Why is "Confusing gravitational field strength g with the universal constant G" a common mistake in Gravitational fields?

Why it happens: They share a letter and both appear in gravitation equations, so students swap them (and their units). How to avoid it: G is the fixed universal constant (6.67×10⁻¹¹ N m² kg⁻²); g = F/m = GM/r² is the field strength (N kg⁻¹) that varies with position. Learn the two units — they are different — and check which one the question needs. Source: IB Physics Theme D subject reports

Why is "Using 1/r instead of 1/r² (forgetting the inverse-SQUARE)" a common mistake in Gravitational fields?

Why it happens: Students remember the force gets weaker with distance but forget the square, or forget to square r after substituting. How to avoid it: Both F = Gm₁m₂/r² and g = GM/r² have r SQUARED in the denominator. Doubling r divides the result by 4, not 2. Always square the distance in your calculator step. Source: IB Physics Theme D subject reports

Why is "Measuring r from the surface instead of the centre of the planet" a common mistake in Gravitational fields?

Why it happens: A question gives an altitude ('400 km above the surface') and students use that number directly as r. How to avoid it: r is always the distance from the CENTRE. For a satellite, add the planet's radius to the altitude: r = R_planet + h. Only then substitute into F = Gm₁m₂/r² or g = GM/r². Source: IB Physics Theme D subject reports

Why is "Thinking the orbital speed depends on the satellite's mass" a common mistake in Gravitational fields?

Why it happens: Heavier objects feel a bigger force, so students assume they must orbit differently. How to avoid it: In GMm/r² = mv²/r the satellite mass m cancels, leaving v = √(GM/r). Orbital speed depends only on the central mass M and the radius r — a light and a heavy satellite at the same height move at the same speed. Source: IB Physics Theme D subject reports

Why is "Saying orbiting astronauts experience 'no gravity' or 'zero gravity'" a common mistake in Gravitational fields?

Why it happens: Astronauts are seen floating, so students conclude gravity has switched off. How to avoid it: The field strength in low orbit is close to its surface value (≈ 8.7 N kg⁻¹ for the ISS) — gravity is what keeps them in orbit. They feel weightless because they are in FREE FALL with no contact force, not because g is zero. Source: IB Physics Theme D subject reports

Why is "Power-of-ten and unit slips when substituting G and large masses" a common mistake in Gravitational fields?

Why it happens: G is 10⁻¹¹ and planetary masses and radii are huge powers of ten, so exponents get mishandled and units dropped. How to avoid it: Enter each value in standard form carefully, keep track of the powers of ten (especially after squaring r), and always attach the correct unit — N for force, N kg⁻¹ for field strength. Source: IB Physics Theme D subject reports

IB DP Physics (100452)

Gravitational fields

0% Complete

Detailed Study Notes

Detailed notes on Fields for IB DP Physics, covering key concepts, explanations, examples, and exam-focused revision points.

D.1 Gravitational fields — IB Physics SL Study Notes (Theme D: Fields)

Newton's law of universal gravitation as an inverse-square law, the idea of a gravitational field and its strength g = F/m, radial and uniform field lines, weight, and orbital motion — where the gravitational force provides the centripetal force, giving orbital speed v = √(GM/r) and Kepler's third law T² ∝ r³.

At a glance

Newton's law of universal gravitation: every mass attracts every other mass with a force $F = \dfrac{Gm_1m_2}{r^2}$ , where $G = 6.67\times10^{-11}\,\text{N m}^2\,\text{kg}^{-2}$ .
It is an inverse-square law — double the separation and the force falls to a quarter. $r$ is always measured centre-to-centre.
A gravitational field is a region where a mass feels a gravitational force. Its strength is $g = \dfrac{F}{m}$ (unit $\text{N kg}^{-1}$ , a vector pointing towards the mass).
Around a point or spherical mass the field strength is $g = \dfrac{GM}{r^2}$ , so it also falls with the square of the distance from the centre.
Field lines are radial (pointing inward) for a point mass, but look uniform (parallel, equally spaced) over a small region near a planet's surface. Weight is $W = mg$ .
For an orbit the gravitational force provides the centripetal force: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ , giving orbital speed $v = \sqrt{\dfrac{GM}{r}}$ — independent of the satellite's mass.
Combining this with $v = \dfrac{2\pi r}{T}$ gives Kepler's third law $T^2 = \dfrac{4\pi^2}{GM}r^3$ , i.e. $T^2 \propto r^3$ .
Orbiting astronauts are in free fall, not in zero gravity — the field strength $g$ where the ISS orbits is still about $8.7\,\text{N kg}^{-1}$ .

What you’ll learn

Mapped to the 100452 subject guide (2025-onwards).

State and apply Newton's law of universal gravitation $F = Gm_1m_2/r^2$ , recognising it as an inverse-square law with $r$ measured centre-to-centre.
Define a gravitational field and gravitational field strength $g = F/m$ , and calculate the field around a point or spherical mass using $g = GM/r^2$ .
Represent gravitational fields with field lines — radial around a point mass and uniform near a planet's surface — and relate weight to field strength through $W = mg$ .
Explain that for a satellite or planet the gravitational force provides the centripetal force, and derive the orbital speed $v = \sqrt{GM/r}$ , noting it is independent of the orbiting mass.
Derive and use Kepler's third law $T^2 = (4\pi^2/GM)r^3$ by combining gravitation with circular motion, and explain the apparent weightlessness of orbiting astronauts as free fall.

Newton's law of universal gravitation

Every mass attracts every other mass with F = Gm₁m₂/r² — an inverse-square law.

Every object with mass attracts every other object with mass. This is gravitation, and Newton captured it in one equation — the law of universal gravitation:

$F = \frac{Gm_1m_2}{r^2}$

$F$ is the attractive force on each mass (they pull on each other equally and oppositely, Newton's third law), in newtons (N).
$m_1$ and $m_2$ are the two masses, in kilograms (kg).
$r$ is the distance between the centres of the two masses, in metres (m).
$G$ is the universal gravitational constant, $G = 6.67\times10^{-11}\,\text{N m}^2\,\text{kg}^{-2}$ . It is the same everywhere in the Universe — that is why it is called universal.

Why "universal"? The same equation governs an apple falling from a tree, the Moon orbiting the Earth, and a planet orbiting the Sun. Newton's insight was that these are the same force.

It is an inverse-square law. The force depends on $1/r^2$ , not $1/r$ . This has a dramatic consequence:

Separation	Force
$r$	$F$
$2r$	$F/4$
$3r$	$F/9$
$10r$	$F/100$

So if you double the distance, the force drops to a quarter — not a half. Getting this square right is worth easy marks.

Measure $r$ from the centres. For two spheres (like planets, or a planet and a satellite), $r$ is the distance between their centres, not the gap between their surfaces. A satellite "400 km above the Earth" is at $r = R_{\text{Earth}} + 400\,\text{km} = 6.37\times10^6 + 0.40\times10^6\,\text{m}$ from the centre.

The two masses attract each other with forces that are equal in size and opposite in direction, separated by a distance r measured from centre to centre.

