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.

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 $T = \frac◆LB◆2\pi r◆RB◆◆LB◆v◆RB◆$So $v = \frac◆LB◆2\pi r◆RB◆◆LB◆T◆RB◆$:

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

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

Since $\frac◆LB◆4\pi^2◆RB◆◆LB◆GM◆RB◆$ 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:

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

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

For Earth: $v_e = \sqrt◆LB◆\frac{2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.37 \times 10^6}◆RB◆ = \sqrt◆LB◆1.25 \times 10^8◆RB◆ \approx 11.2$ 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

Property	Gravitational	Electric
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	$V = -GM/r$	$V = Q/(4\pi\varepsilon_0 r)$
Always attractive?	Yes	No (depends on charge signs)
Shielding possible?	No	Yes

Problem Set

Problem 1

Calculate the gravitational field strength at the surface of Mars, given its mass is $6.42 \times 10^{23}$ kg and its radius is $3.39 \times 10^6$ m.

Answer. $g = \frac{GM}{r^2} = \frac◆LB◆6.67 \times 10^{-11} \times 6.42 \times 10^{23}◆RB◆◆LB◆(3.39 \times 10^6)^2◆RB◆ = \frac◆LB◆4.28 \times 10^{13}◆RB◆◆LB◆1.149 \times 10^{13}◆RB◆ = 3.73$ N kg$^{-1}$.

If you get this wrong, revise: Gravitational Field Strength

Problem 2

A satellite orbits Earth at an altitude of $400$ km. Calculate: (a) its orbital speed, (b) its period, (c) its centripetal acceleration.

Answer. $r = 6370 + 400 = 6770$ km $= 6.77 \times 10^6$ m. $GM = 3.976 \times 10^{14}$.

(a) $v = \sqrt{GM/r} = \sqrt◆LB◆3.976 \times 10^{14}/6.77 \times 10^6◆RB◆ = \sqrt◆LB◆5.87 \times 10^7◆RB◆ = 7660$ m S$^{-1}$.

(b) $T = 2\pi r/v = 2\pi \times 6.77 \times 10^6/7660 = 5550$ s $= 92.5$ min.

(c) $a = v^2/r = GM/r^2 = 3.976 \times 10^{14}/(6.77 \times 10^6)^2 = 8.67$ m s$^{-2}$.

If you get this wrong, revise: Kepler’s Laws

Problem 3

Calculate the escape velocity from the Moon, given $M_{\mathrm{Moon}} = 7.35 \times 10^{22}$ kg and $R_{\mathrm{Moon}} = 1.74 \times 10^6$ m.

Answer. $v_e = \sqrt{2GM/R} = \sqrt◆LB◆2 \times 6.67 \times 10^{-11} \times 7.35 \times 10^{22}/1.74 \times 10^6◆RB◆ = \sqrt◆LB◆5.64 \times 10^6◆RB◆ = 2370$ M s$^{-1} = 2.37$ km s$^{-1}$.

If you get this wrong, revise: Escape Velocity

Problem 4

Two stars of mass $2.0 \times 10^{30}$ kg each are separated by $1.0 \times 10^{11}$ m. They orbit their common centre of mass. Find the orbital period.

Answer. Each star is at distance $r = 5.0 \times 10^{10}$ m from the centre. The gravitational Force provides the centripetal force:

$\frac{Gm^2}{(2r)^2} = \frac{mv^2}{r}$. $\frac{Gm}{4r} = v^2$. $v = \sqrt{Gm/(4r)}$.

$T = 2\pi r/v = 2\pi r\sqrt{4r/(Gm)} = 4\pi\sqrt{r^3/(Gm)} = 4\pi\sqrt◆LB◆(5 \times 10^{10})^3/(6.67 \times 10^{-11} \times 2 \times 10^{30})◆RB◆ = 4\pi\sqrt◆LB◆1.25 \times 10^{33}/1.334 \times 10^{20}◆RB◆ = 4\pi\sqrt◆LB◆9.37 \times 10^{12}◆RB◆ = 4\pi \times 3.06 \times 10^6 = 3.85 \times 10^7$ S $\approx 1.2$ years.

If you get this wrong, revise: Derivation of Kepler’s Third Law

Problem 5

A satellite of mass $500$ kg is in a circular orbit of radius $7.0 \times 10^6$ m around Earth. Calculate: (a) its total energy, (b) the energy needed to move it to an orbit of radius $1.4 \times 10^7$ m.

Answer. (a) $E = -GMm/(2r) = -3.976 \times 10^{14} \times 500/(2 \times 7.0 \times 10^6) = -1.42 \times 10^{10}$ J $= -14.2$ GJ.

(b) New energy: $E' = -3.976 \times 10^{14} \times 500/(2 \times 1.4 \times 10^7) = -7.10 \times 10^9$ J $= -7.10$ GJ.

