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Physics (IAL)

Gravitational Fields

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Edexcel IAL Physics · Fields & Forces

Gravitational Fields

Any two masses in the universe pull on each other — and the closer or heavier they are, the stronger that pull, following one elegant inverse-square rule that governs everything from a dropped apple to a satellite in orbit.

Quick Overview

  • A gravitational field is a region where any mass feels a force — field lines point towards the mass causing the field because gravity is always attractive.
  • Gravitational field strength g = F/m tells you the force per kg at a point.
  • Newton's Law of Gravitation: F = GMm/r² — the force between two masses, obeying an inverse-square law.
  • Near a point mass (or spherical planet), g = GM/r² — this also follows inverse-square (g ∝ 1/r²).
  • Gravitational potential V = −GM/r is always negative and follows a 1/r relationship, not 1/r².
  • Gravity and electric fields are mathematically similar (both inverse-square), but gravity is mass-based and always attractive, while electric fields are charge-based and can attract or repel.
  • For orbits, gravitational force provides centripetal force, giving v² = GM/r and T² = 4π²r³/GM (T² ∝ r³).

1. What Is a Gravitational Field?

Think of a gravitational field as an invisible "zone of influence" that surrounds anything with mass. Step into that zone with your own mass, and you'll feel a pull — even if nothing is touching you. The Earth's field is why you stay glued to the ground; the Sun's field is why the Earth doesn't go flying off into deep space.

The direction of a gravitational field at any point is shown using field lines (arrows). Because gravity only ever attracts — it never pushes things apart — every field line points towards the centre of the mass creating the field. If you dropped a test mass at any point in the field, the field line tells you which way it would accelerate.

Radial fields

Around a single point mass (or any spherical object, like a planet), the field is radial — the lines spread out symmetrically in all directions, like the spokes of a wheel, all converging on the centre.

↖ ↑ ↗ \ | / ←----( M )----→ All arrows point INWARD / | \ toward the mass M ↙ ↓ ↘
Why this makes sense
Imagine standing anywhere around a planet — no matter which side you're on, "down" always means "toward the centre." That's exactly what a radial field is describing.
Practice Question 1

Why do gravitational field lines always point towards a mass, and never away from it?

2. Gravitational Field Strength (g)

Gravitational field strength answers a very specific question: "If I put a 1 kg mass at this exact point, how much force would it feel?" That's literally what the formula says — force per unit mass.

Key Formula
g = F / m
g = gravitational field strength (N kg⁻¹) · F = force/weight (N) · m = mass (kg)

A crucial thing to internalise: an object's mass never changes, no matter where it is in the universe. What changes from planet to planet is g — and since weight = mg, your weight changes even though your mass doesn't. Stand on Earth (g = 9.81 N kg⁻¹) and you weigh a certain amount; stand on Jupiter (g ≈ 25 N kg⁻¹) and that same mass would feel roughly 2.5× heavier — potentially not even able to stand upright!

g vs G — Don't Mix Them Up
g (little g) = gravitational field strength — it changes depending on the mass creating the field and your distance from it.
G (big G) = Newton's Gravitational Constant — this is a fixed, universal number: 6.67 × 10⁻¹¹ N m² kg⁻².

What affects g at a planet's surface?

Two things only: the planet's mass (or density) and its radius. A bigger or denser planet has stronger gravity; a planet with a larger radius (pushing you further from its centre) has weaker surface gravity.

Practice Question 2

Calculate the mass of an object that has a weight of 10 N on Earth's surface (g = 9.81 N kg⁻¹).

3. Newton's Law of Universal Gravitation

This is the master equation of the whole chapter. It tells you the actual force between any two masses anywhere in the universe — from two atoms to two galaxies.

In words: The gravitational force between two masses is proportional to the product of their masses, and inversely proportional to the square of the distance between their centres.

