Gravitational Fields

A gravitational field is a region of space in which a mass experiences a gravitational force. Its strength at a point is the gravitational field strength $g$ — the force per unit mass on a small mass placed there: $g = \frac{F}{m}$ with unit $\text{N kg}^{-1}$ (numerically identical to $\text{m s}^{-2}$, the free-fall acceleration). It is a vector, and because gravity only ever pulls, $g$ always points towards the mass creating the field. Rearranged, $F = mg$ is just the weight of a mass.

The force itself is given by Newton's law of gravitation. The gravitational force between two point masses is directly proportional to the product of the masses and inversely proportional to the square of their separation: $F = \frac{Gm_1m_2}{r^2}$ where $G = 6.67\times10^{-11}\ \text{N m}^2\text{ kg}^{-2}$ is the universal gravitational constant. Two features are examined relentlessly:

• The force is always attractive — mass has only one sign, so (unlike charges) masses can never repel.
• $r$ is the distance between the centres of the two masses, not between their surfaces. A uniform sphere behaves gravitationally as if all its mass were concentrated at its centre.

Because $G$ is so tiny, the force between everyday objects is negligible — gravitation matters only when at least one mass is astronomically large (a planet, star or moon).

Set one of the masses in Newton's law to a test mass of $1\ \text{kg}$ and you get the field strength produced by a single spherical mass $m$ at a distance $r$ from its centre: $g = \frac{Gm}{r^2}$ This is a radial field: the field lines point straight inward towards the centre and spread apart with distance. Note the crucial difference from Newton's law — the field-strength formula contains only the one mass creating the field (unit $\text{N kg}^{-1}$), whereas the force formula contains both masses (unit $\text{N}$).

Where the field lines are close together the field is strong; where they spread out it is weak. This is why $g$ obeys an inverse-square law — double the distance and the same field lines are spread over four times the area, so the field is a quarter as strong.

Near a planet's surface the field is approximately uniform. Over the small range of heights encountered in everyday life, $r$ barely changes, so $g$ is essentially constant in both magnitude and direction. In that region the radial pattern looks locally like a set of parallel, equally spaced vertical lines — which is exactly the uniform-field approximation that justifies using a constant $g = 9.81\ \text{N kg}^{-1}$ in mechanics.

While field strength tells you the force on a mass, gravitational potential tells you the energy. The gravitational potential $V$ at a point is the work done per unit mass in bringing a small mass from infinity to that point: $V = -\frac{Gm}{r}$ Three features are examined:

• $V$ is a scalar (no direction), unit $\text{J kg}^{-1}$.
• $V$ is always negative, approaching zero at infinity. It is negative because gravity is attractive: energy is released as a mass falls inward from infinity, so the potential energy at any finite $r$ is below the zero at infinity.
• $V$ falls off as $1/r$ — more slowly than the field ($1/r^2$).

The gravitational potential energy of a mass $M$ placed at a point of potential $V$ is simply $E = MV = -\frac{GmM}{r}.$ To raise a mass (increase its distance) you do work against gravity, so its potential energy increases towards zero — the change $\Delta E = M\,\Delta V$ is positive. This is how you find the energy needed to lift a spacecraft into orbit.

Equipotentials are surfaces (lines in 2D) joining points of equal potential. For a spherical mass they are concentric spheres centred on the mass, and they are always perpendicular to the field lines (see the dashed circles in the field diagram above). No work is done moving a mass along an equipotential — the potential does not change — whereas work is done moving between them. The field strength equals the potential gradient, $g = -\dfrac{\Delta V}{\Delta r}$, so closely spaced equipotentials indicate a strong field.

For a satellite in a circular orbit, one force keeps it turning: gravity. The gravitational attraction provides the centripetal force needed for circular motion. Setting the two equal for a satellite of mass $m$ orbiting a central mass $M$ at radius $r$: $\frac{GMm}{r^2} = \frac{mv^2}{r}.$ The satellite mass $m$ cancels, giving the orbital speed: $v = \sqrt{\frac{GM}{r}}.$ A larger orbit means a slower satellite. The period follows from the fact that in one orbit the satellite travels a circumference $2\pi r$ at speed $v$, so $v = \dfrac{2\pi r}{T}$.

Deriving Kepler's third law. Combine $v^2 = \dfrac{GM}{r}$ with $v = \dfrac{2\pi r}{T}$ (so $v^2 = \dfrac{4\pi^2 r^2}{T^2}$): $\frac{4\pi^2 r^2}{T^2} = \frac{GM}{r} \quad\Rightarrow\quad T^2 = \frac{4\pi^2}{GM}\,r^3.$ Since $G$, $M$ and $4\pi^2$ are constants, $\boxed{T^2\propto r^3}$ — Kepler's third law. A graph of $T^2$ against $r^3$ for several satellites of the same central body is a straight line through the origin with gradient $\dfrac{4\pi^2}{GM}$, from which the central mass $M$ can be found.

Geostationary orbit. A geostationary satellite stays fixed above one point on the equator. To do so it must satisfy all of these conditions:

1. period of exactly 24 hours (equal to the Earth's rotation period);
2. orbit directly above the equator;
3. orbit in the same direction as the Earth's rotation (west to east);
4. move in a circular orbit of the one radius ($\approx 4.2\times10^{7}\ \text{m}$ from the centre, about $36{,}000\ \text{km}$ up) that gives this period.

Such orbits are ideal for communications and TV relays because a ground aerial can point permanently in one direction.

Escape considerations (qualitative). To leave a planet's field entirely, an object must be given enough kinetic energy to reach infinity (where potential energy is zero). Setting $\tfrac{1}{2}mv^2$ equal to the magnitude of the surface potential energy $\dfrac{GMm}{R}$ gives the escape velocity $v = \sqrt{\dfrac{2GM}{R}}$ — independent of the escaping object's mass, and $\sqrt{2}$ times the surface circular-orbit speed.

This comparison is a classic 6-mark synoptic question linking Unit 5 (gravitational fields) with Unit 4 (electric fields). Learn the parallels and the one decisive difference.

Feature	Gravitational field	Electric field
Force law	$F = \dfrac{Gm_1m_2}{r^2}$	$F = \dfrac{kQ_1Q_2}{r^2}$
Field strength	$g = \dfrac{F}{m}$ (per unit mass)	$E = \dfrac{F}{Q}$ (per unit positive charge)
Radial field	$g = \dfrac{Gm}{r^2}$	$E = \dfrac{kQ}{r^2}$
Potential	$V = -\dfrac{Gm}{r}$ (always $-$)	$V = \dfrac{kQ}{r}$ ($+$ or $-$)
Constant	$G = 6.67\times10^{-11}$	$k = \dfrac{1}{4\pi\varepsilon_0} = 8.99\times10^{9}$
Source property	mass	charge
Direction	always attractive	attractive or repulsive

Similarities

• Both are inverse-square fields: the force between two point sources $\propto 1/r^2$, and the two force laws have exactly the same algebraic form.
• Both define field strength as force per unit test quantity (per unit mass / per unit positive charge), and in both the potential of a point source falls off as $1/r$.

Differences

• The decisive one: mass has only one sign, so the gravitational force is always attractive (and gravitational potential is always negative); charge has two signs, so the electric force can be attractive or repulsive (and electric potential can be positive or negative).
• Gravity is vastly weaker than the electric force between the same two particles (about $10^{36}$ times weaker for two protons), and the two fields respond to different properties of matter — mass versus charge.