Q: What does A-Level Physics: 8) Gravitational Fields Guide cover? A: From Newton's law to geostationary satellites, this post unpacks Section I Topic 8 of the 2026 H2 Physics syllabus for IP students and parents.
TL;DR Newton's inverse-square law is the only new “tool” in this topic; every other result (g, φ, UG, escape speed, orbit radius) is crafted by algebra and energy bookkeeping that you already know from mechanics. Nail those derivations once, and the WA1-to-A-Level questions collapse into four recurring templates.
Concrete example: how to use this page
If the question asks for force between two masses, use Newton's law of gravitation. If it asks for field strength at a point, divide force by test mass and use g=GM/r2
. That distinction stops force and field answers from blurring.
Decision map - choose the gravitational quantity first
Most errors in gravitational fields start from using the right formula for the wrong quantity. Read the command word and the noun in the question before choosing the route.
Question wording
Quantity to find
First setup
Common trap
"force on", "attraction between two masses"
force F
use F=Gm1m2/r2
treating F and g as the same answer
"field strength at a point"
field strength g
use g=GM/r2, then state direction
forgetting that g is a vector
"potential at a point"
potential φ
use φ=−GM/r and add scalar contributions
making potential positive near a mass
"minimum speed to escape"
escape speed ve
set total energy at infinity to zero
using circular-orbit speed without 2
"circular orbit" or "satellite period"
orbit speed or period
set gravitational force equal to centripetal force
using surface g instead of GM/r2
Misconception check: zero field and zero potential are different conditions. Between two masses, field vectors can cancel at one point, but the potentials from both masses are still negative and add together.
Step through the rest of the mechanics/electromagnetism refresh with our free H2 Physics notes so this gravitation set feeds smoothly into circular motion, SHM, and field comparisons.
1 Newton's law of gravitation
Isaac Newton modelled gravity as a mutual, attractive, central force:
F=Gr2m1m2.
Proportional to mass product: doubling either mass doubles the force.
Inverse-square with separation: the force drops by a factor of 4 when distance doubles - the geometry of spreading field lines over a sphere.
Universal constant G:6.67×10−11N⋅m2⋅kg−2, first measured by Cavendish in 1798.
1.1 IP exam cue
List all three features (“attractive”, “inverse-square”, “proportional to masses”) for a 2-mark definition: one IP tuition classic.
2 Gravitational field strength g
Field strength is force per unit mass: g=F/m. Combining with Newton's law gives
g=r2GM.(1)
Near Earth's surface, r≈ Earth's radius, so g≈9.81m⋅s−2 and is directionally “down”.
Parent insight: “Why is g 'constant'?” - because r changes by under one-tenth of a percent across school-lab altitudes, so the drift in g stays below three-tenths of a percent.
Radius-from-centre checkpoint
In gravitational field questions, r is measured from the centre of the planet or star, not from the surface. Convert any altitude first, then choose the formula.
Question clue
Use for r
Why it matters
Common trap
"At the surface of Earth"
RE
The point is one Earth radius from Earth's centre.
Substituting r=0 because altitude is zero.
"At height h above the surface"
RE+h
The field point is farther from the centre by the altitude.
Using h alone in GM/r2
"At altitude 400km"
RE+400km
Altitude is not the orbital radius.
Comparing satellites using only their heights above ground.
"At distance d from the centre"
d
The question has already given the correct centre-to-point distance.
Adding the planet radius again.
Worked check: a satellite at height h above Earth has field strength
g=(RE+h)2GME.
Misconception check: altitude tells you how far the object is above the surface. Gravitational formulae need how far the object is from the centre of mass.
3 Gravitational potential φ
Definition: work done per unit mass by an external agent in bringing a small test mass from infinity to the point.
For a point mass M:
φ=−rGM.(2)
The negative sign encodes that gravity is attractive; zero potential is set at infinity.
