Gravitational Fields

Info: Board Coverage AQA Paper 2 | Edexcel CP3 | OCR (A) Paper 2 | CIE P4

Newton’s Law. Every point mass attracts every other point mass with a force directed along the line Joining them, whose magnitude is:

$\boxed{F = \frac{Gm_1 m_2}{r^2}}$

Where $G = 6.67 \times 10^{-11}$ N m$^2$ kg$^{-2}$ is the gravitational constant.

Consider a point mass emitting gravitational flux uniformly in all directions. The flux through a Sphere of radius $r$ is constant (by Gauss’s law for gravity):

$\oint \mathbf{g} \cdot d\mathbf{A} = -4\pi G M_{\mathrm{enc}}$

Since the surface area is $4\pi r^2$The flux density (field strength) must be $g = GM/r^2$. The Force on a test mass $m$ is then $F = mg = GMm/r^2$. This inverse square law is a direct geometric Consequence of flux conservation in three-dimensional space.

• Always attractive (no negative mass exists).
• Infinite range — the force extends to arbitrarily large distances, weakening as $1/r^2$.
• Acts on all objects with mass — it is the universal force binding large-scale structures.
• Extremely weak compared to electromagnetism: $F_e/F_g \sim 10^{36}$ for elementary particles.

The weakness of gravity means it is only dominant at macroscopic scales where the near-perfect Cancellation of positive and negative electric charges renders electromagnetic forces negligible.

Definition. The gravitational field strength $\mathbf{g}$ at a point is the force per unit mass on a Small test mass placed at that point:

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

SI units: N kg$^{-1}$Equivalent to m s$^{-2}$.

For a point mass $M$ at distance $r$:

$\boxed{g = \frac{GM}{r^2}}$

Directed radially inward towards $M$.

Proof. From Newton’s law: $F = GMm/r^2$. Dividing by $m$: $g = F/m = GM/r^2$. $\square$

Theorem. A uniform spherical shell of mass $M$ and radius $R$ produces: (a) the same field as a Point mass $M$ at its centre for all external points ($r \gt R$), and (b) zero field at all internal Points ($r \lt R$).

Proof (external, $r \gt R$). Consider a test mass $m$ at distance $r$ from the centre. Divide the Shell into thin annular rings perpendicular to the line from the centre to $m$. A ring at polar angle $\theta$ has radius $R\sin\theta$Width $R\,d\theta$And mass:

$dM = \frac{M}{4\pi R^2} \cdot 2\pi R^2 \sin\theta\,d\theta = \frac{M}{2}\sin\theta\,d\theta$

Every element of the ring is at distance $s = \sqrt{r^2 + R^2 - 2rR\cos\theta}$ from $m$. By the Axial symmetry of the ring, the transverse components of force cancel, leaving only the component Along the axis. The angle $\alpha$ between the force direction and the axis satisfies:

$\cos\alpha = \frac{r - R\cos\theta}{s}$

The axial force contribution from the ring is:

$dF = \frac{Gm\,dM}{s^2}\cos\alpha = \frac{GmM}{2}\cdot\frac{(r - R\cos\theta)\sin\theta\,d\theta}{(r^2 + R^2 - 2rR\cos\theta)^{3/2}}$

Substitute u = \cos\theta$$du = -\sin\theta\,d\theta. When \theta = 0$$u = 1; when \theta = \pi$$u = -1:

$F = \frac{GmM}{2}\int_{1}^{-1}\frac{(r - Ru)(-du)}{(r^2 + R^2 - 2rRu)^{3/2}}$

The integral evaluates to $\frac{2r}{r^2 - R^2} \cdot \frac{1}{r^2} \cdot (r^2 - R^2) \cdot \frac{1}{r}$ After careful algebra, giving:

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

This is identical to the field of a point mass $M$ at the centre. $\square$

Proof (internal, $r \lt R$). The same integral with $r \lt R$ evaluates to zero. Physically, for Every mass element pulling the test mass in one direction, there is a compensating element on the Opposite side. The nearer element pulls more strongly (shorter distance) but is subtended by a smaller Solid angle, and these two effects cancel exactly. $\square$

### Field Strength at Altitude

At height $h$ above a planet of radius $R$ and surface field $g_0$:

$\boxed{g = g_0\left(\frac{R}{R + h}\right)^2}$

For $h \ll R$The binomial approximation gives $g \approx g_0(1 - 2h/R)$.

Definition. 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:

$\boxed{V = -\frac{GM}{r}}$

SI units: J kg$^{-1}$.

