Revise Gravitational fields for IB Physics (D.1)

Newton proposed that every particle of matter attracts every other particle. The force is directly proportional to the product of the two masses and inversely proportional to the square of the distance between their centres. For point masses, or for spherical masses treated as points at their centres, this gives a single equation.

$F = G \frac{Mm}{r^2}$

Two features do most of the work in exam questions. First, the force is an inverse-square law: double the separation and the force falls to a quarter; triple it and it falls to a ninth. Second, because a uniform sphere produces the same external field as a point mass at its centre, we measure $r$ from the centre — so for an object at height $h$ above a planet of radius $R$, we must use $r = R + h$.

Rather than think of a force reaching mysteriously across empty space, we say a mass $M$ fills the space around it with a gravitational field. Any second mass placed in that field feels a force. We define the gravitational field strength $g$ at a point as the force per unit mass on a small test mass placed there.

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

Its units are N kg⁻¹. To find the field due to a source mass, substitute Newton's law $F = G\frac{Mm}{r^2}$ into $g = F/m$; the test mass $m$ cancels, leaving a formula for the field created by $M$ alone.

$g = G \frac{M}{r^2}$

The symbol $g$ carries two ideas that happen to share a value. As a field strength, $g = F/m$ is a property of a point in space — it is defined there whether or not any mass sits at the point, and it is measured in N kg⁻¹. As an acceleration, $g$ is what a body actually experiences when gravity is the only force acting on it: its weight is $W = mg$, so by Newton's second law its acceleration is $a = W/m = g$, measured in m s⁻². The two are numerically identical because N kg⁻¹ and m s⁻² are the same combination of units. The distinction matters when other forces are present: a parachutist held back by drag, or a book resting on a table, still sits in a field of about $9.81$ N kg⁻¹, but neither accelerates at $9.81$ m s⁻² because gravity is not the only force. Field strength describes the field; acceleration of free fall describes the motion that field produces when nothing else interferes.

We picture a field with field lines that show the direction of the force on a test mass. For a single point or uniform spherical mass the lines are radial, pointing straight inwards towards the centre, because that is the direction of the pull everywhere around it. The spacing of the lines encodes the strength: near the surface the lines are crowded together where the field is strong, and they fan out with distance as the field weakens according to the inverse-square law.

A satellite in a circular orbit moves at constant speed but constantly changes direction, so it is accelerating towards the centre of the circle. Something must supply that centripetal force — and for a satellite the only force acting is gravity. The key step in every orbit problem is therefore to set the gravitational force equal to the required centripetal force.

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

The orbiting mass $m$ cancels from both sides. Rearranging gives the orbital speed, and the period then follows because the satellite covers one circumference $2\pi r$ in one period at speed $v$.

$v = \sqrt{\frac{GM}{r}} \qquad T = \frac{2\pi r}{v} = 2\pi\sqrt{\frac{r^3}{GM}}$

Field strength tells you the force per unit mass; gravitational potential tells you the energy per unit mass. The gravitational potential $V$ at a point is the work done per unit mass in bringing a small test mass from infinity to that point. Taking infinity as the zero of energy and remembering that gravity is attractive, the potential is always negative.

$V = -G\frac{M}{r} \qquad E_p = mV = -G\frac{Mm}{r}$

Here $V$ is a scalar in J kg⁻¹, and the gravitational potential energy $E_p$ of a mass $m$ in the field of $M$ is simply $mV$. Both are zero at infinity and become more negative as the masses approach, reflecting the fact that you must do positive work to pull them apart. Do not confuse this full-field expression $E_p = -GMm/r$ with the near-surface change $\Delta E_p = mg\,\Delta h$, which is only its approximation for small height changes where $g$ is effectively constant.

Escape speed is the minimum speed at which an object must be launched from a mass's surface so that it just reaches infinity, arriving there with no kinetic energy to spare. Equating the launch kinetic energy to the magnitude of the gravitational potential energy at the surface, $\tfrac{1}{2}mv_{esc}^2 = G\frac{Mm}{r}$, the object's mass cancels and rearranging gives the escape speed.

$v_{esc} = \sqrt{\frac{2GM}{r}}$

For Earth ($M = 6.0 \times 10^{24}$ kg, $r = 6.4 \times 10^6$ m) this is about $1.1 \times 10^4$ m s⁻¹, roughly 11 km s⁻¹. Notice it is independent of the escaping object's mass and of the direction of launch — only the speed matters.

In Paper 2 the marks are analytic: each is tied to a specific line of working — a method mark (M) or an answer mark (A) — and error-carried-forward (ECF) means a wrong number early on need not cost you the marks that follow. But that protection only exists if the method is written down. Study how each mark below is earned by a specific line in a standard orbital-speed calculation.

Gravitational fields are the template for the fields that follow. The same inverse-square structure reappears in electrostatics, where Coulomb's law mirrors Newton's law and electric field strength mirrors $g$; the field-line and potential ideas carry straight across, with the crucial difference that electric charges can repel as well as attract. The orbit analysis — setting the central force equal to the centripetal requirement — is exactly the reasoning used for charged particles in magnetic fields and for electrons in atoms. Master the habit here — pick the right equation, measure $r$ from the centre, balance the forces, show every line — and the rest of the fields topic becomes variations on a method you already own.