When more than one mass is present, decide whether the quantity is a vector or a scalar before adding contributions. This is where many zero-field mistakes begin.
Situation
Field strength g
Potential φ
What to write first
One point lies between two masses
Add field vectors with direction
Add scalar potentials
Draw arrows for g; keep both φ terms negative
Midpoint between two equal masses
Equal and opposite field vectors cancel
Potentials add to a more negative value
g=0, but φ=−GM/r−GM/r
A point is closer to one mass
Net field points toward the stronger nearby pull
Potential is still the scalar sum from both masses
Compare GM/r2 for direction; add −GM/r terms for potential
Common trap: zero resultant field does not mean zero potential. Field can cancel because directions oppose; gravitational potential has no direction and remains negative near masses.
4 Gravitational potential energy UG
For two point masses M and m:
UG=−rGMm.(3)
Think of UG as the shared energy store of the pair; separating them to infinity requires positive work equal to ∣UG∣.
Potential-energy bookkeeping checkpoint
When a mass moves between two radii, compare the initial and final gravitational potential energies before deciding whether energy is supplied or released.
Movement
What happens to UG
Energy statement
Common trap
Move farther from the planet
UG becomes less negative.
External work must be supplied if no other energy source is involved.
Saying energy decreases because the force weakens.
Move closer to the planet
UG becomes more negative.
Gravitational potential energy is released into kinetic energy or another store.
Dropping the negative signs and reversing the conclusion.
Compare two circular orbits
Higher orbit has less negative UG, but lower kinetic energy.
Check total mechanical energy, not only speed.
Assuming the faster low orbit has greater total energy.
Escape to infinity
Final UG=0.
The minimum kinetic energy must raise total energy to zero.
Treating infinity as very far away but still negative in potential.
Worked check: moving a satellite of mass m from radius r1 to a larger radius r2 gives
ΔUG=(−r2GMm)−(−r1GMm).
Since r2>r1, the final value is less negative, so ΔUG>0. Energy must be added to raise the satellite to the higher orbit.
Misconception check: "less negative" means a larger value. For example, −2 is greater than −8, so moving outward increases gravitational potential energy even though the gravitational pull becomes weaker.
5 Escape velocity
Set “initial kinetic energy + potential energy = 0 at infinity”:
21mve2−rGMm=0⇒ve=r2GM.(4)
At Earth's surface ve≈11.2km⋅s−1 - roughly 450 times the expressway speed limit.
6 Circular orbits & centripetal acceleration
Equate gravitational force to the required centripetal force mv2/r:
r2GMm=rmv2⇒v=rGM.(5)
Key takeaway: orbital speed halves when radius quadruples - a fast elimination step in MCQs.
Escape versus orbit checkpoint
Before substituting, decide whether the object is staying in a circular path or leaving the field completely. The two routes start from different physics.
Question cue
First principle
What the final condition means
Common trap
"circular orbit", "satellite speed", or "period"
Gravitational force provides centripetal force.
The object stays at the same radius while changing direction.
Using escape speed because both formulas contain GM/r.
"minimum speed to escape"
Total mechanical energy is zero at infinity.
The object just reaches infinity with no kinetic energy left.
Using orbit speed without the 2 factor.
"additional speed from a circular orbit"
Compare speeds at the same radius.
Add only the difference between escape speed and current orbit speed.
Adding the full escape speed on top of the orbital speed.
Worked check: at the same radius r, vorbit=GM/r and vescape=2GM/r. A probe already in circular orbit needs an extra speed of
Δv=r2GM−rGM.
Misconception check: escape does not mean "move in a larger circle". It means the final gravitational potential energy tends to zero at infinity, so the calculation must use energy.
7 Geostationary satellites
A geostationary satellite has
Orbital period = 24 h (synchronised with Earth's spin),
Zero inclination & eccentricity (lies above the equator),
Altitude ≈ 35 786 km.
Applications span weather monitoring, TV broadcast and VSAT internet.
