Gravitational fields: field strength, potential and orbits
Gravitational fields is the topic where answers come back with the wrong sign, the wrong power of r, or an orbit worked out from the Earth’s surface instead of its centre. None of those is a hard mistake to fix. They’re the same three mistakes, made again and again by students who understand the ideas perfectly well, and most of this guide is about stopping them.
In this guide
Field strength is force per kilogram
Gravitational field strength, g, is the force on each kilogram of mass placed in the field: g = F/m. Its unit is newtons per kilogram, N kg−1. Near the Earth’s surface it’s about 9.81 N kg−1, and the specification expects your child to know that this is numerically equal to the acceleration of free fall, the g they’ve used since GCSE.
Close to the ground the field is uniform: the same strength and direction everywhere you’re likely to stand. Further out, it isn’t. For a planet, the course models the whole mass as if it sat at a single point in the centre, and then g = −GM/r², where M is the planet’s mass, G is the gravitational constant and r is the distance from the centre.
That “from the centre” is the first trap. A question about a satellite 400 km up wants r to be the Earth’s radius plus 400 km, roughly 6,770 km in all. A student who uses the height alone gets an answer hundreds of times too large, and often doesn’t stop to ask whether it makes sense.
The 1/r² makes this an inverse square law. Double the distance from the centre and the field strength falls to a quarter. Triple it and it falls to a ninth.
Why potential is negative
Gravitational potential is the awkward one, and the minus sign is the reason.
The definition is the work done in bringing a unit mass, one kilogram, from infinity to that point. Infinity is the zero, because out there the planet’s pull has faded to nothing. Now think about bringing that kilogram in. Gravity pulls it towards the planet the whole way, so you never have to push it; if anything, you’d have to hold it back. The work done is negative, so the potential is negative everywhere, and it gets more negative the closer you get.
The equation is V = −GM/r. Look at the r. It’s 1/r, not 1/r².
Field strength falls with the square of the distance; potential falls with the distance itself. Mixing up those two powers is the most common algebra slip in the whole topic, partly because both equations sit close together in the booklet and look alike at a glance.
Multiply potential by a mass and you get gravitational potential energy, E = −GMm/r, again negative and again 1/r. There’s a graph to understand as well: the area under a graph of force against distance is the work done moving between two points, which is how the course joins the force and energy versions of the same idea.
Orbits are circular motion with gravity doing the work
This is where the topic pays off, and where my circular motion guide becomes essential reading.
A satellite in a circular orbit needs a resultant force of mv²/r towards the centre. Gravity provides it, so GMm/r² = mv²/r. Two things happen in that line. The satellite’s mass m cancels, which is why a small satellite and the International Space Station at the same height go round at the same speed. And after some rearranging, using v = 2πr/T for one orbit, it becomes T² = (4π²/GM)r³.
That’s Kepler’s third law: the square of the period is proportional to the cube of the orbit radius. The booklet prints the final equation, but questions regularly ask students to derive it, and the derivation is worth practising until it takes under a minute. It works for any system, planets round the Sun, moons round Jupiter, satellites round the Earth, with M being whatever sits in the middle. Kepler’s first two laws, about elliptical orbits and equal areas swept in equal times, are on the specification too, and they come up as short written questions.
The specification also names geostationary orbits. A geostationary satellite sits above the equator, orbits in the same direction as the Earth spins and takes one day to go round, so from the ground it seems to hang still in the sky. That’s why a satellite dish on a house can point at one fixed spot and never move. Your child should be able to state those conditions and use T² ∝ r³ to work out the orbit’s height.
Escape velocity
Escape velocity is the launch speed an object would need at a planet’s surface to coast away and never come back, with no engine running after the launch. It’s an energy balance. The kinetic energy at launch has to be enough to climb all the way out to infinity, where the potential energy reaches zero. Setting ½mv² = GMm/r gives v = √(2GM/r).
The escape velocity equation isn’t in the booklet, so this is one to learn as a derivation rather than look up. And the mass cancels again: a pebble and a spacecraft need the same escape velocity.
A one-page table worth making
The quickest win in this topic is a table, and you can check it without knowing any physics.
Ask your child to draw a grid with four rows: field strength, force, potential and potential energy. For each row, three columns: the equation, the unit, and whether it goes with 1/r or 1/r². Then compare it with OCR’s Data, Formulae and Relationships booklet, which is a free download from OCR’s website. If the 1/r² column holds field strength and force, and the 1/r column holds potential and potential energy, the most common mistake in the topic is already dealt with.
Then one more question for the table: which of these has a direction? Field strength and force do. Potential and energy don’t. It sounds like a small distinction, and it’s exactly what decides how to combine the pull of two planets in the harder questions at the end of a paper.
Common questions
Why is gravitational potential always negative?
Potential is defined as the work done bringing one kilogram from infinity, where it is zero, to the point. Gravity attracts the mass the whole way in, so the work done on it is negative and the potential is below zero everywhere. It becomes more negative closer to the planet: V = −GM/r.
What is the difference between gravitational field strength and gravitational potential?
Field strength is the force per kilogram, measured in N kg⁻¹, and it has a direction. Potential is the work done per kilogram to bring a mass in from infinity, measured in J kg⁻¹, and it has no direction. For a point mass, field strength falls with 1/r² and potential with 1/r.
How do you derive Kepler’s third law in A-Level Physics?
Set the gravitational force equal to the centripetal force, GMm/r² = mv²/r, then substitute v = 2πr/T for one orbit. Rearranging gives T² = (4π²/GM)r³, so the square of the period is proportional to the cube of the orbit radius.