Gravity is astonishingly weak. Because $G$ is so tiny, the attraction between everyday objects is negligible — two 5 kg masses 10 cm apart attract with a force of only about $10^{-7}\,\text{N}$ . Gravity only becomes important when at least one of the masses is enormous, like a planet or a star.

F = Gm₁m₂/r²; G = 6.67×10⁻¹¹ N m² kg⁻² is the same everywhere.
Inverse-square: double the separation → force falls to a quarter.
r is always the centre-to-centre distance, not the surface gap.

See the full worked example for gravitational fields →

Gravitational fields and field strength g = F/m

A field is a region where a mass feels a force; its strength is g = F/m = GM/r².

Rather than always talking about the force between two specific masses, physicists use the idea of a field. A gravitational field is a region of space in which a mass experiences a gravitational force. The Earth is surrounded by its gravitational field; step into it (which we always are) and you feel a force — your weight.

Gravitational field strength $g$ measures how strong the field is at a point. It is the force per unit mass on a small test mass placed there:

$g = \frac{F}{m}$

Its unit is the newton per kilogram ( $\text{N kg}^{-1}$ ).
It is a vector — it points in the direction of the force, i.e. towards the mass creating the field.

The field of a point mass or a sphere. Combine $g = F/m$ with Newton's law. The force on a small mass $m$ a distance $r$ from a large mass $M$ is $F = \dfrac{GMm}{r^2}$ , so:

$g = \frac{F}{m} = \frac{GMm/r^2}{m} = \frac{GM}{r^2}$

Notice the small mass $m$ cancels — the field strength depends only on the mass $M$ producing it and the distance $r$ . Like the force, $g$ obeys an inverse-square law: it falls to a quarter when you double the distance from the centre.

Check with the Earth. With $M = 5.97\times10^{24}\,\text{kg}$ and $r = R_{\text{Earth}} = 6.37\times10^6\,\text{m}$ : $g = \frac{(6.67\times10^{-11})(5.97\times10^{24})}{(6.37\times10^6)^2} = 9.81\,\text{N kg}^{-1}$ This is exactly the familiar "acceleration of free fall" — because $g = F/m$ has the same value and units as an acceleration ( $\text{N kg}^{-1} = \text{m s}^{-2}$ ). Gravitational field strength and free-fall acceleration are two names for the same quantity.

Outside a planet the field strength g falls as 1/r². At twice the surface radius (2R) the field is only a quarter of its surface value — the signature of an inverse-square law.

Two different g's, two different r's. The surface value ( $9.81\,\text{N kg}^{-1}$ for Earth) is just $g = GM/r^2$ evaluated at $r = R_{\text{Earth}}$ . Go higher and $r$ increases, so $g$ decreases. A satellite in orbit is in a weaker field than the ground, but the field is still very much present.

Gravitational field = region where a mass feels a gravitational force.
Field strength g = F/m (N kg⁻¹), a vector pointing towards the mass.
For a point/spherical mass g = GM/r² — the test mass cancels; inverse-square.

See the full worked example for gravitational fields →

Field lines, radial and uniform fields, and weight

Field lines are radial for a point mass but look uniform near a surface; W = mg.

We picture a gravitational field with field lines. The rules:

Field lines point in the direction of the force on a mass — always towards the mass creating the field (gravity only attracts, never repels).
Where the lines are closer together, the field is stronger.

Radial field (around a point mass or a whole planet). The lines point radially inwards, like spokes towards the centre. They get closer together as you approach the mass, showing that $g$ increases (as $1/r^2$ ) the nearer you get.

Uniform field (near a planet's surface). Over a small region — say, a lab, a football pitch, the height of a building — the field lines are so nearly parallel and equally spaced that we treat the field as uniform: $g$ has the same value ( $9.81\,\text{N kg}^{-1}$ ) and the same direction (straight down) everywhere. This is why, in everyday mechanics, we just say " $g = 9.81$ " and forget that it varies with height.

Zoomed out, the field is radial — lines point inward and crowd together near the surface. Zoomed in to a small patch of the surface, the same lines look parallel and equally spaced: a uniform field of strength g pointing straight down.

Weight is a force, and it comes from the field. The weight of an object is the gravitational force the field exerts on it:

$W = mg$

where $m$ is the object's mass (a scalar, the same everywhere) and $g$ is the local field strength. Because $g$ is smaller on the Moon ( $1.6\,\text{N kg}^{-1}$ ) than on Earth ( $9.81\,\text{N kg}^{-1}$ ), the same object weighs about six times less on the Moon even though its mass is unchanged. Mass and weight are not the same thing: mass is how much matter there is (kg); weight is the gravitational force on it (N).

Field lines point towards the mass; closer lines mean a stronger field.
Radial field around a whole planet; uniform (parallel) field near the surface.
Weight W = mg — a force in newtons; mass in kg is unchanged by location.

See the full worked example for gravitational fields →

Orbital motion: gravity provides the centripetal force

For a circular orbit GMm/r² = mv²/r, giving v = √(GM/r), independent of the orbiting mass.

A satellite (or a planet, or the Moon) moving in a circular orbit is constantly changing direction, so it is accelerating towards the centre — it needs a centripetal force. What provides that force? Gravity. There is nothing else out there pulling it. This single idea unlocks every SL orbit problem:

The gravitational force provides the centripetal force.

Set the gravitational force equal to the required centripetal force ( $F_c = mv^2/r$ ):

$\frac{GMm}{r^2} = \frac{mv^2}{r}$

Here $M$ is the mass of the central body (e.g. the Earth) and $m$ is the mass of the orbiting satellite. Now the beautiful part — the satellite's mass $m$ appears on both sides and cancels, and one power of $r$ cancels:

$\frac{GM}{r} = v^2 \quad\Rightarrow\quad v = \sqrt{\frac{GM}{r}}$

Orbital speed depends only on the central mass $M$ and the orbital radius $r$ — not on the mass of the satellite. A tiny CubeSat and the enormous International Space Station, orbiting at the same height, travel at exactly the same speed. This is a favourite examiner point.

The gravitational pull on the satellite points straight towards the centre and supplies exactly the centripetal force needed to keep it circling. Its velocity is at right angles, along the tangent.

Higher orbit → slower speed. Because $v = \sqrt{GM/r}$ , a satellite in a higher orbit (larger $r$ ) actually moves more slowly. Low-Earth-orbit satellites race round at about $7.5\,\text{km s}^{-1}$ (a full orbit in roughly 90 minutes), while the distant Moon ambles along at about $1\,\text{km s}^{-1}$ .

The period. The orbital period $T$ is the time for one full circle. The satellite covers the circumference $2\pi r$ at speed $v$ , so: $T = \frac{2\pi r}{v}$ This link between $v$ , $r$ and $T$ is the bridge to Kepler's third law in the next section.

In orbit, gravity IS the centripetal force: GMm/r² = mv²/r.
The satellite mass cancels → v = √(GM/r), independent of the orbiting mass.
Higher orbit (larger r) means slower speed; T = 2πr/v for one orbit.