Key Formula
FG = Gm₁m₂ / r²
FG = gravitational force (N) · G = 6.67×10⁻¹¹ N m² kg⁻² · m₁, m₂ = the two masses (kg) · r = distance between their centres (m)

The inverse-square law, unpacked

The "1/r²" part of the formula is called an inverse square law, and it has a dramatic effect: if you double the distance between two masses, the force doesn't halve — it drops to a quarter. Triple the distance, and the force falls to a ninth. Distance is brutally punishing to gravitational force.

r ----> Force = F 2r ----> Force = F/4 (double distance = quarter the force) 3r ----> Force = F/9 (triple distance = one-ninth the force)
Common Mistake #1
Forgetting that r is measured between the centres of the two masses — not the distance between their surfaces! If a satellite orbits 2000 km above Earth's surface, and Earth's radius is 6400 km, then r = 2000 + 6400 = 8400 km, NOT just 2000 km.
Common Mistake #2
Forgetting to square the distance r. It's incredibly easy under exam pressure to just divide by r instead of r² — always double check this step.
Practice Question 3

A satellite of mass 6500 kg orbits Earth at 2000 km above the surface. The gravitational force between them is 37 kN. Given Earth's radius = 6400 km, calculate the mass of the Earth.

4. Gravitational Field Due to a Point Mass

Here's a neat trick: you can actually derive the g = GM/r² formula just by combining the two equations you already know. Watch how it flows:

Start: g = F/m Substitute: F with F_G = GMm/r² Result: g = (GMm/r²) / m Cancel m: g = GM/r²
Key Formula
g = GM / r²
M = mass of the body creating the field · r = distance from the centre of M to the point you're measuring · notice m (the "test mass") cancels out completely!

This cancelling of m is actually profound — it tells you that g at a given point doesn't depend on what you're measuring it with. A feather and a bowling ball dropped from the same height experience the exact same g (which is why, ignoring air resistance, they fall at the same rate — Galileo's famous result).

Keep M and r straight
M is always the mass causing the field (e.g. the planet). m (lowercase, in earlier formulas) is the "test mass" experiencing that field. Mixing these up is one of the most common slip-ups in exams.
Practice Question 4

Earth's g at its surface is 9.81 N kg⁻¹. Using Earth's mass = 6.0×10²⁴ kg and radius = 6400 km, calculate g at a distance of 1 million km from Earth's surface.

5. Gravitational Potential in a Radial Field

You've probably used G.P.E = mgΔh before — but that formula has a hidden catch: it only works near Earth's surface, where the field is approximately uniform (parallel field lines, constant g). Once you zoom out far enough that Earth's curvature matters, the field becomes radial, and you need a new tool: gravitational potential.

Gravitational potential (V) is defined as the work done per unit mass to bring a test mass from infinity (where the field's influence is zero) to a specific point in the field.

Key Formula
Vgrav = − GM / r
Vgrav = gravitational potential (J kg⁻¹) · always negative · notice this is 1/r, NOT 1/r²

Why is potential always negative?

Here's the logic: gravity is always attractive, so it always pulls — it never needs help pulling a mass in. But to move a mass away from another mass (against that pull), you have to do work against gravity. Since "infinity" (zero potential, by definition) is the point of maximum separation, and everywhere else requires less energy than getting to infinity, every real point in a gravitational field sits at a potential below zero — hence negative.

The single most important distinction in this chapter
g ∝ 1/r² (inverse-square law — for field strength)
V ∝ 1/r (simple inverse law — for potential, and it's negative)
These two formulas look almost identical on the page but behave completely differently as r changes. Exam writers love testing this exact confusion.
Practice Question 5

Calculate the gravitational potential at the surface of Mars. Radius of Mars = 3400 km, Mass of Mars = 6.4×10²³ kg.

6. Comparing Electric & Gravitational Fields

Gravity and electric fields are mathematical cousins — both follow inverse-square laws, both have radial fields around point sources, and both have near-identical-looking formulas. But they come from fundamentally different sources and behave differently in one key way: direction.

Gravitational FieldsElectric Fields
Origin of forceMassCharge
Force between two pointsF = Gm₁m₂/r²F = Q₁Q₂/4πε₀r²
Type of forceAlways attractiveAttractive (opposite charges) or repulsive (like charges)
Field strengthg = F/ME = F/Q
Field strength (point source)g = GM/r²E = Q/4πε₀r²
PotentialV = −GM/r (always negative)V = Q/4πε₀r (positive or negative)
Work doneΔW = MΔVΔW = QΔV
What's identical
Both fields obey inverse-square laws for force and field strength (1/r²), both have a 1/r relationship for potential, both produce radial fields around point sources, and both have parallel, evenly-spaced field lines in a uniform field (like near a planet's surface, or between charged parallel plates).
Practice Question 6

State one key difference between gravitational and electric fields regarding the direction of the force they produce.

