Gravitational Fields

Gravitational Fields

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

1. Newton’s Law of Universal Gravitation

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.

Geometric Origin of the Inverse Square Law

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.

Properties of the Gravitational Force

• 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.

2. Gravitational Field Strength

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:

$\boxed{\mathbf{g} = \frac◆LB◆\mathbf{F}◆RB◆◆LB◆m◆RB◆}$

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 $m$: $g = F/m = GM/r^2$. $\square$

Proof of the Shell Theorem

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:

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

Every element of the ring is at distance $s = \sqrt◆LB◆r^2 + R^2 - 2rR\cos\theta◆RB◆$ 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:

$\cos\alpha = \frac◆LB◆r - R\cos\theta◆RB◆◆LB◆s◆RB◆$

The axial force contribution from the ring is:

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

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. $\square$

:::caution Warning Earth is approximately but not perfectly spherical (equatorial bulge), so $g$ varies slightly with Latitude even at sea level. :::

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 $g \approx g_0(1 - 2h/R)$.

3. Gravitational Potential

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$^{-1}$.

Derivation

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

$\square$

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.

Field—Potential Relationship

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

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.

Gravitational Potential Energy

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.

4. Escape Velocity

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 $R$.

Derivation from Energy Conservation

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:

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

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

$\square$

Relationship to Orbital Speed

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

:::caution Common Pitfall Escape velocity is independent of the mass and direction of launch of the Projectile. A 1 kg ball and a $10^6$ kg rocket both need the same speed. However, the required kinetic Energy $E_k = \frac{1}{2}mv_e^2$ scales with mass. :::

5. Orbital Mechanics

Circular Orbit Equations

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).

The Vis-Viva Equation

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.

Orbital Speed is Mass-Independent

$v = \sqrt◆LB◆\frac{GM}{r}◆RB◆$

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$^{-1}$.

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

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

6. Kepler’s Laws

First Law: Law of Ellipses

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:

$r = \frac◆LB◆a(1 - e^2)◆RB◆◆LB◆1 + e\cos\theta◆RB◆$

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. $\square$

Second Law: Law of Equal Areas

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 $v_\perp$.

Third Law: Law of Periods

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

Proof for circular orbits. Equating gravitational and centripetal force:

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

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

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

$T^2 = \frac◆LB◆4\pi^2 r^3◆RB◆◆LB◆GM◆RB◆$

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.

7. Geostationary Orbits

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.

Derivation of Orbital Radius

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

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

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

Orbital Speed

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

Three Necessary and Sufficient Conditions

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.

:::caution Warning Geostationary is a stricter condition: geosynchronous AND equatorial AND circular. GPS satellites are Neither — they use medium Earth orbits at 20,200 km altitude with 12-hour periods.

Applications

• Communications: Constant line of sight to a ground station, ideal for broadcast and relay.
• Weather monitoring: Continuous hemispheric observation (e.g., Meteosat, GOES).
• Early warning: Persistent surveillance of fixed geographical regions.

8. Comparison with Electric Fields

Property	Gravitational	Electric
Source property	Mass $M$	Charge $Q$
Force law	$F = Gm_1m_2/r^2$	$F = q_1q_2/(4\pi\varepsilon_0 r^2)$
Field strength	$g = GM/r^2$	$E = Q/(4\pi\varepsilon_0 r^2)$
Potential	$V = -GM/r$ (always $\lt 0$)	$V = Q/(4\pi\varepsilon_0 r)$ (sign of $Q$)
Attractive/repulsive	Always attractive	Both possible
Screening	Impossible	Possible (Faraday cage)
Relative strength	$\sim 10^{-36}$	$\sim 1$

Both fields obey inverse square force laws, possess $1/r$ potentials, and satisfy a Gauss’s law. The Structural parallel is exact, differing only in the source property (mass vs. Charge) and the Existence of negative charge enabling screening and repulsion.

Problem Set

Problem 1

Calculate the gravitational field strength at the surface of Jupiter ($M = 1.90 \times 10^{27}$ kg, $R = 6.99 \times 10^7$ m).