See the full worked example for gravitational fields →

Deriving Kepler's third law (T² ∝ r³)

Combining gravity = centripetal with T = 2πr/v gives T² = (4π²/GM)r³.

This is the derivation IB loves to set as a "show that" question, so learn to reproduce it. It combines everything from the previous section with the definition of the period — no new physics, just algebra.

Step 1 — gravity provides the centripetal force (as before): $\frac{GMm}{r^2} = \frac{mv^2}{r} \quad\Rightarrow\quad v^2 = \frac{GM}{r}$

Step 2 — express the speed using the period. The satellite travels one circumference $2\pi r$ in one period $T$ : $v = \frac{2\pi r}{T} \quad\Rightarrow\quad v^2 = \frac{4\pi^2 r^2}{T^2}$

Step 3 — set the two expressions for $v^2$ equal: $\frac{4\pi^2 r^2}{T^2} = \frac{GM}{r}$

Step 4 — rearrange for $T^2$ . Multiply both sides by $\dfrac{r\,T^2}{GM}$ : $\boxed{T^2 = \frac{4\pi^2}{GM}\,r^3}$

This is Kepler's third law. The whole factor $\dfrac{4\pi^2}{GM}$ is a constant for a given central mass $M$ , so:

$T^2 \propto r^3$

In words: the square of the orbital period is proportional to the cube of the orbital radius. Double the radius and the period grows by a factor of $2^{3/2} \approx 2.83$ ; a planet four times further from the Sun takes $4^{3/2} = 8$ times as long to go round.

Why it matters — weighing the Universe. Rearranged as $M = \dfrac{4\pi^2 r^3}{G T^2}$ , this equation lets astronomers work out the mass of a planet or star just from the radius and period of something orbiting it. Measuring the Moon's orbit gives the mass of the Earth; measuring a planet's orbit gives the mass of the Sun. Kepler first spotted the $T^2 \propto r^3$ pattern in observations of the planets; Newton's gravitation explains it.

Combine GMm/r² = mv²/r with v = 2πr/T.
Eliminate v to get T² = (4π²/GM)r³, so T² ∝ r³.
Rearranged, M = 4π²r³/(GT²) — how we measure the mass of planets and stars.

See the full worked example for gravitational fields →

Apparent weightlessness: astronauts are in free fall

Orbiting astronauts feel weightless because they and their craft fall together — g is not zero.

Photos from the International Space Station (ISS) show astronauts floating, so it is tempting to say "there is no gravity in space." This is wrong, and it is one of the most heavily penalised misconceptions in the topic.

There is plenty of gravity where the ISS orbits. The ISS is only about 400 km up. Its orbital radius is $r = 6.37\times10^6 + 0.40\times10^6 = 6.77\times10^6\,\text{m}$ , so: $g = \frac{GM}{r^2} = \frac{(6.67\times10^{-11})(5.97\times10^{24})}{(6.77\times10^6)^2} \approx 8.7\,\text{N kg}^{-1}$ That is about 89% of the surface value — the astronauts are very much inside the Earth's field. Gravity is exactly what holds them in orbit.

So why do they float? Because the astronaut and the station are both in free fall — they are both accelerating towards the Earth at the same rate $g$ , so there is no contact force between them. On the ground, the floor pushes up on you (a normal reaction) and it is that push you actually feel as "weight". In orbit there is no such push, because the floor is falling away just as fast as you are. This is apparent weightlessness: your true weight $W = mg$ is still large, but there is no supporting force, so you feel weightless.

The same effect on Earth. You feel the identical sensation for a moment at the top of a roller-coaster drop, or in a lift whose cable is cut, or on a "zero-g" aircraft that flies a falling curve. In each case you and your surroundings fall together, the contact force vanishes, and you feel weightless — even though gravity is acting the whole time.

An orbit is just falling that keeps missing the ground. A satellite moves sideways so fast that, as it falls towards the Earth, the curved surface of the Earth falls away beneath it at the same rate. It is in perpetual free fall — which is exactly why orbital speed came out independent of mass earlier: everything falls at the same rate.

Where the ISS orbits, g ≈ 8.7 N kg⁻¹ — gravity is strong, not absent.
Astronauts float because they and the station are in free fall together (no contact force).
Apparent weightlessness: true weight W = mg is still there; only the supporting force is gone.

Quick recap

Newton's law of universal gravitation: F = Gm₁m₂/r², an inverse-square law with G = 6.67×10⁻¹¹ N m² kg⁻² and r measured centre-to-centre.
A gravitational field is a region where a mass feels a force; its strength g = F/m (N kg⁻¹, a vector) equals GM/r² around a point or spherical mass.
Field lines are radial around a whole planet but uniform (parallel) over a small region near the surface; weight W = mg.
For an orbit the gravitational force provides the centripetal force: GMm/r² = mv²/r, so v = √(GM/r) — independent of the satellite's mass; higher orbits are slower.
Combining gravity = centripetal with T = 2πr/v gives Kepler's third law T² = (4π²/GM)r³, so T² ∝ r³.
Orbiting astronauts are in free fall, not zero gravity — g where the ISS orbits is still about 8.7 N kg⁻¹; they float because there is no supporting contact force.

Memorise this

Verbatim phrases, formulae and definitions IB DP mark schemes credit (key for AO1 knowledge marks on Paper 1).

Newton's law: F = Gm₁m₂/r² (inverse-SQUARE; r is centre-to-centre).
G = 6.67×10⁻¹¹ N m² kg⁻² — a universal constant, the same everywhere.
Field strength g = F/m (N kg⁻¹, a vector pointing towards the mass).
Field of a point/spherical mass: g = GM/r² (also inverse-square).
Weight: W = mg (a force in N); mass in kg is unchanged by location.
Orbit: gravity provides the centripetal force → GMm/r² = mv²/r → v = √(GM/r).
Kepler's third law: T² = (4π²/GM)r³, so T² ∝ r³.
Orbiting astronauts are in free fall (g ≠ 0), not in zero gravity.

How it’s examined

D.1 sits at the start of Theme D (Fields) and is examined in every paper. Paper 1A (MCQ): recognising the inverse-square dependence (e.g. how F or g changes when r is doubled or tripled), distinguishing g from G, and identifying that orbital speed is independent of satellite mass — numbers are kept clean because there is no calculator. Paper 1B (data-based): interpreting a g-against-r graph or orbital data, and confirming the T² ∝ r³ relationship from a table or a straight-line graph of T² against r³. Paper 2: structured calculations using F = Gm₁m₂/r², g = GM/r² and v = √(GM/r); 'show that' derivations of orbital speed or Kepler's third law; and extended-response items combining field strength at altitude, orbital speed, period and an explanation of apparent weightlessness (3–10 marks). Command terms: state, define, determine, calculate, show that, describe, explain. Examiner reports repeatedly flag: confusing g with G, forgetting the square in the inverse-square law, measuring r from the surface instead of the centre, claiming orbital speed depends on the satellite's mass, and saying orbiting astronauts have 'no gravity'. Always show full working and quote units — method marks are awarded even when the final arithmetic slips.

