In simple terms
A friendly intro before the formal notes — no formulas yet.
Gravitational field of a point mass
Cambridge 9702 Paper 4 — Gravitational field of a point mass (13.3). Senpai Corner diagram-backed pilot with premium structure and live visuals.
- 1
Force () is directly proportional to the product of masses ().
- 2
Force () is inversely proportional to the square of the distance ().
- 3
is the universal gravitational constant ().
- 4
The force is always attractive.
What this topic covers
The official Cambridge syllabus points this lesson works through.
- 13.3.1
Derive, from Newton's law of gravitation and the definition of gravitational field, the equation for the gravitational field strength due to a point mass
- 13.3.2
Recall and use
- 13.3.3
Understand why g is approximately constant for small changes in height near the Earth's surface
Explore the concept
Use the live diagram, PhET or GeoGebra sim, and synced steps — play it, drag controls, or tap a step.
Step-synced diagram — highlights what to look for in the simulation above.
A point mass M creates a gravitational…
A point mass M creates a gravitational field in the space around it.
12 more simulations for this topic — run them in the Simulations section below
Simulations
Every simulation here runs the real model — try the steps on a card, then check what you see against the notes.
12 simulations · 4 to start with
Start herein this order — each one shows a different piece of the topic
- The Physics ClassroomStart here · 19702 13.3 · IB D.1
The Value of g
Explore how g varies with location around the globe and with altitude
Why this one: Climb to higher altitude and watch g fall away from 9.81 N/kg as 1/r² from Earth's centre.
Try this
- Compare g at the equator and the poles.
- Raise the altitude and watch g.
Look for g falls with altitude and is slightly smaller at the equator than at the poles.
Physics Interactives by The Physics Classroom · Licensed to MarkScheme (site terms otherwise permit linking only)
- 3JCN PhysicsStart here · 29702 13.3 · 9702 13.2 · IB D.1
Kepler's Third Law
Change orbital radius and read the period; verify T^2 proportional to r^3
Why this one: Change the orbital radius, read the period and check T² ∝ r³ follows from g = GM/r².
Try this
- Set an orbital radius and read the period.
- Quadruple the radius and read the period again.
- Check T² / r³ for both orbits.
Look for T² is proportional to r³, so quadrupling the radius multiplies the period by eight.
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
- 3JCN PhysicsStart here · 39702 13.3 · 9702 13.2 · 9702 2.1
Newton's Cannon
Increase a cannonball's launch speed from a mountain until it orbits Earth
Why this one: Raise the launch speed until the ball falls around the Earth instead of onto it — that is an orbit.
Try this
- Launch at a low speed and watch the cannonball fall to Earth.
- Increase the launch speed step by step until the ball completes an orbit.
- Increase the speed further and compare the shape of the orbit.
Look for At the right speed the ball falls toward Earth at the same rate the surface curves away, giving a circular orbit; faster gives an ellipse.
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
- The Physics ClassroomStart here · 49702 13.3 · IB D.1
The Value of g on Other Planets
Compare the value of g on other planets from each planet's mass and radius
Why this one: Compare g on each planet from its mass and radius using g = GM/R².
Try this
- Pick a planet and read g.
- Compare its mass and radius with Earth's.
Look for g equals GM divided by R².
Physics Interactives by The Physics Classroom · Licensed to MarkScheme (site terms otherwise permit linking only)
More simulations8 more on this topic — core ones first
- oPhysicsCore9702 13.2 · 9702 13.3 · IB D.1
Elliptical Orbits & Kepler's 2nd Law
Planet orbiting a sun: set initial speed, distance and masses; see elliptical orbit and equal-area sweeps
Try this
- Set an initial speed that gives a nearly circular orbit.
- Lower the initial speed and watch the orbit become elliptical.
- Increase the sun’s mass with the same starting distance.
Look for The planet moves fastest nearest the sun, and the areas swept in equal times stay equal.
Simulation by Tom Walsh, oPhysics.com — made with GeoGebra · Licensed to MarkScheme (site: free for non-profit educational use; applets made with GeoGebra)
- 3JCN PhysicsCore9702 13.3 · 9702 13.1 · 9702 2.1
Newton's Gravity: Jumping on Planets
Jump on different planets; compare jump height and hang time against surface gravity
Try this
- Jump on Earth and note the jump height and hang time.
