Gravitational Fields

Gravitational Fields

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

Explore the simulation above to develop intuition for this topic.

Intuition

Gravity is the weakest force, but it holds the universe together: Every mass attracts every other mass — you are gravitationally attracted to your phone, your desk, and the person sitting next to you. But the force is so tiny that you only notice it when at least one mass is enormous (like a planet or star). This is why you feel weight (Earth’s gravity) but not the gravitational pull of the building you’re in.

Why it matters: Gravity explains why planets orbit the Sun, why tides exist, why astronauts float in orbit (they’re actually falling), and why black holes trap light. It’s the force that shapes the large-scale structure of the universe.

The key insight: Orbits are a balance between gravity and inertia. The Moon is constantly falling toward Earth, but it’s also moving sideways fast enough that it keeps missing. This is what an orbit is — perpetual free fall that never hits the ground. The same principle applies to satellites and the International Space Station.

1. Newton’s Law of Gravitation

Newton’s Law of Universal Gravitation. Every particle in the universe attracts every other Particle with a force that is:

1. Directly proportional to the product of their masses
2. Inversely proportional to the square of the distance between them

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

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

Intuition. The inverse square law arises from the geometry of three-dimensional space. The Gravitational “flux” spreads over a sphere of area $4\pi r^2$So the field strength decreases as $1/r^2$. This is the same geometric reason that the intensity of light decreases as $1/r^2$.

2. Gravitational Field Strength

Definition. The gravitational field strength $g$ at a point is the force per unit mass Experienced by a small test mass placed at that point:

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

Units. N kg$^{-1}$Which is equivalent to m s$^{-2}$.

For a point mass (or spherical body):

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

Proof. From $F = \frac{GMm}{r^2}$Dividing by $m$: $g = F/m = GM/r^2$. $\square$

Properties:

• $g$ is a vector quantity, directed towards the mass creating the field.
• Near Earth’s surface, $g \approx 9.81$ N kg$^{-1}$ (constant, since $r \approx R_E$).
• Inside a uniform spherical shell, $g = 0$ (shell theorem).

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:

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

Units. J kg$^{-1}$.

Derivation from Work Done

The work done to move a mass $m$ from infinity to distance $r$ from mass $M$:

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

The potential (work per unit mass) is:

$V = \frac{W}{m} = -\frac{GM}{r} \quad \square$

Intuition: Why is gravitational potential negative? We define $V = 0$ at infinity. To bring a Mass from infinity towards $M$Gravity does positive work (the mass accelerates), meaning the System loses potential energy. Alternatively, an external agent would need to do negative work (i.e., the system does work) to move the mass in. Hence $V < 0$ everywhere. The potential becomes More negative as you approach the mass.

Proof that $g = -\frac{dV}{dr}$

$-\frac{dV}{dr} = -\frac{d}{dr}\left(-\frac{GM}{r}\right) = -\left(\frac{GM}{r^2}\right) \cdot (-1) = \frac{GM}{r^2} = g \quad \square$

The negative sign ensures that the field points in the direction of decreasing potential (towards the mass).

4. Gravitational Potential Energy

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

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

This is the energy required to separate the masses to infinity (or equivalently, the energy released When bringing them together from infinity).

Connection to $E_p = mgh$ near the surface. For height $h \ll R_E$:

$\Delta E_p = -\frac{GMm}{R_E + h} - \left(-\frac{GMm}{R_E}\right) = GMm\left(\frac{1}{R_E} - \frac{1}{R_E + h}\right) = \frac{GMmh}{R_E(R_E + h)} \approx \frac{GMmh}{R_E^2} = mgh$

Since $g = GM/R_E^2$. This shows that $E_p = mgh$ is the small-height approximation of the full Gravitational potential energy.

5. Kepler’s Laws

First Law: Law of Orbits

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

Second Law: Law of Areas

A line joining a planet to the Sun sweeps out equal areas in equal times. This means the planet Moves faster when closer to the Sun (near perihelion) and slower when farther (near aphelion).

Proof from conservation of angular momentum. $L = mrv_\perp = \mathrm{const}$. When $r$ is Small, $v_\perp$ must be large, and vice versa.

Third Law: Law of Periods

The square of the orbital period is proportional to the cube of the semi-major axis:

$\boxed{T^2 \propto a^3}$

Derivation of Kepler’s Third Law (for circular orbits)

For a circular orbit of radius $r$The centripetal force is provided by gravity:

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

The period is $T = \frac{2\pi r}{v}$So $v = \frac{2\pi r}{T}$:

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

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

Since $\frac{4\pi^2}{GM}$ is constant for a given central body, $T^2 \propto r^3$. $\square$

6. Escape Velocity

Definition. The escape velocity $v_e$ is the minimum speed needed for an object to escape a Gravitational field (i.e., reach infinity with zero speed).