Sources: IB Diploma Programme Physics Guide (first assessment 2025) — Theme D: Fields (D.1 Gravitational fields); IB Physics Data Booklet (2025); IB Physics subject reports and specimen papers (2023–2025). Last reviewed 2026-07-21.

Take this whole topic with you

Step-by-step worked examples — Gravitational fields

Step-by-step solutions to past-paper-style questions on gravitational fields, written exactly the way a tutor would explain them at the board.

Question type:

1Force between two masses (Newton's law)
Getting startedDirect calculation• Newton's law, AO2
Question
Two small masses of 2.0 kg and 5.0 kg have their centres 0.10 m apart. Calculate the gravitational force of attraction between them. (G = 6.67×10⁻¹¹ N m² kg⁻²) (2 marks)
Step-by-step solution
1. Step 1
  Use Newton's law of universal gravitation with $m_1 = 2.0\,\text{kg}$ , $m_2 = 5.0\,\text{kg}$ , $r = 0.10\,\text{m}$ . Remember to square the separation.
  $F = \frac{Gm_1m_2}{r^2} = \frac{(6.67\times10^{-11})(2.0)(5.0)}{(0.10)^2}$
2. Step 2
  Evaluate: numerator $= 6.67\times10^{-10}$ ; denominator $= 0.010$ .
  $F = \frac{6.67\times10^{-10}}{0.010} = 6.7\times10^{-8}\,\text{N}$
Answer
F ≈ 6.7×10⁻⁸ N (attractive).
Examiner tip
Mark scheme: (1) correct substitution into F = Gm₁m₂/r² with r squared; (2) F = 6.7×10⁻⁸ N with unit. The tiny result is a reminder that gravity between everyday objects is negligible — only large masses matter.
2Sketching and describing gravitational field lines
Getting startedGraph or diagram• field lines, AO1
Question
(a) Sketch the gravitational field lines around an isolated spherical planet. (b) State the difference between this field and the field over a small region at the planet's surface. (c) In which direction does a gravitational field line point? (3 marks)
Step-by-step solution
1. Step 1
  (a) Draw straight lines pointing radially inward towards the centre of the planet, evenly spaced around it, getting closer together nearer the surface (stronger field).
2. Step 2
  (b) Around the whole planet the field is radial (lines converge on the centre and $g$ varies as $1/r^2$ ). Over a small region at the surface the lines are so nearly parallel and equally spaced that the field is uniform — $g$ is effectively constant in size and direction.
3. Step 3
  (c) A gravitational field line points in the direction of the force on a mass, i.e. towards the mass creating the field (gravity is always attractive).
Answer
(a) radial lines pointing inward, closer together near the surface; (b) radial (varying) far out vs uniform (parallel) over a small surface region; (c) towards the mass creating the field.
Examiner tip
Mark scheme: (1) radial lines pointing inward; (2) radial-vs-uniform distinction; (3) lines point towards the mass. A frequent error is drawing the lines pointing outward — gravity only attracts, so lines always point inward.
3Distinguishing g from G
Getting startedIdentify & classify• definitions, AO1
Question
A student writes 'g and G are the same thing'. Explain why this is wrong by stating, for each, (a) what it is, (b) its unit, and (c) whether it changes from place to place. (3 marks)
Step-by-step solution
1. Step 1
  G is the universal gravitational constant in Newton's law $F = Gm_1m_2/r^2$ . Unit: $\text{N m}^2\,\text{kg}^{-2}$ . It has the same fixed value ( $6.67\times10^{-11}$ ) everywhere in the Universe.
2. Step 2
  g is the gravitational field strength (force per unit mass), $g = F/m = GM/r^2$ . Unit: $\text{N kg}^{-1}$ . It is a vector that changes with position — 9.81 N kg⁻¹ at Earth's surface, 1.6 N kg⁻¹ on the Moon, smaller still up in orbit.
3. Step 3
  So they differ in meaning, unit and behaviour: $G$ is a constant of nature; $g$ describes how strong the field is at a particular place.
Answer
G = universal constant (N m² kg⁻², fixed everywhere); g = field strength (N kg⁻¹, a vector that varies with location). They are completely different quantities.
Examiner tip
Mark scheme: (1) G is the universal constant with its unit; (2) g is field strength (N kg⁻¹) that varies; (3) clear statement that they differ. Mixing up g and G — including their units — is one of the most common examiner-flagged errors in Theme D.
4Surface gravitational field strength of a planet
Building confidenceDirect calculation• field strength, AO2
Question
Mars has mass 6.42×10²³ kg and radius 3.39×10⁶ m. Calculate the gravitational field strength at its surface. (G = 6.67×10⁻¹¹ N m² kg⁻²) (3 marks)
Step-by-step solution
1. Step 1
  At the surface the distance from the centre is the planet's radius. Use $g = GM/r^2$ with $M = 6.42\times10^{23}\,\text{kg}$ and $r = 3.39\times10^6\,\text{m}$ .
  $g = \frac{GM}{r^2} = \frac{(6.67\times10^{-11})(6.42\times10^{23})}{(3.39\times10^6)^2}$
2. Step 2
  Numerator $= 4.28\times10^{13}$ ; denominator $= (3.39\times10^6)^2 = 1.15\times10^{13}$ .
  $g = \frac{4.28\times10^{13}}{1.15\times10^{13}}$
3. Step 3
  Evaluate the quotient.
  $g = 3.7\,\text{N kg}^{-1}$
Answer
g ≈ 3.7 N kg⁻¹ (about 38% of Earth's surface value).
Examiner tip
Mark scheme: (1) select g = GM/r² with r = radius; (2) correct substitution and squaring of r; (3) g = 3.7 N kg⁻¹ with unit. The value is sensible — Mars is smaller and less massive than Earth, so weaker gravity.
5How g changes with distance (inverse-square)
Building confidenceWord problem• inverse-square, weight, AO2
Question
At the surface of a planet (radius R) the gravitational field strength is 9.6 N kg⁻¹. (a) Find the field strength at a height R above the surface (i.e. at a distance 2R from the centre). (b) A 50 kg astronaut stands on the surface; what is her weight there? (3 marks)
Step-by-step solution
1. Step 1
  (a) $g = GM/r^2$ , so $g \propto 1/r^2$ . Moving from $r = R$ to $r = 2R$ multiplies $r$ by 2, so $g$ is divided by $2^2 = 4$ .
  $g_{2R} = \frac{g_R}{4} = \frac{9.6}{4} = 2.4\,\text{N kg}^{-1}$
2. Step 2
  (b) Weight is the gravitational force on the astronaut: $W = mg$ at the surface, where $g = 9.6\,\text{N kg}^{-1}$ .
  $W = mg = 50 \times 9.6 = 480\,\text{N}$
Answer
(a) g at 2R = 2.4 N kg⁻¹; (b) weight = 480 N.
Examiner tip
Mark scheme: (1) recognise g ∝ 1/r² and that 2R gives a quarter; (2) g = 2.4 N kg⁻¹; (3) W = mg = 480 N with unit. The classic trap is halving g instead of quartering it — the field falls as the SQUARE of the distance.
6Is there gravity on the ISS?
Building confidenceMulti-step problem• field strength, weightlessness, AO2, AO3
Question
The International Space Station orbits at a height of 400 km above the Earth's surface. Earth's mass is 5.97×10²⁴ kg and its radius is 6.37×10⁶ m. (a) Calculate the gravitational field strength at this height. (b) Astronauts on board appear weightless. Explain why, given your answer to (a). (4 marks)
Step-by-step solution
1. Step 1
  (a) The orbital radius is measured from the centre of the Earth: $r = R_{\text{Earth}} + h = 6.37\times10^6 + 0.40\times10^6 = 6.77\times10^6\,\text{m}$ .
  $r = 6.77\times10^6\,\text{m}$
2. Step 2
  Apply $g = GM/r^2$ .
  $g = \frac{(6.67\times10^{-11})(5.97\times10^{24})}{(6.77\times10^6)^2} = 8.7\,\text{N kg}^{-1}$
3. Step 3
  (b) The field strength is about 8.7 N kg⁻¹ — nearly 90% of the surface value — so gravity is still strong; it is what keeps the ISS in orbit.
4. Step 4
  The astronauts feel weightless because they and the station are both in free fall, accelerating towards the Earth at the same rate. There is no contact (supporting) force between them, so they float — but their true weight $W = mg$ is not zero.
Answer
(a) g ≈ 8.7 N kg⁻¹; (b) gravity is still large, so weightlessness is apparent — the astronauts and station are in free fall together, with no contact force between them.
Examiner tip
Mark scheme: (1) r measured from the centre (add radius to height); (2) g = 8.7 N kg⁻¹; (3) state gravity is still present/strong; (4) apparent weightlessness explained by free fall / no contact force. Saying there is 'no gravity in space' scores zero for part (b).
7Orbital speed and period of a satellite
StretchMulti-step problem• orbital motion, centripetal, AO2
Question
A satellite orbits the Earth in a circular path of radius 7.0×10⁶ m. Earth's mass is 5.97×10²⁴ kg. (a) Show that the gravitational force provides the centripetal force, and hence find the orbital speed. (b) Calculate the orbital period. (5 marks)
Step-by-step solution
1. Step 1
  (a) The only force on the satellite is gravity, and a circular orbit needs a centripetal force, so gravity is the centripetal force. Set them equal:
  $\frac{GMm}{r^2} = \frac{mv^2}{r}$
2. Step 2
  The satellite's mass $m$ cancels (and one power of $r$ ), leaving $v^2 = GM/r$ , so $v = \sqrt{GM/r}$ .
  $v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{(6.67\times10^{-11})(5.97\times10^{24})}{7.0\times10^6}}$
3. Step 3
  Evaluate the orbital speed.
  $v = \sqrt{5.69\times10^{7}} = 7.5\times10^3\,\text{m s}^{-1}$
4. Step 4
  (b) The satellite travels one circumference $2\pi r$ in one period $T$ , so $T = 2\pi r/v$ .
  $T = \frac{2\pi r}{v} = \frac{2\pi(7.0\times10^6)}{7.54\times10^3}$
5. Step 5
  Evaluate the period (≈ 97 minutes).
  $T = 5.8\times10^3\,\text{s}$
Answer
(a) v ≈ 7.5×10³ m s⁻¹; (b) T ≈ 5.8×10³ s (≈ 97 min).
Examiner tip
Mark scheme: (1) equate gravitational and centripetal forces; (2) cancel m → v = √(GM/r); (3) v = 7.5×10³ m s⁻¹; (4) T = 2πr/v; (5) T = 5.8×10³ s. Note the satellite mass never enters — the orbital speed is independent of it.
8Deriving Kepler's third law
StretchShow that / prove• Kepler's third law, AO2, derivation
Question
For a satellite in a circular orbit of radius r around a mass M, show that the orbital period T satisfies T² = (4π²/GM)r³. Hence state how T depends on r. (4 marks)
Step-by-step solution
1. Step 1
  The gravitational force provides the centripetal force for the circular orbit:
  $\frac{GMm}{r^2} = \frac{mv^2}{r} \;\Rightarrow\; v^2 = \frac{GM}{r}$
2. Step 2
  The satellite covers one circumference $2\pi r$ in one period $T$ , so express $v$ (and $v^2$ ) using $T$ :
  $v = \frac{2\pi r}{T} \;\Rightarrow\; v^2 = \frac{4\pi^2 r^2}{T^2}$
3. Step 3
  Set the two expressions for $v^2$ equal:
  $\frac{4\pi^2 r^2}{T^2} = \frac{GM}{r}$
4. Step 4
  Rearrange for $T^2$ (multiply both sides by $rT^2/GM$ ). Since $4\pi^2/GM$ is constant, $T^2 \propto r^3$ .
  $T^2 = \frac{4\pi^2}{GM}\,r^3$
Answer
T² = (4π²/GM)r³, so T² is proportional to r³ (Kepler's third law).
Examiner tip
Mark scheme: (1) equate gravitational and centripetal forces → v² = GM/r; (2) v = 2πr/T; (3) equate the two v² expressions; (4) rearrange to T² = 4π²r³/GM and state T² ∝ r³. A clean 'show that' answer states every step of the algebra — do not skip the substitution for v.