- Jump on the Moon and compare both against its surface gravity.
- Jump on Jupiter and compare again.
Look for For the same take-off speed, jump height and hang time both scale inversely with surface gravity g.
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
- 3JCN PhysicsCore9702 13.3 · 9702 13.2 · IB D.1
Sun and Earth
Vary Earth's orbital speed around the Sun and see circular, elliptical or escape trajectories
Try this
- Set the orbital speed for a circular orbit and watch one revolution.
- Reduce the speed and compare the shape of the orbit.
- Increase the speed until the Earth escapes.
Look for One speed gives a circle; lower or higher speeds give ellipses and beyond escape speed the path is open.
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
- PhET9702 13.3 · IB D.1
Kepler’s Laws
Drag a planet’s orbit and watch equal areas sweep in equal times; compare T² with a³.
Try this
- On “Second Law”, press play — the two swept areas are equal although the arcs differ.
- On “Third Law”, change the semi-major axis — read T and check T²/a³ stays fixed.
Look for Planets move faster near the star; T² ∝ a³ for every orbit round the same mass.
Simulation by PhET Interactive Simulations, University of Colorado Boulder · Licensed to MarkScheme (public licence CC BY-NC 4.0 since 2026-03-30)
- The Physics Classroom9702 13.3 · IB D.1
Your Weight on Other Planets
Hop on the scales on other planets and see how your weight changes with location
Try this
- Enter your mass.
- Read your weight on each planet.
Look for Weight equals mass multiplied by the local value of g.
Physics Interactives by The Physics Classroom · Licensed to MarkScheme (site terms otherwise permit linking only)
- 3JCN Physics9702 13.3 · IB D.1
Gravitational Assist - Sling Shot
Fly a probe past a moving planet to gain speed via gravitational assist
Try this
- Fly the probe past the moving planet and compare its speed before and after.
- Approach from the other side of the planet and compare.
Look for Passing behind a moving planet adds some of the planet's orbital speed to the probe.
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
Key formulas
Tap any symbol to reveal exactly what it means and its units.
Tap a symbol — great for exam definitions
Tap a symbol — great for exam definitions
Tap a symbol — great for exam definitions
Tap a symbol — great for exam definitions
Tap a symbol — great for exam definitions
Full topic notes
Formal explanation with the rigour you need for the exam.
Gravitational Fields: An Invisible Influence
A gravitational field is essentially a region of space surrounding a mass where another mass would experience an attractive force. For most external calculations, especially with large, uniform objects like planets or stars, we simplify things by treating them as if all their mass is concentrated at a single point – known as a point mass at their centre. This makes calculations much more manageable.
Newton's Universal Law of Gravitation
Sir Isaac Newton's groundbreaking law describes the fundamental attraction between any two masses. It states that the attractive force between them is directly proportional to the product of their masses and inversely proportional to the square of the distance separating their centres. This inverse square law is crucial for understanding how gravity weakens with distance.
Force () is directly proportional to the product of masses ().
Force () is inversely proportional to the square of the distance ().
is the universal gravitational constant ().
The force is always attractive.
Gravitational Field Strength ($g$)
Gravitational field strength () quantifies how strong the gravitational pull is at a particular point. It's defined as the gravitational force per unit mass acting on a small test mass placed at that point. Importantly, also represents the acceleration experienced by a free-falling object in that field. Field lines are radial, point inwards, and their density indicates the field's strength.
We can derive the formula for gravitational field strength by combining Newton's Law of Gravitation with the definition of a field.
Derivation:
- Start with the definition of gravitational field strength: the force per unit test mass, .
- Recall Newton's Law of Gravitation for the force between the source mass and the test mass .
- Substitute the expression for into the equation for .
- The test mass cancels out, leaving the formula for the field strength created by the source mass . This shows that the field strength at a point depends only on the source mass and the distance from it, not on the test mass placed there.
Gravitational field strength is force per unit mass ().
It also represents the acceleration of free fall.
Follows an inverse square law ().
Field lines are radial, pointing inwards, and closer lines mean a stronger field.