7.1 Deriving the GEO radius
Set centripetal period T=v2πr and combine with Eq. (5):
r=34π2GMT2≈4.22×104km(6)
Geostationary radius checkpoint
For geostationary questions, separate the orbit condition from the height above Earth's surface. The formula gives the centre-to-satellite radius first; altitude comes only after subtracting Earth's radius.
Step
What to write
Why it matters
Common trap
1
The satellite must orbit above the equator with the same angular speed as Earth.
This fixes the direction and period condition before calculation.
Saying any 24 h orbit is geostationary.
2
Convert the period to seconds before using r=3GMT2/(4π2).
G, M, and r are in SI units.
Substituting T=24 instead of 24×60×60.
3
Interpret the result as orbital radius r, measured from Earth's centre.
Gravitational formulae always use centre-to-centre distance.
Calling 4.22×104km the altitude.
4
Find altitude from h=r−RE.
Altitude is height above the surface.
Adding Earth's radius instead of subtracting it.
Worked check: if r≈4.22×104km and RE≈6.37×103km, then
h≈4.22×104−6.37×103≈3.58×104km.
Misconception check: a geostationary satellite is a special geosynchronous satellite. Matching Earth's period is necessary, but the orbit must also stay above the equator and in the same direction so it appears fixed over one point on Earth.
8 Negative potential gradient shortcut
Many WA problems ask for g at a point off-axis or between bodies. Instead of re-drawing vectors, evaluate φ and differentiate - one line, fewer sign errors.
9 WA timing hacks (tested on IP papers)
Derivations first: write Eqs. (4)-(6) from memory, circle any required answer - anchors marks early.
Unit tagging: copy SI units alongside numbers before punching the calculator.
An early mastery of gravitational fields lets students pre-learn circular motion and satellite communication, compounding advantage in Term 3. We schedule a 60-min clinic right after the Topic 8 lecture - seats fill fast each semester.
11 Mini-drill (do now!)
A probe is already in a 400 km-high circular orbit. What minimum additional speed Δv (applied in the direction of motion) is needed for it to escape Earth? Hint: at the same radius r, orbital speed is v=rGM and escape speed is ve=r2GM, so Δv=ve−v. Answer (Earth):Δv≈3.2km⋅s−1 using r≈(6.37×106+4.00×105)m
Need structured practice on Gravitational Fields? Our H2 Physics tuition Singapore programme covers this topic with weekly problem sets and Paper 4 practical drills.
Comprehensive revision pack
9478 Section II, Topic 8 Syllabus outcomes
Candidates should be able to:
(a) recall and use Newton's law of gravitation in the form F=Gr2m1m2.
(b) derive, from Newton's law of gravitation and the definition of gravitational field strength, the field strength due to a point mass, g=Gr2M.
(c) recall and use g=Gr2M for the gravitational field strength due to a point mass to solve problems.
(d) show an understanding that near the surface of the Earth, gravitational field strength is approximately constant and equal to the acceleration of free fall.
(e) define gravitational potential at a point as the work done per unit mass by an external force in bringing a small test mass from infinity to that point.
(f) solve problems using the equation ϕ=−GrM for the gravitational potential in the field due to a point mass.
(g) show an understanding that the gravitational potential energy of a system of two point masses is UG=−GrMm.
(h) recall that gravitational field strength at a point is equal to the negative potential gradient at that point and use this to solve problems.
(i) analyse problems related to escape velocity by considering energy stores and transfers.
(j) analyse circular orbits in inverse square law fields by relating the gravitational force to the centripetal acceleration it causes.
(k) show an understanding of satellites in geostationary orbit and their applications.
Concept map (in words)
Start from the inverse-square law. Differentiate once to obtain field strength, integrate to recover potential energy. Combine with conservation of energy to handle escape and transfer problems. Link to circular motion by equating gravitational and centripetal forces. Graphs of g(r) and φ(r) reveal behaviour between masses, so practise reading them.