$V = \frac{W}{m} = \frac{1}{m}\int_{\infty}^{r} \frac{GMm}{r'^2}\,dr' = GM\int_{\infty}^{r}\frac{dr'}{r'^2} = GM\left[-\frac{1}{r'}\right]_{\infty}^{r} = -\frac{GM}{r}$

$\square$

Why negative? We define $V = 0$ at infinity. As the test mass moves inward, gravity does positive Work, so the potential decreases. The potential at any finite $r$ is therefore negative, becoming more Negative as $r$ decreases. An external agent must supply energy $GMm/r$ to move the mass from $r$ back To infinity.

$\boxed{g = -\frac{dV}{dr}}$

Proof. $V = -GM/r = -GM r^{-1}$. Then:

$\frac{dV}{dr} = GM\,r^{-2} = \frac{GM}{r^2}$

The field (directed inward, i.e. In the negative radial direction) is:

$g_r = -\frac{GM}{r^2} = -\frac{dV}{dr}$

$\square$

The minus sign confirms that the field points in the direction of decreasing potential.

For two masses $M$ and $m$ separated by $r$:

$\boxed{E_p = -\frac{GMm}{r}}$

Connection to $E_p = mgh$. For height $h \ll R_E$The Taylor expansion gives:

$\Delta E_p = -\frac{GMm}{R_E + h} + \frac{GMm}{R_E} = \frac{GMmh}{R_E(R_E + h)} \approx \frac{GMmh}{R_E^2} = mgh$

Since $g = GM/R_E^2$. The linear formula $mgh$ is the first-order approximation of the full Gravitational potential energy.

Definition. The escape velocity $v_e$ is the minimum launch speed for an object to reach infinity With zero residual speed from the surface of a body of mass $M$ and radius $R$.

At launch: E_k = \frac{1}{2}mv_e^2$$E_p = -GMm/R. At infinity: E_k = 0$$E_p = 0. By energy Conservation:

$\frac{1}{2}mv_e^2 - \frac{GMm}{R} = 0$

$\boxed{v_e = \sqrt{\frac{2GM}{R}}}$

$\square$

The circular orbital speed at radius $r$ is $v_{\mathrm{orb}} = \sqrt{GM/r}$. Therefore:

$v_e = \sqrt{2}\,v_{\mathrm{orb}}$

Escape requires exactly twice the kinetic energy of a circular orbit: $\frac{1}{2}mv_e^2 = 2 \times \frac{1}{2}mv_{\mathrm{orb}}^2$. A spacecraft in circular orbit needs a speed increase of $(\sqrt{2} - 1) \times 100\% \approx 41.4\%$ to escape.

Worked Example: Escape from Mars

Calculate the escape velocity from Mars ($M = 6.42 \times 10^{23}$ kg, $R = 3.39 \times 10^6$ m).

Answer. $v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 6.42 \times 10^{23}}{3.39 \times 10^6}} = \sqrt{\frac{8.56 \times 10^{13}}{3.39 \times 10^6}} = \sqrt{2.53 \times 10^7} = 5020$ m s$^{-1}$ $= 5.02$ km s$^{-1}$.

## 5. Orbital Mechanics

For a satellite of mass $m$ in a circular orbit of radius $r$ around mass $M$:

Quantity	Expression
Orbital speed	$v = \sqrt{GM/r}$
Orbital period	$T = 2\pi\sqrt{r^3/(GM)}$
Centripetal acceleration	$a = GM/r^2 = g$
Kinetic energy	$E_k = GMm/(2r)$
Potential energy	$E_p = -GMm/r$
Total energy	$E_{\mathrm{total}} = -GMm/(2r)$

Proof of energy relations. From $\frac{GMm}{r^2} = \frac{mv^2}{r}$: $v^2 = GM/r$.

$E_k = \frac{1}{2}mv^2 = \frac{GMm}{2r}$. $E_p = -GMm/r$. Total: $E_k + E_p = -\frac{GMm}{2r}$.

Note: $E_k = \frac{1}{2}|E_p|$ and $E_{\mathrm{total}} = E_k$. This is the virial theorem for bound Gravitational systems: $2E_k + E_p = 0$. $\square$

Key insight. The total energy is negative — the satellite is gravitationally bound. To move to a Higher orbit, energy must be added (the orbit becomes less negative). The kinetic energy decreases With increasing $r$But the total energy increases (potential energy increase dominates).