$\Delta E = E' - E = -7.10 - (-14.2) = 7.10$ GJ.

If you get this wrong, revise: Orbital Energy

Problem 6

Calculate the gravitational potential at a point $3.0 \times 10^7$ m from the centre of Earth ($M_E = 5.97 \times 10^{24}$ kg).

Answer. $V = -GM/r = -6.67 \times 10^{-11} \times 5.97 \times 10^{24}/3.0 \times 10^7 = -1.33 \times 10^7$ J Kg$^{-1}$.

If you get this wrong, revise: Gravitational Potential

Problem 7

Prove that the escape velocity from a planet of radius $R$ and surface gravitational field strength $g$ is $v_e = \sqrt{2gR}$.

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

If you get this wrong, revise: Escape Velocity

Problem 8

A geostationary satellite orbits above the equator with a period of 24 hours. Calculate its orbital radius.

Answer. $T = 86400$ s. $T^2 = \frac◆LB◆4\pi^2◆RB◆◆LB◆GM◆RB◆r^3$. $r^3 = \frac◆LB◆GMT^2◆RB◆◆LB◆4\pi^2◆RB◆ = \frac◆LB◆3.976 \times 10^{14} \times 7.46 \times 10^9◆RB◆◆LB◆39.48◆RB◆ = 7.52 \times 10^{22}$. $r = 4.22 \times 10^7$ m $= 42\,200$ km.

If you get this wrong, revise: Derivation of Kepler’s Third Law

Problem 9

Show that the orbital speed of a satellite is independent of the satellite’s mass.

Answer. $v = \sqrt{GM/r}$. The satellite’s mass $m$ does not appear — it cancels in the Derivation: $\frac{GMm}{r^2} = \frac{mv^2}{r} \implies v^2 = GM/r$. The orbital speed depends only On the mass of the central body and the orbital radius. $\square$

If you get this wrong, revise: Derivation of Kepler’s Third Law

Problem 10

Explain why a satellite in a higher orbit moves more slowly than one in a lower orbit, even though the higher satellite has greater total energy.

Answer. From $v = \sqrt{GM/r}$: as $r$ increases, $v$ decreases — the satellite moves slower. However, the total energy $E = -GMm/(2r)$ is less negative (greater) at larger $r$. This is because The potential energy increases more than the kinetic energy decreases. The satellite has more total Energy but is moving slower — the extra energy is in the form of gravitational potential energy.

If you get this wrong, revise: Orbital Energy

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:

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

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).

:::caution Common Pitfall Satellites in low Earth orbit are NOT in “zero gravity.” The gravitational Field strength at 300 km altitude is still about 89% of its surface value. Astronauts experience Weightlessness because they are in free fall, not because gravity is absent. :::

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.

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

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

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

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

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

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

Orbital Speed

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

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.

:::caution Common Pitfall A geostationary orbit is NOT the same as a geosynchronous orbit. A Geosynchronous orbit has period 24 hours but can be inclined or elliptical, so the satellite appears To trace a figure-eight in the sky. Geostationary implies geosynchronous AND equatorial AND Circular. :::

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.

:::tip Exam Technique When asked why $E_p = -GMm/r$ is negative, explain: we define $V = 0$ at Infinity, so bringing masses together releases energy (the system loses potential energy). The Negative sign reflects this. The formula $E_p = mgh$ is only a special case for small heights near The surface. :::

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):

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

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

Numerical Values

Body	$v_{\mathrm{esc}}$ (km s$^{-1}$)
Earth	11.2
Moon	2.4
Mars	5.0
Jupiter	59.5

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.

:::info Info

Problem 11

Calculate the gravitational field strength at an altitude of 500 km above Earth’s surface. What percentage decrease is this compared to the surface value? ($R_E = 6370$ km, $M_E = 5.97 \times 10^{24}$ kg).

Answer. $r = 6370 + 500 = 6870$ km $= 6.87 \times 10^6$ m. $g = GM/r^2 = 6.67 \times 10^{-11} \times 5.97 \times 10^{24}/(6.87 \times 10^6)^2 = 3.98 \times 10^{14}/4.72 \times 10^{13} = 8.43$ N kg$^{-1}$.

Percentage decrease: $(9.81 - 8.43)/9.81 \times 100 = 14.1\%$.

Using the approximation: $\Delta g/g_0 \approx 2h/R_E = 2 \times 500/6370 = 0.157 = 15.7\%$. The Approximation overestimates slightly because $h/R_E$ is not very small.

If you get this wrong, revise: Variation of $g$ with Altitude

Problem 12

Show that the orbital radius of a geostationary satellite is approximately 42,000 km. Calculate its orbital speed and verify using $v = \omega r$.

