H2 Physics Gravitational Fields Notes | A-Level 9478

Study guideUpdated 21 Aug 2026
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 (gg, φ\varphi, UGU_G, 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/r2g=GM/r^2. 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 wordingQuantity to findFirst setupCommon trap
"force on", "attraction between two masses"force FFuse F=Gm1m2/r2F = Gm_1m_2/r^2

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=Gm1m2r2. F = G \dfrac{m_1 m_2}{r^2}.

  • Proportional to mass product: doubling either mass doubles the force.
  • Inverse-square with separation: the force drops by a factor of 44 when distance doubles - the geometry of spreading field lines over a sphere.
  • Universal constant GG: 6.67×1011 Nm2kg26.67 \times 10^{-11} \space \pu{N.m2.kg-2}

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 gg

Field strength is force per unit mass: g=F/mg = F/m. Combining with Newton's law gives

g=GMr2.(1) g = \dfrac{GM}{r^2}. \tag{1}

Near Earth's surface, rr \approx Earth's radius, so g9.81 ms2g \approx 9.81 \space \pu{m.s-2} and is directionally “down”.

Parent insight: “Why is gg 'constant'?” - because rr changes by under one-tenth of a percent across school-lab altitudes, so the drift in gg stays below three-tenths of a percent.

Radius-from-centre checkpoint

In gravitational field questions, rr is measured from the centre of the planet or star, not from the surface. Convert any altitude first, then choose the formula.

Question clueUse for rrWhy it mattersCommon trap
"At the surface of Earth"RER_EThe point is one Earth radius from Earth's centre.Substituting r=0r=0

Worked check: a satellite at height hh above Earth has field strength

g=GME(RE+h)2. g = \frac{GM_E}{(R_E+h)^2}.

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\varphi

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 MM:

φ=GMr.(2) \varphi = -\dfrac{GM}{r}. \tag{2}

The negative sign encodes that gravity is attractive; zero potential is set at infinity.

Gradient link: g=dφdr g = -\dfrac{\mathrm{d}\varphi}{\mathrm{d} r} - differentiating Eq. (2) regenerates Eq. (1).

3.1 Field versus potential checkpoint

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.

SituationField strength ggPotential φ\varphiWhat to write first
One point lies between two massesAdd field vectors with directionAdd scalar potentialsDraw arrows for gg; keep both φ\varphi terms negative

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 UGU_G

For two point masses MM and mm:

UG=GMmr.(3) U_G = -\dfrac{GMm}{r}. \tag{3}

Think of UGU_G as the shared energy store of the pair; separating them to infinity requires positive work equal to UG|U_G|.

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.

MovementWhat happens to UGU_GEnergy statementCommon trap
Move farther from the planetUGU_G becomes less negative.External work must be supplied if no other energy source is involved.

Worked check: moving a satellite of mass mm from radius r1r_1 to a larger radius r2r_2 gives

ΔUG=(GMmr2)(GMmr1). \Delta U_G = \left(-\frac{GMm}{r_2}\right)-\left(-\frac{GMm}{r_1}\right).

Since r2>r1r_2 > r_1, the final value is less negative, so ΔUG>0\Delta U_G > 0

Misconception check: "less negative" means a larger value. For example, 2-2 is greater than 8-8, 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”:

12mve2GMmr=0ve=2GMr.(4) \dfrac{1}{2} m v_e^2 - \dfrac{GMm}{r} = 0 \quad \Rightarrow \quad v_e = \sqrt{\dfrac{2GM}{r}}. \tag{4}

At Earth's surface ve11.2 kms1v_e \approx 11.2 \space \pu{km.s-1} - roughly 450 times the expressway speed limit.


6 Circular orbits & centripetal acceleration

Equate gravitational force to the required centripetal force mv2/rm v^2 / r:

GMmr2=mv2rv=GMr.(5) \dfrac{GMm}{r^2} = \dfrac{m v^2}{r} \quad \Rightarrow \quad v = \sqrt{\dfrac{GM}{r}}. \tag{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 cueFirst principleWhat the final condition meansCommon 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/rGM/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\sqrt{2}

Worked check: at the same radius rr, vorbit=GM/rv_\text{orbit} = \sqrt{GM/r}

Δv=2GMrGMr. \Delta v = \sqrt{\frac{2GM}{r}} - \sqrt{\frac{GM}{r}}.

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

  1. Orbital period = 24 h (synchronised with Earth's spin),
  2. Zero inclination & eccentricity (lies above the equator),
  3. Altitude ≈ 35 786 km.

Applications span weather monitoring, TV broadcast and VSAT internet.

7.1 Deriving the GEO radius

Set centripetal period T=2πrv T = \dfrac{2\pi r}{v} and combine with Eq. (5):

r=GMT24π234.22×104 km(6) r = \sqrt[3]{\dfrac{GMT^2}{4\pi^2}} \approx 4.22 \times 10^4 \space \pu{km} \tag{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.

StepWhat to writeWhy it mattersCommon trap
1The 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.
2Convert the period to seconds before using r=GMT2/(4π2)3r = \sqrt[3]{GMT^2/(4\pi^2)}

Worked check: if r4.22×104 kmr \approx 4.22 \times 10^4 \space \pu{km} and RE6.37×103 kmR_E \approx 6.37 \times 10^3 \space \pu{km}

h4.22×1046.37×1033.58×104 km. h \approx 4.22 \times 10^4 - 6.37 \times 10^3 \approx 3.58 \times 10^4 \space \pu{km}.

