Year 12 Physics Module 5 ⏱ ~50 min 5 MC · 3 Short Answer Lesson 18 of 18 IQ3: Gravitational Fields Consolidation

Gravitational Fields Consolidation

Apollo 13's return depended on careful trajectory planning within the combined Earth–Moon gravitational system. This consolidation revisits the field, potential, energy, escape-speed and Kepler models used to reason about idealised orbital systems—without pretending that a real three-body mission reduces to one formula.

Today's challenge: A spacecraft travelling through the Earth–Moon system is influenced by more than one body and may also use engine burns, so the single-mass equations in this lesson are models, not a complete mission simulation. Before starting, write down which gravitational formula is hardest to remember and the condition that makes it valid.
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Diagnose what needs consolidating

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Two printable resources support this gravitational-fields consolidation.

Build Foundations & guided practice Exam Consolidation worksheet
Before you begin, reflect

Which gravitational formula is hardest to remember? Write it down three times along with when to use it.

Warm-up, what is the correct expression for gravitational potential energy at distance $r$ from a planet of mass $M$?

Learning Intentions
goals

Know, Recall all Phase 3 Formulae

  • State and explain all five Phase 3 gravitational field formulae
  • Identify the trap in each formula and when it applies

Understand, Identify and Correct Common Errors

  • Diagnose each of the six common exam errors
  • Distinguish gravitational potential $V$ from potential energy $U$

Can Do, Link Graphs, Signs and Conditions

  • Interpret the gradient of a potential–radius graph
  • Choose and justify an IQ3 model before calculating
Key Terms, Gravitational Fields
vocab
Universal gravitation$F = GMm/r^2$; all masses attract all other masses.
Field strength$g = GM/r^2$; force per unit mass at a point in the field.
Potential energy$U = -GMm/r$ for two isolated point or spherical masses, with $U=0$ as $r\to\infty$.
Gravitational potential$V = U/m = -GM/r$ (J/kg): the external work per unit mass required to bring a test mass quasistatically from infinity.
Escape velocity$v_e = \sqrt{2GM/r}$; minimum speed to escape a gravitational field permanently.
Kepler's Third Law$T^2 \propto a^3$ for bodies orbiting the same central mass; $a$ is semi-major axis and equals $r$ only for a circle.
Cross-lesson links: L14–L17 built the gravitational-field toolkit. L18 consolidates that IQ3 strand: potential energy and work, potential and field gradient, escape-speed conditions, and Kepler/orbital-energy relationships. Whole-module synthesis remains in the module quiz.
Misconceptions to fix before you review
Wrong: Using $g = 9.8 \text{ m/s}^2$ in orbital or altitude problems.
Right: $g = 9.8 \text{ m/s}^2$ is only valid at Earth's surface. At altitude $h$, use $g = GM/(R+h)^2$ or the ratio $g' = g(R/(R+h))^2$.
Wrong: Confusing gravitational potential $V$ (J/kg) with potential energy $U$ (J).
Right: $V = -GM/r$ is per unit mass; $U = -GMm/r$ is total energy. They are related by $U = mV$ and $\Delta U = m\Delta V$.

The formula $g = 9.8 \text{ m/s}^2$ can be used to calculate gravitational field strength at any altitude above Earth's surface.

Gravitational potential $V$ and gravitational potential energy $U$ have the same SI units.

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Map each formula to its conditions

IQ3: Gravitational Fields Formula and Condition Map

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Phase 3 Formula Sheet & Sprint Cards
+5 XP

Five formulae, when to use each, key variables, and common traps. Cover the right side, recall, then click to check.

Full Module Summary, all Phase 3 gravitational field formulae

Module-wide reference map. This lesson assesses only the gravitation, orbital energy, potential, escape velocity and Kepler branches; conditions such as “circular orbit” and “same launch/landing height” are part of each formula.

Newton's law of gravitation

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

Use: Attractive force between two masses. $G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$. Always: $r$ = centre-to-centre distance.

Gravitational field strength

$g = \dfrac{GM}{r^2}$

Use: Force per unit mass at distance $r$ from a mass $M$. At altitude $h$: $r = R + h$. Ratio form: $g' = g\!\left(\dfrac{R}{R+h}\right)^2$.