Answer. $g = \frac{GM}{R^2} = \frac◆LB◆6.67 \times 10^{-11} \times 1.90 \times 10^{27}◆RB◆◆LB◆(6.99 \times 10^7)^2◆RB◆ = \frac◆LB◆1.267 \times 10^{17}◆RB◆◆LB◆4.886 \times 10^{15}◆RB◆ = 25.9$ N kg$^{-1}$.

Problem 2

Two stars of mass $3.0 \times 10^{30}$ kg each orbit their common centre of mass with separation $2.0 \times 10^{11}$ m. Find the orbital period.

Answer. Each star orbits at radius $r = 1.0 \times 10^{11}$ m.

$T^2 = \frac◆LB◆4\pi^2 r^3◆RB◆◆LB◆G(2M)◆RB◆ = \frac◆LB◆4\pi^2 \times 10^{33}◆RB◆◆LB◆6.67 \times 10^{-11} \times 6.0 \times 10^{30}◆RB◆ = \frac◆LB◆3.95 \times 10^{34}◆RB◆◆LB◆4.00 \times 10^{20}◆RB◆ = 9.88 \times 10^{13}$

$T = 9.94 \times 10^6$ s $= 115$ days.

Problem 3

Show that for a satellite in circular orbit, the ratio of kinetic energy to the magnitude of Potential energy is exactly $1:2$.

Answer. E_k = GMm/(2r)$$|E_p| = GMm/r. Therefore $E_k / |E_p| = 1/2$. This follows from the Virial theorem: $2E_k + E_p = 0$. $\square$

Problem 4

Prove that the gravitational field inside a uniform solid sphere of radius $R$ at distance $r$ from The centre is $g = GMr/R^3$.

Answer. By the shell theorem, only the mass within radius $r$ contributes. For uniform density $\rho = 3M/(4\pi R^3)$The enclosed mass is $M_{\mathrm{enc}} = \rho \cdot 4\pi r^3/3 = Mr^3/R^3$.

$g = \frac◆LB◆GM_{\mathrm{enc}}◆RB◆◆LB◆r^2◆RB◆ = \frac{GMr^3/R^3}{r^2} = \frac{GMr}{R^3}$.

$\square$

Problem 5

A comet approaches the Sun from very far away with speed $v_0 = 5.0$ km s$^{-1}$ and perihelion Distance $r_p = 1.0 \times 10^{10}$ m. Find its speed at perihelion. ($M_{\odot} = 1.99 \times 10^{30}$ kg.)

Answer. Energy conservation: $\frac{1}{2}mv_p^2 - \frac{GMm}{r_p} = \frac{1}{2}mv_0^2$.

$v_p = \sqrt{v_0^2 + 2GM/r_p} = \sqrt◆LB◆2.5 \times 10^7 + 2.65 \times 10^{10}◆RB◆ = \sqrt◆LB◆2.653 \times 10^{10}◆RB◆ = 1.63 \times 10^5$ m s$^{-1}$ $= 163$ km s$^{-1}$.

Problem 6

Calculate the gravitational potential energy of the Earth--Moon system. ($M_E = 5.97 \times 10^{24}$ Kg, $M_M = 7.35 \times 10^{22}$ kg, $r = 3.84 \times 10^8$ m.)

Answer. $E_p = -\frac{GM_E M_M}{r} = -\frac◆LB◆6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 7.35 \times 10^{22}◆RB◆◆LB◆3.84 \times 10^8◆RB◆ = -7.63 \times 10^{28}$ J.

Problem 7

Show that the work required to move a satellite from a circular orbit of radius $r$ to a circular Orbit of radius $2r$ is $GMm/(4r)$.

Answer. E_1 = -GMm/(2r)$$E_2 = -GMm/(4r). Work $= E_2 - E_1 = -GMm/(4r) + GMm/(2r) = GMm/(4r)$. $\square$

Problem 8

Derive $v_e = \sqrt{2gR}$ for a planet of radius $R$ and surface field strength $g$.

Answer. $g = GM/R^2$So $GM = gR^2$. $v_e = \sqrt{2GM/R} = \sqrt{2gR^2/R} = \sqrt{2gR}$. $\square$

Problem 9

A geostationary satellite has mass 1200 kg. Calculate (a) its total orbital energy, (b) the energy Needed to escape Earth's gravity from its orbit.