Answer. $\omega = 2\pi/(24 \times 3600) = 7.272 \times 10^{-5}$ rad s$^{-1}$. $r^3 = GM/\omega^2 = 3.976 \times 10^{14}/(7.272 \times 10^{-5})^2 = 7.52 \times 10^{22}$. $r = 4.22 \times 10^7$ m $= 42\,200$ km.

$v = \sqrt{GM/r} = \sqrt◆LB◆3.976 \times 10^{14}/4.22 \times 10^7◆RB◆ = 3070$ m s$^{-1}$.

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

If you get this wrong, revise: Geostationary Orbits

Problem 13

Use the Taylor expansion to show that $E_p = -GMm/(R+h)$ reduces to $mgh$ for $h \ll R$And find the percentage error when $h = 10$ km for Earth.

Answer. $E_p(R + h) = -\frac{GMm}{R}(1 + h/R)^{-1} \approx -\frac{GMm}{R}(1 - h/R)$. $\Delta E_p = GMmh/R^2 = mgh$. $\square$

Error term: $\delta = h/R = 10000/6.37 \times 10^6 = 1.57 \times 10^{-3} = 0.157\%$. So $mgh$ is Accurate to about 0.16% at 10 km altitude.

If you get this wrong, revise: Gravitational Potential Energy — Taylor Expansion

Problem 14

A spacecraft is in a circular orbit of radius $r$ around Earth. What speed increase is needed for it to escape? Express your answer as a fraction of its current orbital speed.

Answer. Current speed: $v_{\mathrm{orbit}} = \sqrt{GM/r}$. Escape speed: $v_{\mathrm{esc}} = \sqrt{2GM/r} = \sqrt{2}\,v_{\mathrm{orbit}}$.

Required speed increase: $\Delta v = v_{\mathrm{esc}} - v_{\mathrm{orbit}} = (\sqrt{2} - 1)v_{\mathrm{orbit}} \approx 0.414\,v_{\mathrm{orbit}}$.

The spacecraft needs a speed increase of approximately 41.4% of its orbital speed.

If you get this wrong, revise: Escape Velocity — Extended Discussion

Problem 15

A satellite of mass 800 kg is moved from a circular orbit of radius $7.0 \times 10^6$ m to a circular orbit of radius $1.05 \times 10^7$ m. Calculate: (a) the total energy in each orbit, (b) the energy that must be added, (c) the escape velocity from the lower orbit.

Answer. $GM = 3.976 \times 10^{14}$ N m$^2$ kg$^{-1}$.

(a) $E_1 = -GMm/(2r_1) = -3.976 \times 10^{14} \times 800/(2 \times 7.0 \times 10^6) = -2.27 \times 10^{10}$ J.

$E_2 = -3.976 \times 10^{14} \times 800/(2 \times 1.05 \times 10^7) = -1.52 \times 10^{10}$ J.

(b) $\Delta E = E_2 - E_1 = -1.52 \times 10^{10} - (-2.27 \times 10^{10}) = 7.5 \times 10^9$ J $= 7.5$ GJ.

(c) $v_{\mathrm{esc}} = \sqrt{2GM/r_1} = \sqrt◆LB◆2 \times 3.976 \times 10^{14}/7.0 \times 10^6◆RB◆ = \sqrt◆LB◆1.136 \times 10^8◆RB◆ = 10660$ M s$^{-1} = 10.7$ km s$^{-1}$.

If you get this wrong, revise: Escape Velocity — Extended Discussion and Orbital Energy

---

::: :::tip Tip Ready to test your understanding of Gravitational Fields? The contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Gravitational Fields with other physics topics to test synthesis under exam conditions.

See for instructions on self-marking and building a personal test matrix.

::: :::danger Danger

• Confusing gravitational field strength g with gravitational force: g = GM/r squared is the field strength (force per unit mass, N/kg). The gravitational force on an object is F = mg = GMm/r squared. Field strength does not depend on the test mass; force does. This distinction matters in exam questions about orbits and satellites.

• Forgetting that gravitational potential is always negative: By convention, gravitational potential is defined as zero at infinity. Since work must be done to move a mass from infinity to any finite distance, potential is negative at all finite distances. The potential is most negative at the surface of a planet and approaches zero far away.

• Confusing geostationary orbit conditions: A geostationary satellite must orbit in the equatorial plane (not any inclination), in the same direction as Earth’s rotation, and with a period of exactly 24 hours. A polar orbit satellite with period 24 hours is NOT geostationary because it does not remain above the same point on the equator.

• Assuming gravitational force shields or cancels inside a shell: Inside a uniform spherical shell, the gravitational field strength is ZERO at every point (not just the centre). This is a consequence of the shell theorem. Between two concentric shells, only the mass of the inner shell contributes to the field at that point.

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. Forgetting to include units in final answers, especially when working with derived units like $\text{N}\,\text{kg}^{-1}\,\text{m}^2$.

5. Confusing displacement with distance, or velocity with speed, particularly in graphs and calculations.

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

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

::: :::