Gravitational potential energy

$U = -\dfrac{GMm}{r}$   ·   $W_{\mathrm{ext}} = \Delta U = GMm\!\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$   ·   $W_g=-\Delta U$

Use: Potential energy of an isolated two-body system, taking $U\to0$ as $r\to\infty$. The external-work equality assumes a quasistatic transfer with no kinetic-energy change. Use $mgh$ only for small height changes where $g$ is approximately uniform.

Gravitational potential

$V = -\dfrac{GM}{r}$   ·   $g_r = -\dfrac{dV}{dr} = -\dfrac{GM}{r^2}$

Use: Potential energy per unit mass (J/kg), with outward radial direction positive. The signed field component $g_r$ is negative (inward); its magnitude is $|\vec g|=GM/r^2=dV/dr$. Related to $U$ by $U=mV$.

Escape velocity

$v_e = \sqrt{\dfrac{2GM}{r}}$

Use: Minimum speed to escape a gravitational field permanently. Factor of $\sqrt{2}$ is mandatory, from energy conservation. $v_e = \sqrt{2}\,v_{\text{orbital}}$.

Sprint Cards, click to reveal traps

$F = \dfrac{GMm}{r^2}$
Click to reveal when to use, trap, and connection
Use whenTwo masses, distance between centres known. Finding force of gravitational attraction.
Trap$r$ is centre-to-centre, not surface-to-surface. Always write $r = R + h$ explicitly.
Connects to$g = F/m = GM/r^2$; for circular orbits, equate with $F = mv^2/r$.
$U = -\dfrac{GMm}{r}$
Click to reveal when to use, trap, and connection
Use whenEnergy calculations, work done moving between radii, total orbital energy.
TrapThe negative sign is mandatory. $U = 0$ at $r = \infty$. Never use $U = mgh$ for orbital problems.
Connects to$W_{\mathrm{ext}}=\Delta U=GMm(1/r_1-1/r_2)$ for a quasistatic transfer; $W_g=-\Delta U$. Total energy is $E=-GMm/(2r)$ for circular orbits.
$V = -\dfrac{GM}{r}$
Click to reveal when to use, trap, and connection
Use whenPotential per unit mass (J/kg). Equipotential surfaces. Relating $g$ and $V$.
Trap$V$ is potential (J/kg), NOT potential energy $U$ (J). Do not confuse units. To get energy: $U = mV$ and $\Delta U = m\Delta V$.
Connects toWith outward positive, $g_r=-dV/dr$ while $|\vec g|=dV/dr$ for this spherical field; $\Delta U=m\Delta V$.
$v_e = \sqrt{\dfrac{2GM}{r}}$
Click to reveal when to use, trap, and connection
Use whenMinimum speed to escape a gravitational field permanently (total energy = 0).
TrapFactor of 2 is mandatory (from energy conservation). $v_e = \sqrt{2} \times v_{\text{orbital}}$. Omitting gives ~41% error.
Connects toTotal orbital energy $E = -GMm/2r$; binding energy $|E|$.
$T^2 = \dfrac{4\pi^2}{GM}\,a^3$
Click to reveal when to use, trap, and connection
Use whenRelating orbital period to semi-major axis for a bound body orbiting a dominant central mass.
TrapUse semi-major axis $a$ for an ellipse. Equating $GMm/r^2=mv^2/r$ with $v=2\pi r/T$ derives only the circular special case, where $a=r$.
Connects to$E_{\text{total}}=-GMm/(2a)$ for a bound Keplerian ellipse; in a circle $a=r$.

IQ3 formula set: $F=GMm/r^2$; $|\vec g|=GM/r^2$ (use $r=R+h$); $U=-GMm/r$; $W_{\mathrm{ext}}=\Delta U$ and $W_g=-\Delta U$; $V=-GM/r$; $g_r=-dV/dr$ (outward positive); $v_e=\sqrt{2GM/r}$; $T^2=(4\pi^2/GM)a^3$; $E=-GMm/(2a)$ for a bound Keplerian orbit, with $a=r$ for a circle.