Model Answers — Gravitational fields

High-scoring sample answers for gravitational fields on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). State Newton's law of universal gravitation in words, and write down the equation, defining each symbol.
Model answer
In words: The gravitational force between two point masses is directly proportional to the product of the masses and inversely proportional to the square of the distance between their centres, and acts along the line joining them.

Equation: $F = \frac{Gm_1m_2}{r^2}$ where $F$ is the attractive force, $m_1$ and $m_2$ are the two masses, $r$ is the distance between their centres, and $G = 6.67\times10^{-11}\,\text{N m}^2\,\text{kg}^{-2}$ is the universal gravitational constant.
Why this scores
Why this scores 2/2. (1) statement: proportional to product of masses AND inversely proportional to the square of the separation; (2) correct equation with all symbols defined. Omitting the word 'square' (inverse-square) or forgetting to define G costs the second mark.
Question 2
3 marks
Q (3 marks). The Moon has mass 7.35×10²² kg and radius 1.74×10⁶ m. Calculate the gravitational field strength at its surface. (G = 6.67×10⁻¹¹ N m² kg⁻²)
Model answer
At the surface, the distance from the centre equals the Moon's radius, so use $g = GM/r^2$ :

$g = \frac{GM}{r^2} = \frac{(6.67\times10^{-11})(7.35\times10^{22})}{(1.74\times10^6)^2}$