Derivation from energy conservation. At launch, the object has kinetic energy $\frac{1}{2}mv_e^2$ and potential energy $-\frac{GMm}{r}$. At infinity, both KE and PE are zero. By Conservation of energy:

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

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

For Earth: $v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.37 \times 10^6}} = \sqrt{1.25 \times 10^8} \approx 11.2$ Km s$^{-1}$.

Intuition. The escape velocity is $\sqrt{2}$ times the circular orbital velocity at the same Radius. This factor of $\sqrt{2}$ comes from the ratio of kinetic energies: escape requires $2 \times$ the orbital KE (since $v_e^2 = 2GM/r = 2v_{\mathrm{orbit}}^2$).

7. Orbital Energy

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

Kinetic energy:

$E_k = \frac{1}{2}mv^2 = \frac{1}{2}m\frac{GM}{r} = \frac{GMm}{2r}$

Potential energy:

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

Total energy:

$\boxed{E_{\mathrm{total}} = E_k + E_p = -\frac{GMm}{2r}}$

Intuition. The total energy is negative — the satellite is bound. To move to a higher orbit, Energy must be added. The total energy is exactly half the potential energy (and the negative of the Kinetic energy).

8. Comparison of Gravitational and Electric Fields

Problem Set

9. Variation of $g$ with Altitude

The gravitational field strength decreases with distance from the centre of a planet:

$g(r) = \frac{GM}{r^2}$

Field Strength at Height $h$ Above the Surface

At a height $h$ above the surface of a planet of radius $R$:

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

Where $g_0 = GM/R^2$ is the field strength at the surface.

Approximate Decrease for Small Altitudes

For $h \ll R$Using the binomial approximation $(1 + h/R)^{-2} \approx 1 - 2h/R$:

$g_h \approx g_0\left(1 - \frac{2h}{R}\right)$

The fractional decrease is approximately $2h/R$.

Numerical Examples for Earth

$R_E = 6370$ km, $g_0 = 9.81$ N kg$^{-1}$.

At $h = 100$ km: $g_{100} = 9.81 \times (6370/6470)^2 = 9.81 \times 0.9693 = 9.51$ N kg$^{-1}$. This Is a decrease of 0.30 N kg$^{-1}$Or approximately 3.1%.

At $h = 300$ km (typical low Earth orbit): $g_{300} = 9.81 \times (6370/6670)^2 = 9.81 \times 0.9120 = 8.95$ N kg$^{-1}$A decrease of about 8.8%.

At $h = 35786$ km (geostationary orbit): $g_{\mathrm{GEO}} = 9.81 \times (6370/42156)^2 = 9.81 \times 0.0228 = 0.224$ N kg$^{-1}$Only about 2.3% of the surface value.

Field Strength Inside the Earth

For a uniform sphere of radius $R$ and mass $M$The field strength at distance $r$ from the centre ($r \lt R$) is:

$g(r) = \frac{GM_{\mathrm{enc}}}{r^2} = \frac{G(Mr^3/R^3)}{r^2} = \frac{GM}{R^3}\,r = g_0 \frac{r}{R}$

Where $M_{\mathrm{enc}} = M(r/R)^3$ is the mass enclosed within radius $r$ (shell theorem).

This shows $g$ increases linearly from $0$ at the centre to $g_0$ at the surface. Maximum $g$ occurs At the surface (for a uniform sphere).

10. Geostationary Orbits

Definition. A geostationary orbit is a circular orbit in the equatorial plane with a period Equal to one sidereal day (approximately 24 hours). A satellite in this orbit appears stationary Relative to a point on Earth’s surface.

Derivation of the Orbital Radius

The centripetal acceleration of the satellite is provided by gravity:

$\frac{GMm}{r^2} = m\omega^2 r$

Where $\omega = 2\pi/T$ and $T = 24 \times 3600 = 86400$ s.

$\frac{GM}{r^2} = \omega^2 r \implies r^3 = \frac{GM}{\omega^2}$

$\boxed{r = \sqrt[3]{\frac{GM}{\omega^2}}}$

Substituting $GM = 3.976 \times 10^{14}$ N m$^2$ kg$^{-1}$ and $\omega = 2\pi/86400 = 7.272 \times 10^{-5}$ rad s$^{-1}$:

$r^3 = \frac{3.976 \times 10^{14}}{(7.272 \times 10^{-5})^2} = \frac{3.976 \times 10^{14}}{5.288 \times 10^{-9}} = 7.52 \times 10^{22}$

$r = \sqrt[3]{7.52 \times 10^{22}} = 4.22 \times 10^7 \mathrm{ m} = 42\,200 \mathrm{ km}$

The altitude above Earth’s surface is $h = r - R_E = 42200 - 6370 = 35\,830$ km.