For any Keplerian orbit (circular or elliptical) with semi-major axis $a$:

$\boxed{v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right)}$

Setting $a = r$ recovers the circular orbit result. For a parabolic escape trajectory ($a \to \infty$): $v^2 = 2GM/r$Giving the escape speed.

$v = \sqrt{\frac{GM}{r}}$

The satellite’s mass $m$ cancels. This is the same reason that all objects fall at the same rate in a Gravitational field (equivalence principle).

Worked Example: Satellite Orbit Change

A 500 kg satellite is in a circular orbit of radius $7.0 \times 10^6$ m. Calculate the energy Required to move it to a circular orbit of radius $1.4 \times 10^7$ m.

Answer. $GM = 3.98 \times 10^{14}$ N m$^2$ kg$^{-1}$.

$E_1 = -\frac{GMm}{2r_1} = -\frac{3.98 \times 10^{14} \times 500}{2 \times 7.0 \times 10^6} = -1.42 \times 10^{10}$ J.

$E_2 = -\frac{3.98 \times 10^{14} \times 500}{2 \times 1.4 \times 10^7} = -7.11 \times 10^9$ J.

$\Delta E = E_2 - E_1 = -7.11 \times 10^9 - (-1.42 \times 10^{10}) = 7.1 \times 10^9$ J $= 7.1$ GJ.

Every planet moves in an elliptical orbit with the Sun at one focus.

Proof sketch. Starting from $\mathbf{F} = -\frac{GMm}{r^2}\hat{\mathbf{r}}$ (central force), the orbit Equation in polar coordinates is:

$r = \frac{a(1 - e^2)}{1 + e\cos\theta}$

Where $a$ is the semi-major axis and $e$ is the eccentricity. For $E \lt 0$ (bound orbit), $e \lt 1$ And the orbit is an ellipse with the central mass at one focus. $\square$

A line joining a planet to the Sun sweeps out equal areas in equal times.

Proof. The gravitational force is central ($\mathbf{F} \parallel \mathbf{r}$), so torque $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F} = 0$. Angular momentum $L = mrv_\perp$ is constant.

Area swept in time $dt$: $dA = \frac{1}{2}r \cdot v_\perp\,dt = \frac{L}{2m}\,dt$.

$\frac{dA}{dt} = \frac{L}{2m} = \mathrm{const}$

$\square$

The planet moves fastest at perihelion (closest approach) and slowest at aphelion (farthest point), Consistent with conservation of angular momentum: small $r$ requires large $v_\perp$.

$\boxed{T^2 = \frac{4\pi^2}{GM}\,a^3}$

Proof for circular orbits. Equating gravitational and centripetal force:

$\frac{GMm}{r^2} = \frac{mv^2}{r} \implies v = \sqrt{\frac{GM}{r}}$

Since $T = 2\pi r/v$:

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

$T^2 = \frac{4\pi^2 r^3}{GM}$

For elliptical orbits, replace $r$ with the semi-major axis $a$. $\square$

The proportionality constant $4\pi^2/(GM)$ depends only on the central body, not on the orbiting Object. This is how Kepler determined the relative distances of the planets from the Sun using only Their observed periods.

Definition. A geostationary orbit is a circular, prograde, equatorial orbit with period equal to One sidereal day ($T = 86164$ s), causing the satellite to remain fixed above a point on the equator.

$r^3 = \frac{GMT^2}{4\pi^2} = \frac{3.98 \times 10^{14} \times (86164)^2}{4\pi^2} = 7.54 \times 10^{22}$

$\boxed{r = 4.22 \times 10^7\ \mathrm{m} = 42\,200\ \mathrm{km}}$

Altitude above Earth’s surface: $h = 42200 - 6370 = 35\,830$ km.

$v = \sqrt{\frac{GM}{r}} = \frac{2\pi r}{T} = 3070\ \mathrm{m\,s}^{-1}$

1. Correct radius: $r \approx 42\,200$ km (from Kepler’s third law with $T = 86164$ s).
2. Equatorial plane: Any inclination causes the satellite to trace a figure-eight (anisotropic pattern) as seen from the ground.
3. Prograde rotation: The satellite must orbit west to east, matching Earth’s rotation.

From Newton’s apple to quantum particles, physics explains how the world works through measurable quantities and testable laws. Forces cause acceleration, energy transforms between forms but is never lost, and waves carry information across vast distances. These concepts form the foundation for engineering, astronomy, and our understanding of reality itself.

• Mechanics
• Waves
• Electricity
• Fields

• Physics
• Mechanics
• Fields