$g = \frac{4.90\times10^{12}}{3.03\times10^{12}} = 1.6\,\text{N kg}^{-1}$

The gravitational field strength at the Moon's surface is about 1.6 N kg⁻¹ — roughly one sixth of Earth's, which is why astronauts could bound across the surface.
Why this scores
Why this scores 3/3. (1) select g = GM/r² with r = radius; (2) correct substitution and squaring of r; (3) g = 1.6 N kg⁻¹ with unit. Using the diameter instead of the radius, or forgetting to square r, are the usual slips.
Question 3
2 marks
Q (2 marks). Define gravitational field strength, and state its SI unit and whether it is a scalar or a vector.
Model answer
Gravitational field strength at a point is the gravitational force per unit mass experienced by a small test mass placed at that point: $g = \frac{F}{m}$

Its SI unit is the newton per kilogram (N kg⁻¹), and it is a vector (it has a direction — towards the mass producing the field).
Why this scores
Why this scores 2/2. (1) force per unit mass, g = F/m; (2) unit N kg⁻¹ and identified as a vector. 'Force per unit mass' is the key phrase — answers that just say 'the force of gravity' do not earn the definition mark.
Question 4
4 marks
Q (4 marks). The Earth (mass 5.97×10²⁴ kg) and the Moon (mass 7.35×10²² kg) are separated by 3.84×10⁸ m (centre to centre). (a) Calculate the gravitational force between them. (b) State the size and direction of the force the Moon exerts on the Earth. (G = 6.67×10⁻¹¹ N m² kg⁻²)
Model answer
(a) Apply Newton's law of universal gravitation: $F = \frac{Gm_1m_2}{r^2} = \frac{(6.67\times10^{-11})(5.97\times10^{24})(7.35\times10^{22})}{(3.84\times10^8)^2}$

$F = \frac{2.93\times10^{37}}{1.47\times10^{17}} = 2.0\times10^{20}\,\text{N}$

(b) By Newton's third law, the Moon pulls on the Earth with a force of the same magnitude, 2.0×10²⁰ N, directed towards the Moon (equal and opposite to the Earth's pull on the Moon).
Why this scores
Why this scores 4/4. (1) correct substitution into F = Gm₁m₂/r²; (2) F = 2.0×10²⁰ N with unit; (3) same magnitude by Newton's third law; (4) direction towards the Moon. Part (b) tests the equal-and-opposite nature of gravitation — the forces on the two bodies are identical in size.
Question 5
5 marks
Q (5 marks). A satellite moves in a circular orbit of radius 8.0×10⁶ m around the Earth (mass 5.97×10²⁴ kg). (a) Explain why the gravitational force acts as the centripetal force. (b) Derive an expression for the orbital speed and calculate it. (c) State whether a heavier satellite in the same orbit would move faster, slower or at the same speed. (G = 6.67×10⁻¹¹ N m² kg⁻²)
Model answer
(a) A satellite in a circular orbit continually changes direction, so it accelerates towards the centre and requires a centripetal force. The only force acting on it is the Earth's gravitational pull, which is directed towards the centre — so gravity provides exactly this centripetal force.

(b) Equate the gravitational force to the centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r}$ The satellite mass $m$ cancels, giving $v^2 = GM/r$ , so: $v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{(6.67\times10^{-11})(5.97\times10^{24})}{8.0\times10^6}} = 7.1\times10^3\,\text{m s}^{-1}$

(c) It would move at the same speed. The satellite's mass cancels in the derivation, so orbital speed depends only on $M$ and $r$ , not on the orbiting mass.
Why this scores
Why this scores 5/5. (1) accelerates towards centre → needs centripetal force; (2) gravity is that force → set GMm/r² = mv²/r; (3) cancel m and rearrange to v = √(GM/r); (4) v = 7.1×10³ m s⁻¹; (5) same speed, mass-independent. Part (c) directly targets the 'orbital speed depends on mass' misconception.
Question 6
4 marks
Q (4 marks). An astronaut inside an orbiting spacecraft appears to float, apparently weightless. A student concludes 'there is no gravity acting on the astronaut'. Explain, with reference to the forces involved, why the student is wrong and why the astronaut nevertheless feels weightless.
Model answer
The spacecraft orbits at a relatively small distance from the Earth, where the gravitational field strength is still large (for a low orbit it is close to its surface value). Gravity is definitely acting on the astronaut — in fact it is the gravitational force that provides the centripetal force keeping both astronaut and craft in orbit. So 'no gravity' is wrong.

The astronaut has a real weight $W = mg$ , which is far from zero.

The astronaut and the spacecraft are both in free fall: they accelerate towards the Earth at the same rate $g$ . Because they fall together, there is no contact (normal) force between the astronaut and the floor of the craft.

What we normally sense as 'weight' is that supporting contact force. With no contact force, the astronaut feels weightless — this is apparent weightlessness, not an absence of gravity.
Why this scores
Why this scores 4/4. (1) gravity is still present/strong (it provides the centripetal force); (2) true weight W = mg is not zero; (3) astronaut and craft in free fall together; (4) no contact/normal force → apparent weightlessness. The answer must contrast true weight with the missing contact force — simply saying 'they are falling' is not enough for full marks.
Question 7
6 marks
Q (6 marks). (a) By combining Newton's law of gravitation with circular motion, derive Kepler's third law T² = (4π²/GM)r³ for a satellite orbiting a mass M. (b) A geostationary satellite has an orbital period of exactly 24 hours (86 400 s). Taking Earth's mass as 5.97×10²⁴ kg, calculate the radius of its orbit. (G = 6.67×10⁻¹¹ N m² kg⁻²)
Model answer
(a) Gravity provides the centripetal force for the circular orbit: $\frac{GMm}{r^2} = \frac{mv^2}{r} \;\Rightarrow\; v^2 = \frac{GM}{r}$ The satellite covers one circumference in one period, so $v = 2\pi r/T$ and $v^2 = 4\pi^2 r^2/T^2$ . Equating the two expressions for $v^2$ : $\frac{4\pi^2 r^2}{T^2} = \frac{GM}{r} \;\Rightarrow\; T^2 = \frac{4\pi^2}{GM}\,r^3$

(b) Rearrange for $r$ : $r^3 = \frac{GMT^2}{4\pi^2} = \frac{(6.67\times10^{-11})(5.97\times10^{24})(86\,400)^2}{4\pi^2}$ $r^3 = \frac{2.97\times10^{24}}{39.5} = 7.53\times10^{22}\,\text{m}^3$ $r = \sqrt[3]{7.53\times10^{22}} = 4.2\times10^7\,\text{m}$

The geostationary orbit radius is about 4.2×10⁷ m (≈ 42 000 km from Earth's centre).
Why this scores
Why this scores 6/6. (a) 3 marks: gravity = centripetal → v² = GM/r (1); v = 2πr/T (1); rearrange to T² = 4π²r³/GM (1). (b) 3 marks: rearrange for r³ (1); substitute correctly, r³ = 7.5×10²² m³ (1); cube-root to r = 4.2×10⁷ m (1). Forgetting to take the cube root — or leaving the answer as r³ — is the common final-step error.
Question 8
10 marks
Q (10 marks — extended response). A satellite of mass 1200 kg is placed in a circular orbit 600 km above the Earth's surface. Earth's mass is 5.97×10²⁴ kg and its radius is 6.37×10⁶ m. (G = 6.67×10⁻¹¹ N m² kg⁻²) (a) Calculate the orbital radius, measured from the centre of the Earth. (b) Calculate the gravitational field strength at this altitude. (c) Show that the gravitational force on the satellite provides the centripetal force, and calculate the orbital speed. (d) Calculate the orbital period. (e) The astronauts on board describe themselves as 'weightless'. State the actual gravitational force (weight) on a 70 kg astronaut at this altitude, and explain in terms of forces why they nevertheless feel weightless.
Model answer
(a) Orbital radius — add the altitude to the Earth's radius: $r = R_{\text{Earth}} + h = 6.37\times10^6 + 0.60\times10^6 = 6.97\times10^6\,\text{m}$