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 gg at a point off-axis or between bodies. Instead of re-drawing vectors, evaluate φ\varphi and differentiate - one line, fewer sign errors.


9 WA timing hacks (tested on IP papers)

  1. Derivations first: write Eqs. (4)-(6) from memory, circle any required answer - anchors marks early.
  2. Unit tagging: copy SI units alongside numbers before punching the calculator.
  3. Escape-vs-orbit confusion check: escape needs 2\sqrt{2} factor; circular orbit does not.

10 Why parents should care

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\Delta v (applied in the direction of motion) is needed for it to escape Earth?
Hint: at the same radius rr, orbital speed is v=GMr v = \sqrt{\dfrac{GM}{r}}

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=Gm1m2r2 F = G\dfrac{m_1 m_2}{r^2}

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) g(r) and φ(r) \varphi(r) reveal behaviour between masses, so practise reading them.

Key definitions & formulae

QuantityExpression / idea
Gravitational field strengthg=GMr2 (towards the mass) g = \dfrac{GM}{r^2} \text{ (towards the mass)}
Gravitational potentialφ=GMr \varphi = -\dfrac{GM}{r}

Term guide

Symbol / termMeaning
G G Universal gravitational constant
M M Mass of the primary body (e.g. planet)
m m Mass of the test object
r r Distance from the centre of mass

Derivations & reasoning to master

  1. Kepler's third law: combine centripetal requirement with Newton's law to show T2r3 T^2 \propto r^3 .
  2. Escape velocity: equate 12mv2 \tfrac{1}{2} m v^2

Worked example 1 - satellite relocation

A weather satellite moves from an orbit of radius 7.0×106 m 7.0 \times 10^{6} \space \pu{m} to the geostationary radius 4.2×107 m 4.2 \times 10^{7} \space \pu{m} . Calculate the work required per unit mass and the change in kinetic energy per unit mass.

Method: compute φ(r) \varphi(r) at both radii to find Δφ \Delta \varphi . Use v=GMr v = \sqrt{\dfrac{GM}{r}}

Taking Earth GM3.99×1014m3s2GM \approx 3.99\times 10^{14}\,\pu{m3.s-2},

Δφ=GM(1r11r2)4.75×107Jkg1. \Delta\varphi = GM\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right) \approx 4.75\times 10^{7}\,\pu{J.kg-1}.

Δ(Ekm)=12(GMr2GMr1)2.37×107Jkg1. \Delta\left(\dfrac{E_k}{m}\right) = \tfrac{1}{2}\left(\dfrac{GM}{r_2} - \dfrac{GM}{r_1}\right) \approx -2.37\times 10^{7}\,\pu{J.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 2.37×107Jkg12.37\times 10^{7}\,\pu{J.kg-1}.

Worked example 2 - zero net force point

Two planets of masses 5.0×1023 kg 5.0 \times 10^{23} \space \pu{kg} and 8.0×1023 kg 8.0 \times 10^{23} \space \pu{kg} are separated by 3.2×108 m 3.2 \times 10^{8} \space \pu{m}

Strategy: set GM1x2=GM2(dx)2 \dfrac{GM_1}{x^2} = \dfrac{GM_2}{(d - x)^2}

x=d1+M2/M1=3.2×1081+8/51.41×108m. x = \dfrac{d}{1 + \sqrt{M_2/M_1}} = \dfrac{3.2\times 10^{8}}{1 + \sqrt{8/5}} \approx 1.41\times 10^{8}\,\pu{m}.

So the point is 1.41×108m\approx 1.41\times 10^{8}\,\pu{m} from M1M_1 (and 1.79×108m\approx 1.79\times 10^{8}\,\pu{m}

φ=GM1xGM2dx5.35×105Jkg1. \varphi = -\dfrac{GM_1}{x} - \dfrac{GM_2}{d-x} \approx -5.35\times 10^{5}\,\pu{J.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 T^2 / r^3 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 \sqrt{2} factor).
  • Neglecting the mass of the orbiting body in energy equations - mass cancels only because m m appears in every term.

Quick self-check quiz

  1. If Earth's mass doubled but radius stayed the same, how would g at the surface change? - It would double.
  2. Where is gravitational potential zero? - At infinity (reference level by definition).
  3. What provides the centripetal force for the Moon's orbit? - Earth's gravitational pull.
  4. 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.
  5. Why is gravitational potential negative? - Work must be done to separate masses to infinity (attractive interaction).

Revision workflow

  1. Re-derive escape velocity and Kepler's law until you can do both within five minutes.
  2. Complete two past-paper questions on satellite motion and one on zero-field points.
  3. Sketch g(r) g(r) and φ(r) \varphi(r) for Earth plus Moon; label key radii and explain features.
  4. Create a mind map linking gravitational results to electrostatic analogues (Topic 14 preview).


12 Further reading


Last updated 14 Jul 2025. We will refresh the guide once the 2027 draft syllabus lands.


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Chee Wei Jie
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Chee Wei Jie·Academic Advisor (Physics)

Sources

  1. SEAB: GCE A-Level H2 Physics (9478) syllabus (first examination 2026) (PDF)