.
Understand the why
Questions, explained
Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.
Why is gravitational potential negative?
With gravitational potential defined as zero at infinity, points near an isolated attracting mass have negative potential. Positive work must be supplied to move a mass from a bound position out to infinity without changing its kinetic energy.
The negative sign refers to the chosen energy reference, not to a negative mass or an outward field. Potential rises towards zero as distance increases, so −1 MJ kg⁻¹ is higher than −4 MJ kg⁻¹. A mass moving outward gains potential energy, whereas one falling inward loses it. The field is related to the negative gradient of potential, making it point inward while the potential curve rises outward. Keep the scalar potential distinct from the vector field strength: their signs and units express different physical ideas.
How do you find the speed of an orbiting satellite?
For a circular orbit governed by one dominant central mass, equate gravitational force to mv²/r. The orbiting mass cancels, leaving v = √(GM/r), with r measured from the central body’s centre rather than its surface.
The cancellation means that small satellites with different masses can share the same circular orbit and speed. It does not mean gravity is absent, nor does it make their kinetic energies equal. A larger orbital radius gives a lower speed but a longer path, so the period increases. The circular assumption is essential: in an elliptical orbit speed changes around the path and the simple expression is not valid at every point. Neglect atmospheric drag and other significant bodies only when that approximation is appropriate to the stated orbit.
A changed circular orbit
Two circular orbits around the same planet have radii r and 4r. Compare their speeds and periods.
- v is proportional to r⁻½, so the outer speed is 1/√4 = 1/2 of the inner speed.
- T² is proportional to r³, so the outer period is √(4³) = 8 times the inner period.
Outer speed is half; outer period is eight times as long.
Why must geostationary satellites orbit over the equator?
A geostationary satellite must remain above one fixed point on Earth. Its orbit must therefore be circular, equatorial, in Earth’s rotational direction and have the same period as Earth’s rotation, approximately one sidereal day.
Gravity points towards Earth’s centre, so the orbital plane of an ideal free satellite passes through that centre. An inclined orbital plane makes the satellite move north and south relative to the equator, preventing a fixed ground position. Matching the rotation period alone gives a geosynchronous orbit, which need not be geostationary if it is inclined or eccentric. The sidereal period is approximately 23 hours 56 minutes, often rounded to 24 hours in school calculations. Follow a supplied period while keeping the geometric conditions explicit.
Why must I add the planet’s radius to a satellite’s altitude?
Altitude is the distance above a planet’s surface; orbital radius is the distance from its centre. In inverse-square gravity formulas use r = R + h, where R is planet radius and h is altitude.
Confusing these distances is particularly serious for a low orbit, because the altitude may be only a small fraction of the planet’s radius. Using h alone makes the calculated field or orbital speed far too large. Draw the centre, surface and satellite on one line and label both distances before substituting. The same centre-to-centre requirement applies when calculating gravitational force between spherical bodies outside one another. The point-mass expression is not generally valid inside a planet by simply using its whole mass at the centre; the mass distribution then matters.
Field at altitude
A planet has radius R and surface field 18 N kg⁻¹. Find field strength at altitude R above its surface.
- The centre-to-satellite distance is 2R.
- g = 18 × (R/2R)² = 18/4.
4.5 N kg⁻¹ towards the centre.
Why do astronauts feel weightless if gravity keeps them in orbit?
Orbiting astronauts are weightless in the sense that they lack a normal support force, not because Earth’s gravity has vanished. The spacecraft and astronauts fall freely together while moving sideways fast enough to remain in orbit.
A scale usually measures a contact force rather than gravity directly. In free fall, the floor does not need to push upward to support an astronaut against gravity, so the scale reading can be nearly zero. Gravity still supplies the inward acceleration that bends the path. In low Earth orbit its magnitude remains a substantial fraction of the surface value. Small relative accelerations can arise from rotation, thrust or field gradients, so real spacecraft conditions are not perfectly force-free. The useful distinction is between gravitational force and apparent weight measured by support.
When does mgh stop being a good approximation?
The expression mgΔh assumes the gravitational field is approximately constant over the height change. For large radial changes, use ΔU = GMm(1/rinitial − 1/rfinal), because gravity weakens with distance.
Near a planet’s surface, a height change small compared with its radius makes the inverse-square field vary little, so mgh is an effective approximation. Extending the surface value of g to a distant orbit overestimates the potential-energy increase, because the force becomes smaller along the outward journey. Keep kinetic energy separate: placing an object into orbit requires both lifting it and giving it the required orbital speed, subject to the starting motion. The potential-energy equation gives only the change in gravitational store, not every energy transfer made by a launch vehicle.
Your learning checklist
Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.
- Use Newton's law of gravitation: F = GMm/r²
- Define gravitational field strength g = F/m; use g = GM/r²
- Define gravitational potential; use φ = −GM/r
- Analyse circular orbits: equate gravitational and centripetal force
- Describe geostationary orbits and derive orbital speed and period relationships
Sources & how to cite these notes
Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.
Reference this page
Dr Desouky Physics Academy. Edexcel IAL Physics: Gravitational fields. Updated 2026-09-25. https://www.drdesouky.com/revision/a-level-physics/gravitational-fields
Use the arrow beside an explanation heading to link straight to that section. Spotted a problem? Contact the academy with the topic and the step you want us to check.