(b) Field strength at this altitude: $g = \frac{GM}{r^2} = \frac{(6.67\times10^{-11})(5.97\times10^{24})}{(6.97\times10^6)^2} = 8.2\,\text{N kg}^{-1}$

(c) Orbital speed — the gravitational force is the only force on the satellite and points to the centre, so it provides the centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r} \;\Rightarrow\; v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{(6.67\times10^{-11})(5.97\times10^{24})}{6.97\times10^6}} = 7.6\times10^3\,\text{m s}^{-1}$ (The satellite's 1200 kg mass cancels, so it is not needed here.)

(d) Orbital period: $T = \frac{2\pi r}{v} = \frac{2\pi(6.97\times10^6)}{7.56\times10^3} = 5.8\times10^3\,\text{s} \;(\approx 97\ \text{minutes})$

(e) Weight and apparent weightlessness. The gravitational field strength there is 8.2 N kg⁻¹, so the true weight of a 70 kg astronaut is: $W = mg = 70 \times 8.2 = 5.7\times10^2\,\text{N}$ This is large — gravity is clearly acting, and it is exactly what holds the astronaut in orbit. The astronaut feels weightless because they and the spacecraft are both in free fall, accelerating towards the Earth at the same rate. There is therefore no contact (normal) force between the astronaut and the craft, and it is that supporting force we normally sense as weight. Hence the weightlessness is only apparent — the true weight of about 570 N remains.
Why this scores
Why this scores 10/10. (a) r = 6.97×10⁶ m, adding altitude to radius (2); (b) g = 8.2 N kg⁻¹ (2); (c) equate gravitational and centripetal forces, cancel m, v = 7.6×10³ m s⁻¹ (2); (d) T = 2πr/v = 5.8×10³ s (2); (e) true weight W = mg ≈ 570 N and correct free-fall / no-contact-force explanation (2). The grade-9 discriminators are measuring r from the centre in (a), recognising the satellite mass cancels in (c), and — in (e) — quoting a large non-zero weight while explaining the sensation through the missing contact force.
Question 9
6 marks
Q (6 marks). Planet X has twice the mass and twice the radius of Earth. Earth's surface gravitational field strength is 9.81 N kg⁻¹. (a) Using g = GM/r², determine the surface field strength on Planet X as a fraction of Earth's, showing your reasoning. (b) Calculate its value in N kg⁻¹. (c) Hence state the weight of a 70 kg astronaut standing on the surface of Planet X.
Model answer
(a) Surface field strength is $g = \dfrac{GM}{R^2}$ . Replacing $M \to 2M$ and $R \to 2R$ : $g_X = \frac{G(2M)}{(2R)^2} = \frac{2GM}{4R^2} = \frac{1}{2}\cdot\frac{GM}{R^2} = \frac{1}{2}g_{\text{Earth}}$ The mass doubles the field, but doubling the radius quarters it (inverse-square), so overall the field is half Earth's.

(b) Numerically: $g_X = \tfrac{1}{2}(9.81) = 4.9\,\text{N kg}^{-1}$

(c) Weight on Planet X: $W = mg_X = 70 \times 4.9 = 3.4\times10^2\,\text{N} \;(\approx 340\,\text{N})$
Why this scores
Why this scores 6/6. (a) 3 marks: substitute 2M and 2R into g = GM/R² (1); show the radius term gives a factor of 1/4 (1); conclude g_X = ½ g_Earth (1). (b) g_X = 4.9 N kg⁻¹ (1). (c) W = mg = 340 N (2). The discriminator is handling the inverse-square correctly — students who forget to square the radius factor wrongly get g_X = g_Earth.

Key Formulae — Gravitational fields

The formulae you need to memorise for gravitational fields on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Newton's law of universal gravitation
$F = \frac{Gm_1m_2}{r^2}$
$F$
gravitational force of attraction (N)
$G$
universal gravitational constant, 6.67×10⁻¹¹ N m² kg⁻²
$m_1, m_2$
the two masses (kg)
$r$
distance between the centres of the masses (m)
When to use
To find the attractive force between any two masses. Remember it is inverse-square and r is centre-to-centre.
Example
2.0 kg and 5.0 kg, 0.10 m apart: F = (6.67×10⁻¹¹)(2.0)(5.0)/(0.10)² = 6.7×10⁻⁸ N.
Gravitational field strength (definition)
$g = \frac{F}{m}$
$g$
gravitational field strength (N kg⁻¹)
$F$
gravitational force on the test mass (N)
$m$
the test mass (kg)
When to use
To relate the force on a mass to the field it sits in — force per unit mass. Also gives weight when rearranged as F = mg.
Example
A 2.0 kg mass feeling 16 N: g = 16/2.0 = 8.0 N kg⁻¹.
Field strength of a point / spherical mass
$g = \frac{GM}{r^2}$
$g$
gravitational field strength (N kg⁻¹)
$G$
universal gravitational constant
$M$
mass producing the field (kg)
$r$
distance from the centre of M (m)
When to use
To find the field strength at a distance r from a planet or star (the test mass cancels). Inverse-square in r.
Example
Earth's surface: g = (6.67×10⁻¹¹)(5.97×10²⁴)/(6.37×10⁶)² = 9.81 N kg⁻¹.
Weight
$W = mg$
$W$
weight — the gravitational force on an object (N)
$m$
mass of the object (kg)
$g$
local gravitational field strength (N kg⁻¹)
When to use
To find the gravitational force on an object in a field. Mass stays the same everywhere; weight changes with g.
Example
50 kg where g = 9.6 N kg⁻¹: W = 50 × 9.6 = 480 N.
Orbital speed (gravity = centripetal force)
$v = \sqrt{\frac{GM}{r}}$
$v$
orbital speed of the satellite (m s⁻¹)
$M$
mass of the central body (kg)
$r$
orbital radius from the centre of M (m)
When to use
For a circular orbit, from equating GMm/r² = mv²/r. Independent of the satellite's mass; larger r gives smaller v.
Example
r = 7.0×10⁶ m around Earth: v = √((6.67×10⁻¹¹)(5.97×10²⁴)/7.0×10⁶) = 7.5×10³ m s⁻¹.
Kepler's third law
$T^2 = \frac{4\pi^2}{GM}r^3$
$T$
orbital period (s)
$r$
orbital radius (m)
$M$
mass of the central body (kg)
$G$
universal gravitational constant
When to use
To link period and radius of a circular orbit, or to find the central mass. Since 4π²/GM is constant, T² ∝ r³.
Example
Geostationary orbit, T = 86 400 s: r = (GMT²/4π²)^(1/3) = 4.2×10⁷ m.