7. Orbital Motion

This is where the whole chapter comes together. A satellite (or planet) in a stable circular orbit isn't "floating" — it's actually constantly falling toward the object it orbits, but moving sideways fast enough that it keeps missing! The gravitational force is exactly what's needed to act as the centripetal force that keeps it curving around in a circle instead of flying off in a straight line.

v (velocity, tangent to orbit) ↗ •---→ / m | \ | ↓ F_g (points toward centre — this | M IS the centripetal force) \ / \ /

Deriving orbital speed

Simply set the gravitational force equal to the centripetal force needed for circular motion, then solve:

Setting F_gravity = F_centripetal
GMm/r² = mv²/r
The satellite's own mass m cancels from both sides — its mass makes no difference to its orbital speed!
Key Formula — Orbital Speed
v² = GM / r
M = mass of the object being orbited · r = orbital radius · notice: this means ALL satellites at the same radius travel at the SAME speed, no matter their own mass.

Deriving the time period relationship

Since orbital speed can also be written as distance ÷ time (circumference ÷ period), i.e. v = 2πr/T, you can combine this with v² = GM/r to link time period directly to orbital radius:

Key Formula — Kepler's Third Law
T² = 4π²r³ / GM
Often summarised simply as T² ∝ r³ — this is one of the most quoted relationships in astrophysics.
Why is a satellite in orbit "accelerating" even at constant speed?
Because velocity is a vector — it has both size AND direction. Even if the speed (magnitude) stays constant in a circular orbit, the direction is constantly changing as it curves around. A changing direction means a changing velocity, and a changing velocity means acceleration — this is called centripetal acceleration, and it always points toward the centre.
Practice Question 7

Two satellites orbit the same planet. Satellite A orbits at radius r, and Satellite B orbits at radius 4r. How does Satellite B's time period compare to Satellite A's?

Practice Question 8 (Harder — Binary Star System)

A binary star system has two stars orbiting a fixed point B. Star M₁ orbits at radius R₁, star M₂ orbits at radius R₂, both with angular speed ω. Show that ω² = GM₂ / [R₁(R₁+R₂)²].

What to Memorise

Gravitational Field Strength
g = F/m  →  g = GM/r²  (N kg⁻¹)
Newton's Law of Gravitation
FG = Gm₁m₂/r²  (N)
Newton's Gravitational Constant
G = 6.67 × 10⁻¹¹ N m² kg⁻² (fixed, universal)
Gravitational Potential
Vgrav = −GM/r  (J kg⁻¹) — always negative
Orbital Speed
v² = GM/r
Time Period of Orbit
T² = 4π²r³/GM  →  T² ∝ r³
Inverse Square Law
Doubling r → force/field strength drops to ¼. Applies to F and g, NOT V.
Radial Field
Field around a point/spherical mass; field lines converge on the centre.
Uniform Field
Approximation near a planet's surface; field lines are parallel & equally spaced.
r in every formula
Always the distance between the CENTRES of the two masses.

Concepts Checklist

Exam Tips & Common Mistakes

Always add radius + altitude. If a satellite orbits "500 km above the surface," you must add the planet's radius to get the true value of r (measured from centre to centre) before plugging into any formula.
Don't forget to square r. In F = Gm₁m₂/r² and g = GM/r², the r² is easy to miss under time pressure — it's one of the most common lost marks in this topic.
Potential is negative — keep the sign. When calculating Vgrav = −GM/r, don't drop the negative sign, and remember increasing r makes V less negative (i.e. it increases towards zero), not "more negative."
g ∝ 1/r² vs V ∝ 1/r. These two relationships look almost identical but behave very differently as distance increases — examiners frequently test whether you can tell them apart in 6-mark comparison questions.
For two-body orbit problems (e.g. binary stars), remember the separation r used in the gravity formula is the SUM of both orbital radii (R₁ + R₂), since that's the true distance between the two masses' centres.
Comparison questions (gravity vs electric fields) often appear as long-answer 6-markers. Practice writing a clean, structured comparison covering: origin of force, direction (attract/repel), inverse-square relationships, and sign of potential.
Density substitution. If a question gives density instead of mass, remember ρ = m/V, so you may need to calculate mass first (using the volume of a sphere, V = 4/3πr³) before applying gravitational formulas.
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  • 6. Comparing Electric & Gravitational Fields
  • Exam Tips & Common Mistakes
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