Orbital Speed

$v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{3.976 \times 10^{14}}{4.22 \times 10^7}} = \sqrt{9.42 \times 10^6} = 3070 \mathrm{ m s}^{-1}$

Verification: $v = \omega r = (7.272 \times 10^{-5})(4.22 \times 10^7) = 3070$ m s$^{-1}$. $\checkmark$

Conditions for a Geostationary Orbit

Three conditions must all be satisfied:

1. Correct radius: $r \approx 42\,200$ km (derived above).
2. Equatorial plane: The orbit must lie in Earth’s equatorial plane; otherwise, the satellite would appear to drift north and south.
3. Same direction as Earth’s rotation: The satellite must orbit from west to east.

Applications. Communications satellites (constant line of sight to a ground station), weather Satellites (continuous monitoring of a hemisphere). GPS satellites are NOT geostationary — they use Medium Earth orbits for better geometric accuracy.

11. Gravitational Potential Energy — Taylor Expansion

The General Form

The gravitational potential energy of two masses $M$ and $m$ separated by distance $r$ is:

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

This is negative because the zero of potential energy is defined at infinity ($r \to \infty$). Work Must be done against gravity to separate the masses, increasing $E_p$ towards zero.

Recovering $mgh$ via Taylor Expansion

For a mass at height $h$ above Earth’s surface ($r = R_E + h$):

$E_p(R_E + h) = -\frac{GMm}{R_E + h} = -\frac{GMm}{R_E}\left(1 + \frac{h}{R_E}\right)^{-1}$

For $h \ll R_E$Expand using the binomial series $(1 + x)^{-1} \approx 1 - x + x^2 - \cdots$ with $x = h/R_E$:

$E_p(R_E + h) \approx -\frac{GMm}{R_E}\left(1 - \frac{h}{R_E} + \frac{h^2}{R_E^2} - \cdots\right)$

The change in potential energy from the surface to height $h$:

$\Delta E_p = E_p(R_E + h) - E_p(R_E) = -\frac{GMm}{R_E}\left(1 - \frac{h}{R_E}\right) - \left(-\frac{GMm}{R_E}\right)$

$\Delta E_p = \frac{GMmh}{R_E^2} = mgh$

Since $g = GM/R_E^2$.

$\boxed{\Delta E_p = mgh \quad \mathrm{(valid for } h \ll R_E\mathrm{)}}$

Proof that this is the first-order approximation. The Taylor expansion gives $\Delta E_p = mgh(1 - h/R_E + h^2/R_E^2 - \cdots)$. The leading term is $mgh$And the correction is Of relative order $h/R_E$. For $h = 1$ km, $h/R_E \approx 1.6 \times 10^{-4}$So the error is about 0.016%. $\square$

Negative Total Energy Means a Bound System

For a satellite in orbit, $E_{\mathrm{total}} = -GMm/(2r) \lt 0$. Negative total energy means the System is gravitationally bound — the satellite cannot escape without an input of energy. To Escape, enough energy must be added to raise $E_{\mathrm{total}}$ to zero.

12. Escape Velocity — Extended Discussion

Derivation from Energy Conservation

At the surface of a planet (radius $r$), an object of mass $m$ has:

• Kinetic energy: $\frac{1}{2}mv^2$
• Gravitational potential energy: $-\frac{GMm}{r}$

To just reach infinity with zero speed (the minimum condition for escape):

$\frac{1}{2}mv_{\mathrm{esc}}^2 - \frac{GMm}{r} = 0$

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

Numerical Values

Relationship to Orbital Speed

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

$v_{\mathrm{esc}} = \sqrt{2} \, v_{\mathrm{orbit}}$

The escape velocity is $\sqrt{2} \approx 1.41$ times the circular orbital speed at the same radius. This factor of $\sqrt{2}$ arises because escaping requires exactly twice the kinetic energy of a Circular orbit: $\frac{1}{2}mv_{\mathrm{esc}}^2 = 2 \times \frac{1}{2}mv_{\mathrm{orbit}}^2$.

Implications

An object in a circular orbit needs a speed increase of $(\sqrt{2} - 1) \times 100\% \approx 41\%$ To escape. This is why spacecraft use gravitational slingshots or multi-stage rockets rather than a Single impulse to escape Earth’s gravity efficiently.

Cross-References

• Dynamics — Gravitational field strength g connects to weight via W = mg, derived from Newton’s second law.
• Circular Motion — Orbital motion requires centripetal acceleration, which gravity provides for satellites and planets.
• Work, Energy and Power — Gravitational potential energy near the surface is a special case of the general gravitational potential.
• Electric Fields — Gravitational and electric fields share the inverse square law structure and analogous mathematical treatments.