Pause, copy the highlighted formula set into your book before moving on.

A satellite orbits at height $h = 2R_E$ above Earth's surface. The gravitational field strength there compared to the surface is:

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Find and repair the reasoning error

Error Clinic

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Six Common Gravitational Errors
+5 XP

Each card describes a common error that costs marks in exams. Find the fix, then reveal the explanation.

We just saw the complete gravitational fields formula set. That raises a question: which of these formulae do students most often apply incorrectly? This card answers it → six common errors: wrong $g$, missing negative, confusing $V$ and $U$, wrong $r$, missing $\sqrt{2}$, and misapplying Kepler's Third Law.

E1
Using $g = 9.8$ for orbital problems
▾
Student working: A satellite orbits 400 km above Earth. Student uses $F = mg = m \times 9.8$ to find gravitational force. Incorrect.
Fix: $g = 9.8 \text{ m/s}^2$ is only valid at Earth's surface. At altitude $h$, always use $g = GM/(R+h)^2$ or the ratio $g' = g(R/(R+h))^2$. Treating $g$ as constant in orbital calculations is the most common error in HSC exams.
E2
Forgetting the negative sign on $U = -GMm/r$
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Student working: External work in a quasistatic transfer from $r_1$ to $r_2$: $W_{\mathrm{ext}}=GMm(1/r_1-1/r_2)$, but the student uses $U=+GMm/r$, giving the wrong sign. Incorrect.
Fix: The negative sign is physically meaningful, gravity is attractive and $U = 0$ is defined at $r = \infty$. All bound systems have negative total energy. Omitting it gives incorrect energy rankings and wrong signs for work calculations.
E3
Confusing $V$ (J/kg) with $U$ (J)
▾
Student working: Question asks for gravitational potential energy of a 2000 kg satellite. Student calculates $V = -GM/r$ and writes that as the answer in joules. Incorrect.
Fix: $V = -GM/r$ has units of J/kg (potential per unit mass). $U = -GMm/r$ has units of J (total potential energy). To get energy from potential: $U = mV$ and $\Delta U = m\Delta V$. Always check units in your final answer.
E4
Using $r = R$ instead of $r = R + h$
▾
Student working: A satellite orbits at 500 km altitude. Student substitutes $r = R_E = 6.37 \times 10^6$ m into $F = GMm/r^2$. Incorrect.
Fix: $r$ is always centre-to-centre distance. For a satellite at height $h$ above the surface, $r = R_{\text{planet}} + h$. Write this explicitly in every problem. Never substitute $R$ when the question gives you an altitude.
E5
Forgetting the $\sqrt{2}$ in escape velocity
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Student working: Escape velocity from Earth's surface: student writes $v_e = \sqrt{GM/R}$. Incorrect.
Fix: $v_e = \sqrt{2GM/r}$, not $\sqrt{GM/r}$. The factor of 2 comes from energy conservation: $\tfrac{1}{2}mv_e^2 = |U| = GMm/r$. $\sqrt{2} \approx 1.414$, so the error is about 41% if you forget it.
E6
Thinking Kepler's Third Law requires circular orbits
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Student working: Student refuses to apply $T^2 \propto a^3$ to a comet with a closed elliptical orbit. Incorrect reasoning.
Fix: Kepler's Third Law $T^2 \propto a^3$ applies to a bound Keplerian ellipse when $a$ is the semi-major axis. For a circle, $a=r$; an unbound parabolic or hyperbolic path has no orbital period.

Six error rules: E1, use $g = GM/r^2$ at altitude (not 9.8); E2, $U = -GMm/r$, negative sign mandatory; E3, $V$ (J/kg) $\neq$ $U$ (J); convert: $U = mV$; E4, always $r = R + h$; E5, $v_e = \sqrt{2GM/r}$, not $\sqrt{GM/r}$; E6, Kepler's 3rd Law applies to ellipses using semi-major axis $a$.

Add the highlighted six error checklist to your notes before the check below.

Three of these statements about gravitational fields are correct. Pick the odd one out (the incorrect statement).