Answer. $r = 4.22 \times 10^7$ m. $GM = 3.98 \times 10^{14}$.

(a) $E = -GMm/(2r) = -3.98 \times 10^{14} \times 1200 / (2 \times 4.22 \times 10^7) = -5.66 \times 10^9$ J.

(b) To escape: $E_{\mathrm{final}} = 0$. Energy needed $= 0 - E = 5.66 \times 10^9$ J $= 5.66$ GJ.

Problem 10

Use the vis-viva equation to find the speed of a satellite at perigee ($r_p = 7.0 \times 10^6$ m) of an Elliptical orbit with apogee $r_a = 4.2 \times 10^7$ m.

Answer. Semi-major axis: $a = (r_p + r_a)/2 = (7.0 + 42.0) \times 10^6 / 2 = 2.45 \times 10^7$ m.

$v_p = \sqrt{GM(2/r_p - 1/a)} = \sqrt◆LB◆3.98 \times 10^{14}(2.857 \times 10^{-7} - 4.082 \times 10^{-8})◆RB◆ = \sqrt◆LB◆3.98 \times 10^{14} \times 2.449 \times 10^{-7}◆RB◆ = \sqrt◆LB◆9.75 \times 10^7◆RB◆ = 9870$ m s$^{-1}$.

Common Pitfalls

1. Confusing gravitational field strength $g$ with gravitational potential $V_g$ — one is force per unit mass, the other is energy per unit mass.

2. Forgetting that field lines point in the direction a positive test charge (or mass) would move.

3. Misidentifying the system boundary when applying conservation laws — define what is included before writing equations.

4. Neglecting air resistance or assuming ideal conditions when the question specifies a real-world scenario.

5. Using the wrong equation from the data sheet — take time to read the full equation, including conditions and variable definitions.

6. Rounding intermediate answers too early, which compounds errors in multi-step calculations.

Worked Examples

Example 1: Gravitational Field Strength at an Altitude

Problem. Calculate the gravitational field strength at $300\ \mathrm{km}$ above the Earth’s surface. (Earth mass $M = 5.97 \times 10^{24}\ \mathrm{kg}$, radius $R = 6.37 \times 10^6\ \mathrm{m}$, $G = 6.67 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$.)

Solution. Distance from centre: $r = 6.37 \times 10^6 + 3 \times 10^5 = 6.67 \times 10^6\ \mathrm{m}$.

$g = \frac{GM}{r^2} = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{(6.67 \times 10^6)^2} = \frac{3.982 \times 10^{14}}{4.449 \times 10^{13}} = 8.95\ \mathrm{m\,s^{-2}}$

This is about 91% of the surface value ($9.81\ \mathrm{m\,s^{-2}}$).

$\blacksquare$

Example 2: Escape Velocity

Problem. Derive and calculate the escape velocity from the surface of the Earth.

Solution. At the surface, total energy: $E_k + E_p = \frac{1}{2}mv_{\mathrm{esc}}^2 - \frac{GMm}{R}$.

At infinity: $E_k = 0$ and $E_p = 0$, so total energy $= 0$.

$\frac{1}{2}mv_{\mathrm{esc}}^2 = \frac{GMm}{R} \implies v_{\mathrm{esc}} = \sqrt{\frac{2GM}{R}}$

$v_{\mathrm{esc}} = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.37 \times 10^6}} = \sqrt{1.25 \times 10^8} \approx 11.2\ \mathrm{km\,s^{-1}}$

$\blacksquare$

Summary

• Newton’s law of gravitation: $F = \frac{GMm}{r^2}$; always attractive.
• Gravitational field strength: $g = \frac{GM}{r^2}$; at the surface $g \approx 9.81\ \mathrm{m\,s^{-2}}$.
• Gravitational potential: $V = -\frac{GM}{r}$; work done to move mass $m$: $W = m\Delta V$.
• Escape velocity: $v_{\mathrm{esc}} = \sqrt{\frac{2GM}{R}}$; for Earth $\approx 11.2\ \mathrm{km\,s^{-1}}$.
• Orbital velocity: $v = \sqrt{\frac{GM}{r}}$; $g = -\frac{dV}{dr}$ links field and potential.

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