Gravitational Potential ($\phi$)
Gravitational potential () is a scalar quantity representing the work done per unit mass to move a test mass from infinity to a specific point within the field. By convention, the gravitational potential at infinity is defined as zero. Since gravity is always attractive and work is done by the field as a mass approaches the source, gravitational potential is always a negative value.
Gravitational potential is work done per unit mass from infinity.
Potential at infinity is conventionally zero.
It is always negative, as energy is released moving towards the mass.
The Link Between Field Strength and Potential
There is a direct mathematical relationship between gravitational field strength () and gravitational potential (). The field strength at a point is equal to the negative of the potential gradient at that point. The potential gradient is the rate of change of potential with distance. This means that a region with a rapidly changing potential (a steep slope on a potential-distance graph) will have a strong gravitational field.
The negative sign is important: it signifies that the gravitational field vector () always points in the direction of decreasing potential. Think of it like a ball rolling downhill – it moves from a higher potential to a lower potential, and the direction of its acceleration is down the steepest slope. For a point mass, potential becomes less negative (increases) as you move away from the mass, and the field points inwards, towards the mass, in the direction of decreasing potential.
This relationship is visualized in graphs. A graph of potential () against distance () for a point mass is a curve in the negative quadrant, approaching zero as approaches infinity. The gradient of this graph at any point is equal to . A graph of field strength () against distance () shows an inverse square relationship, always positive, and also approaching zero as increases.
Gravitational Potential Energy ($E_p$)
Gravitational potential energy () is the energy an object possesses due to its position within a gravitational field. For a system of two masses, it represents the work done to assemble them from infinite separation to their current distance. Like gravitational potential, is also a negative value. To completely remove an object from a field, you'd need to supply energy equal to the negative of its current potential energy.
Gravitational potential energy is the energy of a mass in a gravitational field.
It is always negative for attractive fields, defined relative to zero at infinity.
Energy required to escape the field is the absolute value of .
Worked examples
See the formulas applied — reveal one step at a time, like the exam.
Calculate the gravitational field strength on the surface of Mars, given its mass is and its radius is . Use the universal gravitational constant .
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Identify known values:
A satellite of mass 1200 kg is in a stable circular orbit at an altitude of 500 km above the Earth's surface. Calculate the work done required to move it to a higher stable orbit at an altitude of 2000 km. (Mass of Earth, M = kg; Radius of Earth, m; N m kg)
- 1
Identify the principle: The work done is the change in the satellite's gravitational potential energy (GPE).
How it all connects
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Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.
Glossary
Key terms for this topic — skim now; the Check step will test them.
- point mass
For most external calculations, especially with large, uniform objects like planets or stars, we simplify things by treating them as if all their mass is concentrated at a single point – known as a point mass at their centre. This makes calculations much more manageable.
- inverse square law
This inverse square law is crucial for understanding how gravity weakens with distance.
- Derivation
Derivation:
- Gravitational field
A region surrounding a mass where other objects with mass experience an attractive force.
- Gravitational field strength ()
The gravitational force per unit mass experienced by a small test mass, also equal to the acceleration of free fall.
- Gravitational potential ()
The work done per unit mass to move a test mass from infinity to a specific point within the field.
- gravitational potential
φ is inversely proportional to the distance: φ ∝ -1/r.
Quick check
Write your answer first, then compare it with the model one — the gap is what you would have lost.
Teach it back
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Teach it back
Explain this topic as if teaching a friend. We name the gaps an examiner would still dock.
Revision flashcards
Guess first, then flip — retrieval beats re-reading.
Key takeaways
Review these before you close the topic — retrieval beats re-reading.
Force () is directly proportional to the product of masses ().
Force () is inversely proportional to the square of the distance ().
is the universal gravitational constant ().
The force is always attractive.
Practice — then mark it
The whole point: a real Cambridge question, marked mark-by-mark.
For the position of the probe where x = 1.47 × 10¹¹m: (i) calculate g
Show that, for any particular value of x, the numerical values of g and F are related by g = (4πGM / L) F where M is the mass of the Sun, L is the luminosity of the Sun and G is the gravitational constant.
Extra simulations & links
PhET, GeoGebra and other curated tools — open in a new tab.
Frequently asked
Checkpoint
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