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Reason from graphs, signs and conditions

Graph and sign convention

Take the radial coordinate $r$ as positive outwards. For $V=-GM/r$, the graph rises towards zero as $r$ increases and has positive gradient $dV/dr=GM/r^2$. The gravitational field points inwards, so its signed radial component is $g_r=-dV/dr=-GM/r^2$. The field magnitude is $|\vec g|=GM/r^2$.

Neither the field nor the force is exactly zero at a finite radius around an isolated mass. Both tend to zero only in the limit $r\to\infty$.

Worked exemplar 1, potential and field at the same radius
1. Define the system. At $r=2.0\times10^7$ m from Earth's centre, use the point/spherical-mass model with $M_E=5.97\times10^{24}$ kg.
2. Calculate potential. $V=-GM_E/r=-1.99\times10^7$ J kg$^{-1}$.
3. Calculate the signed field. $g_r=-GM_E/r^2=-0.995$ m s$^{-2}$. The negative sign means inward; the magnitude is $0.995$ N kg$^{-1}$.
Worked exemplar 2, interpret a potential gradient
1. Differentiate. $V=-GMr^{-1}$ gives $dV/dr=+GM/r^2$.
2. Apply the vector relation. With outward positive, $g_r=-dV/dr=-GM/r^2$.
3. Interpret. A positive slope on the $V$–$r$ graph corresponds to an inward field. Do not replace the signed component with the positive magnitude midway through a direction question.
Worked exemplar 3, energy of a circular satellite
1. Check applicability. For a circular orbit only, $K=GMm/(2r)$, $U=-GMm/r$ and $E=-GMm/(2r)$.
2. Substitute. For $m=2000$ kg and $r=8.5\times10^6$ m, $K=4.68\times10^{10}$ J and $U=-9.36\times10^{10}$ J.
3. Reconcile signs. $E=K+U=-4.68\times10^{10}$ J. Negative total energy means the satellite is gravitationally bound.
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Use feedback to choose your next drill

Assessment boundary

This lesson consolidates IQ3 gravitational fields. Complete the Practice bank below for independent checking. Use the module quiz for whole-module summative practice; projectile motion and circular-motion questions belong to their earlier lessons and the module quiz, not this consolidation.

Wrap-up, Misconceptions & Summary

Misconceptions, final check

Wrong: "A lower satellite orbit has more total energy because the satellite moves faster."
Right: A lower orbit has more negative total energy ($E = -GMm/2r$). Although kinetic energy is greater, the potential energy is even more negative, the satellite is more tightly bound. Higher orbits have less negative (greater) total energy.
Wrong: "Gravitational potential is zero at Earth's surface."
Right: Gravitational potential is zero at $r = \infty$ by definition. At Earth's surface, $V = -GM/R \approx -6.25 \times 10^7$ J/kg, a large negative value. The convention of $V = 0$ at infinity is the only physically natural reference point.

Copy into your books

The Five Formulae

  • $F = GMm/r^2$ · $g = GM/r^2$
  • $U = -GMm/r$ · $V = -GM/r$
  • $v_e = \sqrt{2GM/r}$

Derived Results

  • $W_{\mathrm{ext}}=\Delta U$ · $W_g=-\Delta U$
  • $E=-GMm/(2a)$; circle: $a=r$
  • $g_r = -dV/dr$ · $U = mV$

Six Common Errors

  • Using $g = 9.8$ at altitude (E1)
  • Forgetting negative on $U$ (E2)
  • Confusing $V$ and $U$ units (E3)

Constants

  • $G = 6.67 \times 10^{-11}$ N m²/kg²
  • $M_E = 5.97 \times 10^{24}$ kg
  • $R_E = 6.37 \times 10^6$ m

A satellite in a lower orbit has a higher total mechanical energy than one in a higher orbit.

The escape velocity from any radius is exactly $\sqrt{2}$ times the circular orbital speed at that radius.

Gravitational potential $V$ is zero at Earth's surface.