Key definitions & formulae
Quantity
Expression / idea
Gravitational field strength
g=r2GM (towards the mass)
Gravitational potential
φ=−rGM
Gravitational potential energy
UG=−rGMm
Escape velocity
ve=r2GM
Orbital speed (circular)
v=rGM
Orbital period
T=2πGMr3
Field from multiple masses
gtot=∑igi
Potential from multiple masses
φtot=∑iφi
Term guide
Symbol / term
Meaning
G
Universal gravitational constant
M
Mass of the primary body (e.g. planet)
m
Mass of the test object
r
Distance from the centre of mass M
g
Gravitational field strength
φ
Gravitational potential
UG
Gravitational potential energy
ve
Escape velocity at radius r
v
Orbital speed for circular motion
T
Orbital period
gtotal
Vector sum of gravitational fields from multiple masses
φtotal
Scalar sum of potentials from multiple masses
Derivations & reasoning to master
Kepler's third law: combine centripetal requirement with Newton's law to show T2∝r3.
Escape velocity: equate 21mv2 with ∣UG∣ at launch radius, emphasising independence from m.
Energy change for orbital transfers: compute ΔU and ΔEk when moving between radii (e.g., GEO to LEO).
A weather satellite moves from an orbit of radius 7.0×106m to the geostationary radius 4.2×107m. Calculate the work required per unit mass and the change in kinetic energy per unit mass.
Method: compute φ(r) at both radii to find Δφ. Use v=rGM to evaluate the kinetic energy change. Comment on the net energy input required.
Taking Earth GM≈3.99×1014m3⋅s−2,
Δφ=GM(r11−r21)≈4.75×107J⋅kg−1.
Δ(mEk)=21(r2GM−r1GM)≈−2.37×107J⋅kg−1.
So the satellite has higher (less negative) potential but lower kinetic energy in the higher orbit; net energy input per unit mass is about 2.37×107J⋅kg−1.
Worked example 2 - zero net force point
Two planets of masses 5.0×1023kg and 8.0×1023kg are separated by 3.2×108m. Find the point along the line where a small probe experiences zero net gravitational force and determine the potential there.
Strategy: set x2GM1=(d−x)2GM2 to solve for x, then sum the potentials to obtain φ.
x=1+M2/M1d=1+8/53.2×108≈1.41×108m.
So the point is ≈1.41×108m from M1 (and ≈1.79×108m from M2). The potential there is
φ=−xGM1−d−xGM2≈−5.35×105J⋅kg−1.
Practical & data tasks
Plot g vs r using spreadsheet for Earth data; compare near-surface approximation with full expression.
Use NASA orbital databases to compute periods and check T2/r3 consistency.
Model potential wells with elastic sheets or digital simulations; observe how objects move in curved spacetime analogies.
Common misconceptions & exam traps
Thinking gravitational potential is positive; remember zero at infinity, negative elsewhere.
Forgetting that potential is scalar, so contributions add algebraically even when fields oppose.
Confusing escape velocity with orbital velocity (missing 2 factor).
Neglecting the mass of the orbiting body in energy equations - mass cancels only because m appears in every term.
Quick self-check quiz
If Earth's mass doubled but radius stayed the same, how would g at the surface change? - It would double.
Where is gravitational potential zero? - At infinity (reference level by definition).
What provides the centripetal force for the Moon's orbit? - Earth's gravitational pull.
Does a satellite in higher orbit have more or less kinetic energy than one in low orbit? - Less kinetic energy but higher (less negative) total energy.
Why is gravitational potential negative? - Work must be done to separate masses to infinity (attractive interaction).
Revision workflow
Re-derive escape velocity and Kepler's law until you can do both within five minutes.
Complete two past-paper questions on satellite motion and one on zero-field points.
Sketch g(r) and φ(r) for Earth plus Moon; label key radii and explain features.
Create a mind map linking gravitational results to electrostatic analogues (Topic 14 preview).