Key Definitions and Keywords — Gravitational fields

Definitions to memorise and the exact keywords mark schemes credit for gravitational fields answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Newton's law of universal gravitation
Examiner keyword
Every point mass attracts every other with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres: F = Gm₁m₂/r².
Universal gravitational constant (G)
Examiner keyword
The fixed constant of proportionality in Newton's law, G = 6.67×10⁻¹¹ N m² kg⁻², the same everywhere in the Universe. Not to be confused with g.
Inverse-square law
Examiner keyword
A relationship in which a quantity is proportional to 1/r². For gravitation, doubling the separation reduces the force (and the field strength) to a quarter.
Gravitational field
Examiner keyword
A region of space in which a mass experiences a gravitational force. Every mass sets up a field around itself.
Gravitational field strength (g)
Examiner keyword
The gravitational force per unit mass on a small test mass at a point: g = F/m. A vector, measured in N kg⁻¹, pointing towards the mass producing the field. For a point/spherical mass, g = GM/r².
Gravitational field line
Examiner keyword
A line showing the direction of the gravitational force on a mass. Lines point towards the mass creating the field; closer lines indicate a stronger field.
Radial field
Examiner keyword
The field around a point or spherical mass, in which the field lines point radially inward towards the centre and g varies as 1/r².
Uniform gravitational field
Examiner keyword
A field in which g has the same magnitude and direction everywhere, represented by parallel, equally spaced field lines — a good approximation over a small region near a planet's surface.
Weight
Examiner keyword
The gravitational force acting on an object, W = mg. A force in newtons; it changes with the local field strength g, unlike mass.
Mass
A measure of the amount of matter in an object (kg). It is the same everywhere and, unlike weight, does not depend on the gravitational field.
Centripetal force
Examiner keyword
The net inward force required to keep an object moving in a circle, directed towards the centre, of size mv²/r. For an orbit, gravity provides this force.
Orbital motion
Examiner keyword
The circular (to a good approximation) motion of a satellite or planet in which the gravitational force supplies the centripetal force, giving orbital speed v = √(GM/r).
Satellite
Any body in orbit around a larger mass under gravity — natural (e.g. the Moon) or artificial (e.g. the ISS).
Kepler's third law
Examiner keyword
For bodies orbiting the same central mass, the square of the orbital period is proportional to the cube of the orbital radius: T² = (4π²/GM)r³, so T² ∝ r³.
Apparent weightlessness
Examiner keyword
The sensation of having no weight, felt by an object in free fall (e.g. an orbiting astronaut) because there is no contact/supporting force — even though the true weight W = mg is not zero.
Free fall
Examiner keyword
Motion under gravity alone, with acceleration equal to the local field strength g. An orbit is continuous free fall: the satellite falls towards the planet while moving sideways fast enough to keep missing it.

Common Mistakes and Misconceptions — Gravitational fields

The traps other students keep falling into on gravitational fields questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Confusing gravitational field strength g with the universal constant G
IB Physics Theme D subject reports
Why it happens
They share a letter and both appear in gravitation equations, so students swap them (and their units).
How to avoid it
G is the fixed universal constant (6.67×10⁻¹¹ N m² kg⁻²); g = F/m = GM/r² is the field strength (N kg⁻¹) that varies with position. Learn the two units — they are different — and check which one the question needs.
✕Using 1/r instead of 1/r² (forgetting the inverse-SQUARE)
IB Physics Theme D subject reports
Why it happens
Students remember the force gets weaker with distance but forget the square, or forget to square r after substituting.
How to avoid it
Both F = Gm₁m₂/r² and g = GM/r² have r SQUARED in the denominator. Doubling r divides the result by 4, not 2. Always square the distance in your calculator step.
✕Measuring r from the surface instead of the centre of the planet
IB Physics Theme D subject reports
Why it happens
A question gives an altitude ('400 km above the surface') and students use that number directly as r.
How to avoid it
r is always the distance from the CENTRE. For a satellite, add the planet's radius to the altitude: r = R_planet + h. Only then substitute into F = Gm₁m₂/r² or g = GM/r².
✕Thinking the orbital speed depends on the satellite's mass
IB Physics Theme D subject reports
Why it happens
Heavier objects feel a bigger force, so students assume they must orbit differently.
How to avoid it
In GMm/r² = mv²/r the satellite mass m cancels, leaving v = √(GM/r). Orbital speed depends only on the central mass M and the radius r — a light and a heavy satellite at the same height move at the same speed.
✕Saying orbiting astronauts experience 'no gravity' or 'zero gravity'
IB Physics Theme D subject reports
Why it happens
Astronauts are seen floating, so students conclude gravity has switched off.
How to avoid it
The field strength in low orbit is close to its surface value (≈ 8.7 N kg⁻¹ for the ISS) — gravity is what keeps them in orbit. They feel weightless because they are in FREE FALL with no contact force, not because g is zero.
✕Power-of-ten and unit slips when substituting G and large masses
IB Physics Theme D subject reports
Why it happens
G is 10⁻¹¹ and planetary masses and radii are huge powers of ten, so exponents get mishandled and units dropped.
How to avoid it
Enter each value in standard form carefully, keep track of the powers of ten (especially after squaring r), and always attach the correct unit — N for force, N kg⁻¹ for field strength.

Gravitational fields

Detailed Study Notes

At a glance

What you’ll learn

Newton's law of universal gravitation

Gravitational fields and field strength g = F/m

Field lines, radial and uniform fields, and weight

Orbital motion: gravity provides the centripetal force

Deriving Kepler's third law (T² ∝ r³)

Apparent weightlessness: astronauts are in free fall

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Force between two masses (Newton's law)

2Sketching and describing gravitational field lines

3Distinguishing g from G

4Surface gravitational field strength of a planet

5How g changes with distance (inverse-square)

6Is there gravity on the ISS?

7Orbital speed and period of a satellite

8Deriving Kepler's third law

Newton's law of universal gravitation

Gravitational field strength (definition)

Field strength of a point / spherical mass

Weight

Orbital speed (gravity = centripetal force)

Kepler's third law

Newton's law of universal gravitation

Universal gravitational constant (G)

Inverse-square law

Gravitational field

Gravitational field strength (g)

Gravitational field line

Radial field

Uniform gravitational field

Weight

Mass

Centripetal force

Orbital motion

Satellite

Kepler's third law

Apparent weightlessness

Free fall

✕Confusing gravitational field strength g with the universal constant G

✕Using 1/r instead of 1/r² (forgetting the inverse-SQUARE)

✕Measuring r from the surface instead of the centre of the planet

✕Thinking the orbital speed depends on the satellite's mass

✕Saying orbiting astronauts experience 'no gravity' or 'zero gravity'

✕Power-of-ten and unit slips when substituting G and large masses