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Complete the independent IQ3 assessment

Quick recall, Gravitational Fields Consolidation
+5 XP

A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

Short Answer, 10 marks
+5 XP

ApplyBand 4(3 marks) 1. A 1500 kg satellite orbits Earth at $r = 1.0 \times 10^7$ m from Earth's centre. Calculate (a) its orbital speed, (b) its kinetic energy, and (c) its total mechanical energy. $(M_E = 5.97 \times 10^{24} \text{ kg})$

1 mark: correct orbital speed using $v = \sqrt{GM/r}$ · 1 mark: kinetic energy · 1 mark: total energy using $E = -GMm/2r$ or $E = -KE$

AnalyseBand 5(3 marks) 2. Explain why the gravitational potential $V$ is always negative and why it approaches zero as $r \rightarrow \infty$. Include the definition of $V$ in your explanation.

1 mark: defines $V$ as external work per unit mass from infinity · 1 mark: explains the negative value for an attractive field · 1 mark: explains $V \to 0$ only as $r \to \infty$

EvaluateBand 6(4 marks) 3. A student argues: "Since $g = GM/r^2$ and $V = -GM/r$, a point where $V$ is more negative must also have a larger $g$." Evaluate this statement, using a specific numerical example at two different radii to support your reasoning. $(M_E = 5.97 \times 10^{24} \text{ kg})$

1 mark: identifies the trend is correct but the relationship is not proportional · 1 mark: $|\vec g| \propto 1/r^2$ while $V \propto -1/r$ · 1 mark: numerical example at $R_E$ and $2R_E$ · 1 mark: correct signed relation $g_r=-dV/dr$ with outward positive

Show all answers

Multiple Choice, Key

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.

Short Answer, Model Answers

SA1 (3 marks): (a) $v = \sqrt{GM_E/r} = \sqrt{(6.67 \times 10^{-11})(5.97 \times 10^{24})/(1.0 \times 10^7)} = \mathbf{6.31 \times 10^3 \text{ m/s}}$ (1 mark). (b) $KE = \tfrac{1}{2}mv^2 = 0.5 \times 1500 \times (6310)^2 = \mathbf{2.99 \times 10^{10} \text{ J}}$ (1 mark). (c) $E_{\text{total}} = -KE = \mathbf{-2.99 \times 10^{10} \text{ J}}$ (1 mark; or equivalently $-GMm/2r$).

SA2 (3 marks): $V=-GM/r$ and $V=0$ as $r\to\infty$ (1 mark). It is the external work per unit mass required to bring a test mass quasistatically from infinity to $r$; gravity does positive work on the inward displacement, so the external agent does negative work and $V$ is negative (1 mark). At every finite $r$, $V<0$ and the gravitational influence is non-zero; both tend to zero only as $r\to\infty$ (1 mark).

SA3 (4 marks): The trend is correct but the claim is incomplete (1 mark). As $r$ decreases, $V=-GM/r$ becomes more negative while the field magnitude $|\vec g|=GM/r^2$ increases; they are not proportional (1 mark). At $r=R_E$, $V=-6.25\times10^7$ J kg$^{-1}$ and $|\vec g|=9.81$ N kg$^{-1}$; at $2R_E$, $V=-3.13\times10^7$ J kg$^{-1}$ and $|\vec g|=2.45$ N kg$^{-1}$ (1 mark). With outward positive, the signed relation is $g_r=-dV/dr=-GM/r^2$; the positive gradient $dV/dr$ is the field magnitude, not the signed component (1 mark).

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Retrieve the IQ3 relationships and reflect

Check what actually stuck
Take the full module quiz
quiz

A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.

Start the module quiz →
Arcade Practice · Asteroid Blaster

Gravitational orbits, apply your knowledge in the final module challenge. Lighter than the boss, pure practice that hammers home the gravitational field relationships.

How did your thinking change?

At the start, the Apollo 13 context highlighted an important limitation: the equations in this lesson assume an isolated point or spherical source mass, whereas a real Earth–Moon trajectory involves two gravitational sources, changing reference frames and planned engine burns.

The single-mass models are still essential building blocks. They let you check signs, energy scales, local field strength and idealised orbit relationships before a more complete numerical trajectory model is used. After working through the formula map and error clinic, reflect again:

  • Which of the six errors (E1–E6) do you need to watch for most in the exam?
  • Can you now distinguish $V$ (J/kg) from $U$ (J) without looking at your notes?
  • What is your plan to avoid your most common error?
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