Gravitational Fields
Revise Gravitational Fields for Physics (IAL) — revision notes and instant AI marking. Free to start.
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.
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.
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 (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.
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.
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.
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:
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).
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.
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.
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.
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 Fields | Electric Fields | |
|---|---|---|
| Origin of force | Mass | Charge |
| Force between two points | F = Gm₁m₂/r² | F = Q₁Q₂/4πε₀r² |
| Type of force | Always attractive | Attractive (opposite charges) or repulsive (like charges) |
| Field strength | g = F/M | E = F/Q |
| Field strength (point source) | g = GM/r² | E = Q/4πε₀r² |
| Potential | V = −GM/r (always negative) | V = Q/4πε₀r (positive or negative) |
| Work done | ΔW = MΔV | ΔW = QΔV |
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.
Deriving orbital speed
Simply set the gravitational force equal to the centripetal force needed for circular motion, then solve:
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:
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?
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
Concepts Checklist
Exam Tips & Common Mistakes
- 6. Comparing Electric & Gravitational Fields
- Exam Tips & Common Mistakes
Read the full Gravitational Fields notes free
That's the preview — create a free account to read the rest, plus flashcards and practice questions with instant AI marking. No credit card.
Unlock the full notes free →